# data[].attributes.body
Practice combining like terms

# data[].attributes.attachment_link.readable_text
Practise your algebraic simplification skills with this self-marking exercise.

This is level 1; Basic linear expressions.

Simplify each of the following by collecting together like terms.

© Transum Mathematics 1997-2023
Scan the QR code below to visit the online version of this activity.

https://Transum.org/go/?Num=615

Description of Levels

Close

Level 1 - Collect like terms in basic linear expressions

Example \(7d - 3d\)

Level 2 - Collect like terms in more interesting linear expressions

Example \(5n+4−4n+8\)

Level 3 - Collect like terms in quadratic expressions

Example \(6y+9y^2−3y\)

Level 4 - Collect like terms in Expressions including multiplication and division

Example \(6y+9y\times y−3y^2\)

Brackets Level 1 - Collect like terms when 1 term is repeated

Example: \(5d - (2d + 2)\)

Brackets Level 2 - Collecting like terms when 2 terms are repeated

Example: \((9d - 7) - (5d - 2)\)

Brackets Level 3 - Multiplying a single positive integer over a bracket

Example: \(4(2d + 4)\)

Brackets Level 4 - Multiplying a single negative integer over a bracket

Example: \(-5(5d + 4)\)

Brackets Level 5 - Multiplying a variable over a bracket

Example: \(4d(5d - 5)\)

Brackets Level 6 - Expanding products of two simple binomials

Example: \((d + 7)(d - 2)\)

Brackets Level 7 - Expanding products of two binomials

Example: \((4d + 7)(3d + 2)\)

Brackets Level 8 - Squaring a binomial

Example: \((3d + 2)^2\)

Brackets Level 9 - Simplifying more complex expressions involving brackets

Example: \(5(3d+1)-2(1-2d)\)

Collecting Like Terms

Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can double-click the 'Check' button to make it float at the bottom of your screen.

Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don't yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.

Log in Sign up

Close

# data[].attributes.attachment_link.description
Practise your algebraic simplification skills with this self-marking exercise.

# data[].attributes.body
Integer worksheet

# data[].attributes.body
Combining like terms and Distributive Property worksheet

# data[].attributes.attachment_link.title
Combining like terms and Distributive Property worksheet

# data[].attributes.body
Two step equation maze

# data[].attributes.body
Multi-step equations practice

# data[].attributes.attachment_link.title
Multi-step equations practice worksheet- 10 problems

# data[].attributes.body
Getting to know you survey

# data[].attributes.body
Adding integers BINGO

# data[].attributes.body
Adding and subtracting negative numbers

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.attachment_link.title
Adding & subtracting negative numbers (practice) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.body
Multiplying and Dividing Integers

# data[].attributes.body
Combining like terms with negative coefficients & distribution

# data[].attributes.attachment_link.title
Combining like terms with negative coefficients & distribution

# data[].attributes.attachment_link.description
Practice combining like terms using the distributive property. These problems involve adding and subtracting negative numbers.

# data[].attributes.body
The Distributive Property

# data[].attributes.body
Combining like terms with rational coefficients

# data[].attributes.attachment_link.title
Combining like terms with rational coefficients

# data[].attributes.attachment_link.description
Practice combining like terms for expressions that have rational coefficients. This requires adding and subtracting fractions.

# data[].attributes.body
Testing solutions to equations

# data[].attributes.attachment_link.title
Testing solutions to equations (practice) | Khan Academy

# data[].attributes.attachment_link.description
Practice plugging in values to equations and checking to see if the equations hold true.

# data[].attributes.body
Equation flashcards

# data[].attributes.attachment_link.description
Practice solving for y in basic algebraic equations using addition, subtraction and multiplication.

# data[].attributes.body
How does the order of operation relate
to solving equations?

What does the solution of an equation represent?

What does it mean for an equation
to have one solution, no solution, or
infinitely many solutions?

What is the relationship between the
variable and the solution?

How do you know your solution is correct?

# data[].attributes.body
We will apply our knowledge of rational numbers to solve linear equations with rational coefficients and justify the solution.

We will apply our knowledge of distribution and combining like terms to solve linear equations.

We will solve multi-step equations and justify each step.

We will use substitution and the order of operations to verify our solution is correct.

We will identify equations that have one solution, no solution, and infinitely many solutions for each equation.

We will create equations that will have a pre-designated number of solutions.

# data[].attributes.body
PRIORITY STANDARDS:
EE.7​ Solve linear equations in one variable.
a)​ Give examples of linear equations in one variable with one solution, infinitely many
solutions, or no solutions. Show which of these possibilities is the case by successively
transforming the given equation into simpler forms, until an equivalent equation of the form
x = a, a = a, or a = b results (where a and b are different numbers).

b)​ Solve linear equations with rational number coefficients, including equations whose
solutions require expanding expressions using the distributive property and collecting like
terms.

SUPPORTING STANDARDS:
G.5​ Use informal arguments to establish facts about the angle sum and exterior angle of
triangles, about the angles created when parallel lines are cut by a transversal, and the
angle-angle criterion for similarity of triangles. For example, arrange three copies of the same
triangle so that the sum of the three angles appears to form a line, and give an argument in
terms of transversals why this is so.

# data[].attributes.body
Simplify
Substitute
Justify
Validate
Modify
Precise
Model
Criteria
Indicate
Verify
Criteria
Criterion
Model
Modify
Precise
Specific
Formulate
Derive
Contrast
Conjecture
Justify
Equivalent

# data[].attributes.body
Additive Inverse/Opposite
Combine Like Terms
Distributive Property
Equation
Expanding
Integers
Inverse Operation
Multiplicative
Inverse/Reciprocal
Opposite
Solution
Coefficient
Consistent/Tautology/Identity-an equation with infinitely many solutions
Constant
Inconsistent/Self-Contradiction-an equation that does not have a solution

TIER 3-angles
Alternate Interior Angle
Complementary Angles ​Degree
Exterior Angles
Interior Angles
Parallel Lines
Supplementary Angles
Transversals
Triangle Angle Sum
Vertical Angles
​Terms
Variable

# data[].attributes.body
Determine the missing operation in an equation

# data[].attributes.attachment_link.title
Algebra: Missing Operations: StudyJams! Math | Scholastic.com

# data[].attributes.attachment_link.description
There are four operations to try when a sign is missing: addition, subtraction, multiplication, or division. Students will solve problems by trying these.

# data[].attributes.body
Addition and Subtraction One-step Equations

# data[].attributes.attachment_link.title
Addition & Subtraction Equations: StudyJams! Math | Scholastic.com

# data[].attributes.attachment_link.description
How do we solve for a variable in an equation? In this activity, students will learn to solve for variables in equations by using addition and subtraction.

# data[].attributes.body
Multiplication and Division One-step Equations

# data[].attributes.attachment_link.title
Multiplication and Division Equations: StudyJams! Math | Scholastic.com

# data[].attributes.attachment_link.description
Multiplication and division can be used to solve for variables in equations. Students will learn to solve problems that will help them master these operations.

# data[].attributes.body
Variables, Expressions, and Equations

# data[].attributes.attachment_link.readable_text
When we're dealing with basic arithmetic, we see the concrete numbers there. We'll see 23 plus 5. We know what these numbers are right over here, and we can calculate them. It's going to be 28. We can say 2 times 7. We could say 3 divided by 4. In all of these cases, we know exactly what numbers we're dealing with. As we start entering into the algebraic world-- and you probably have seen this a little bit already-- we start dealing with the ideas of variables. And variables, there's a bunch of ways you can think about them, but they're really just values in expressions where they can change. The values in those expressions can change. For example, if I write x plus 5, this is an expression right over here. This can take on some value depending on what the value of x is. If x is equal to 1, then x plus 5, our expression right over here, is going to be equal to 1. Because now x is 1. It'll be 1 plus 5, so x plus 5 will be equal to 6. If x is equal to, I don't know, negative 7, then x plus 5 is going to be equal to-- well, now x is negative 7. It's going to be negative 7 plus 5, which is negative 2. So notice x here is a variable, and its value can change depending on the context. And this is in the context of an expression. You'll also see it in the context of an equation. It's actually important to realize the distinction between an expression and an equation. An expression is really just a statement of value, a statement of some type of quantity. So this is an expression. An expression would be something like what we saw over here, x plus 5. The value of this expression will change depending on what the value of this variable is. And you could just evaluate it for different values of x. Another expression could be something like, I don't know, y plus z. Now everything is a variable. If y is 1 and z is 2, it's going to be 1 plus 2. If y is 0 and z is negative 1, it's going to be 0 plus negative 1. These can all be evaluated, and they'll essentially give you a value depending on the values of each of these variables that make up the expression. An equation, you're essentially setting expressions to be equal to each other. That's why they're called equations. You're equating two things. An equation, you'll see one expression being equal to another expression. For example, you could say something like x plus 3 is equal to 1. And in this situation where you have one equation with only one unknown, you can actually figure out what x needs to be in this scenario. And you might even do it in your head. What plus 3 is equal to 1? Well, you could do that in your head. If I have negative 2, plus 3 is equal to 1. In this context, an equation is starting to constrain what value this variable can take on. But it doesn't have to necessarily constrain it as much. You could have something like x plus y plus z is equal to 5. Now you have this expression is equal to this other expression. 5 is really just an expression right over here. And there are some constraints. If someone tells you what y and z is, then you're going to get an x. If someone tells you what x and y is, then that constrains what z is. But it depends on what the different things are. For example, if we said y is equal to 3 and z is equal to 2, then what would be x in that situation? If y is equal to 3 and z is equal to 2, then you're going to have the left-hand expression is going to be x plus 3 plus 2. It's going to be x plus 5. This part right over here is just going to be 5. x plus 5 is equal to 5. And so what plus 5 is equal to 5? Well, now we're constraining that x would have to be equal to 0. Hopefully you realize the difference between expression and equation. In an equation, essentially you're equating two expressions. The important thing to take away from here is that a variable can take on different values depending on the context of the problem. And to hit the point home, let's just evaluate a bunch of expressions when the variables have different values. For example, if we had the expression x to the y power, if x is equal to 5 and y is equal to 2, then our expression here is going to evaluate to, well, x is now going to be 5. y is going to be 2. It's going to be 5 to the second power, or it's going to evaluate to 25. If we changed the values-- let me do that in that same color-- if we said x is equal to negative 2 and y is equal to 3, then this expression would evaluate to-- let me do it in that-- negative 2. That's what we're going to substitute for x now in this context. And y is now 3, negative 2 to the third power, which is negative 2 times negative 2 times negative 2, which is negative 8. Negative 2 times negative 2 is positive 4, times negative 2 again is equal to negative 8. We could do even more complex things. We could have an expression like the square root of x plus y and then minus x, like that. Let's say that x is equal to 1 and y is equal to 8. Then this expression would evaluate to, well, every time we see an x, we want to put a 1 there. So we would have a 1 there, and you'd have a 1 over there. And every time you see a y, you would put an 8 in its place in this context. We're setting these variables. So you'd see an 8. Under the radical sign, you would have a 1 plus 8. So you'd have the principal root of 9, which is 3. This whole thing would simplify. In this context, when we set these variables to be these things, this whole thing would simplify to be 3. 1 plus 8 is 9. The principal root of that is 3. And then you'd have 3 minus 1, which is equal to 2.

# data[].attributes.attachment_link.title
Variables, expressions, & equations (video) | Khan Academy

# data[].attributes.attachment_link.description
When we're dealing with basic arithmetic, we see the concrete numbers there. We'll see 23 plus 5. We know what these numbers are right over here, and we can calculate them. It's going to be 28. We can say 2 times 7. We could say 3 divided by 4.

# data[].attributes.body
One-step equation intuition

# data[].attributes.attachment_link.readable_text
So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects right over here. And each of these nine objects have a mass of 1 kilograms. So in total, you have 9 kilograms on this side. And over here, you have three objects. They all have the same mass. And we don't know what it is. We're just calling that mass x. And what I want to do here is try to tackle this a little bit more symbolically. In the last video, we said, hey, why don't we just multiply 1/3 of this and multiply 1/3 of this? And then, essentially, we're going to keep things balanced, because we're taking 1/3 of the same mass. This total is the same as this total. That's why the scale is balanced. Now, let's think about how we can represent this symbolically. So the first thing I want you to think about is, can we set up an equation that expresses that we have these three things of mass x, and that in total, their mass is equal to the total mass over here? Can we express that as an equation? And I'll give you a few seconds to do it. Well, let's think about it. Over here, we have three things with mass x. So their total mass, we could write as-- we could write their total mass as x plus x plus x. And over here, we have nine things with mass of 1 kilogram. I guess we could write 1 plus 1 plus 1. That's 3. Plus 1 plus 1 plus 1 plus 1. How many is that? 1, 2, 3, 4, 5, 6, 7, 8, 9. And actually, this is a mathematical representation. If we set it up as an equation, it's an algebraic representation. It's not the simplest possible way we can do it, but it is a reasonable way to do it. If we want, we can say, well, if I have an x plus another x plus another x, I have three x's. So I could rewrite this as 3x. And 3x will be equal to? Well, if I sum up all of these 1's right over here-- 1 plus 1 plus 1. We're doing that. We have 9 of them, so we get 3x is equal to 9. And let me make sure I did that. 1, 2, 3, 4, 5, 6, 7, 8, 9. So that's how we would set it up. And so the next question is, what would we do? What can we do mathematically? Actually, to either one of these equations, but we'll focus on this one right now. What can we do mathematically in order to essentially solve for the x? In order to figure out what that mystery mass actually is? And I'll give you another second or two to think about it. Well, when we did it the last time with just the scales we said, OK, we've got three of these x's here. We want to have just one x here. So we can say, whatever this x is, if the scale stays balanced, it's going to be the same as whatever we have there. There might be a temptation to subtract two of the x's maybe from this side, but that won't help us. And we can even see it mathematically over here. If we subtract two x's from both sides, on the left-hand side you're going to have 3x minus 2x. And on the right-hand side, you're going to have 9 minus 2x. And you're just going to be left with 3 of something minus 2 something is just 1 of something. So you will just have an x there if you get rid of two of them. But on the right-hand side, you're going to get 9 minus 2 x's. So the x's still didn't help you out. You still have a mystery mass on the right-hand side. So that doesn't help. So instead, what we say is-- and we did this the last time. We said, well, what if we took 1/3 of these things? If we take 1/3 of these things and take 1/3 of these things, we should still get the same mass on both sides because the original things had the same mass. And the equivalent of doing that mathematically is to say, why don't we multiply both sides by 1/3? Or another way to say it is we could divide both sides by 3. Multiplying by 1/3 is the same thing as dividing by 3. So we're going to multiply both sides by 1/3. When you multiply both sides by 1/3-- visually over here, if you had three x's, you multiply it by 1/3, you're only going to have one x left. If you have nine of these one-kilogram boxes, you multiply it by 1/3, you're only going to have three left. And over here, you can even visually-- if you divide by 3, which is the same thing as multiplying by 1/3, you divide by 3. So you divide by 3. You have an x is equal to a 1 plus 1 plus 1. An x is equal to 3. Or you see here, an x is equal to 3. Over here you do the math. 1/3 times 3 is 1. You're left with 1x. So you're left with x is equal to 9 times 1/3. Or you could even view it as 9 divided by 3, which is equal to 3.

# data[].attributes.attachment_link.title
One-step equations intuition (video) | Khan Academy

# data[].attributes.attachment_link.description
So once again, we have three equal, or we say three identical objects. They all have the same mass, but we don't know what the mass is of each of them. But what we do know is that if you total up their mass, it's the same exact mass as these nine objects right over here.

# data[].attributes.body
One-step equations

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Video transcript

Now that we're comfortable with the "why" of why we do something to both sides of an equation, let's see if we can apply it to some equations to solve for an unknown variable. So let's say that you have x plus seven is equal to ten, and I want to solve for x. All its saying is something plus seven is equal to ten, and you might be able to figure that out in your head, but if you want to do it a little bit more systematically, you're like well just all I want on the left hand side is an x. Well if all I want on the left hand side is an x I'd want to get rid of the seven. I want to subtract seven from the left-hand side, but if I want to maintain an equality here, whatever I do to the left-hand side I also have to do to the right-hand side going back to our scales that's so that we can keep our scale balanced, so that we can say that the left is still equal to the right. And so what we're going to be left with is x and then the sevens cancel out is equal to ten minus seven is equal to three. So that unknown is three, and you can verify it, three plus seven is indeed equal to ten. Let's try one more. Let's say we have a minus five is equal to negative two. So this is a little bit more interesting since we have all of these negative numbers here, but we can use the exact same logic. We just want an a over here on the left-hand side so we have to get rid of this negative five somehow. Well the best way of getting rid of a negative 5 is to add five to it. So I'll do that. So I will add five to the left-hand side. But if I want the left-hand side to stay equal to the right-hand side, whatever I do to the left I have to do the right. So I'm going to have to add five on the right-hand side as well, and so on the left-hand side I'm left with a, and then the negative five and the positive five cancel out and on the right hand side, and they're going to stay equal because I did the same thing to both sides, we have negative two plus five which is equal to three. So a is equal to three. Once again you can verify it. Three minus five is indeed equal to negative two.

# data[].attributes.attachment_link.title
One-step addition & subtraction equations (video) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.body
One-step equations with fractions and decimals

# data[].attributes.attachment_link.readable_text
- [Voiceover] Let's give ourselves some practice solving equations. So let's say we had the equation 1/3 plus A is equal to 5/3. What is the A that makes this equation true? If I had 1/3 plus this A, what does A need to be in order for 1/3 plus that to be equal to 5/3? So there's a bunch of different ways of doing this, and this is one of the fun things about equations is there's no exactly one right way to do it. But let's think about what at least I think might be the simplest way. And before I work through anything, you should always try to pause the video, and do it on your own. So what I like to think about is can I have just my A on one side of the equation? And since it's already on the left-hand side, let's see if I can keep it on the left-hand side, but get rid of this 1/3 somehow. Well the easiest was I can think of getting rid of this 1/3 is to subtract 1/3 from the left-hand side of the equation. Now I can't just do that from the left-hand side of the equation. If 1/3 plus A is equal to 5/3, and if I just subtract 1/3 from the left-hand side, then they're not going to be equal anymore. Then this thing is going to be 1/3 less, which this thing isn't going to change. So then this thing on the left would become less than 5/3. So in order to hold the equality, whatever I do on the left-hand side I have to do on the right-hand side as well. So I have to subtract 1/3 from both sides. And if I do that, then on the left-hand side, 1/3 minus 1/3, that's the whole reason why I subtracted 1/3 was to get rid of the 1/3, and I am left with A is equal to 5/3 minus 1/3, 5/3 minus 1/3, minus 1/3, and what is that going to be equal to? I have five of something, in this case I have 5/3, and I'm gonna subtract 1/3. So I'm gonna be left with 4/3. So I could write A is equal to 4/3. And you could check to make sure that works. 1/3 plus 4/3 is indeed equal to 5/3. Let's do another one of these. So let's say that we have the equation K minus eight is equal to 11.8. So once again I wanna solve for K. I wanna have just a K on the left-hand side. I don't want this subtracting this eight right over here. So in order to get rid of this eight, let's add eight on the left-hand side. And of course, if I do it on the left-hand side, I have to do it on the right-hand side as well. So we're gonna add eight to both sides. The left-hand side, you are substracting eight and then you're adding eight. That's just going to cancel out, and you're just going to be left with K. And on the right-hand side, 11.8 plus eight. Well, 11 plus eight is 19, so it's going to be 19.8. And we're done, and once again, what's neat about equations, you can always check to see if you got the right answer. 19.8 minus eight is 11.8. Let's do another one, this is too much fun. Alright, so let's say that I had 5/13 is equal to T minus 6/13. Alright, this is interesting 'cause now I have my variable on the right-hand side. But let's just leave it there. Let's just see if we can solve for T by getting rid of everything else on the right-hand side. And like we've done in the past, if I'm subtracting 6/13, so why don't I just add it? Why don't I just add 6/13? I can't just do that on the right-hand side. Then the two sides won't be equal anymore, so I gotta do it on the left-hand side if I wanna hold the equality. So what happens? So what happens? On the left-hand side I have, let me give myself a little bit more space, I have 5/13 plus 6/13, plus 6/13 are equal to, are equal to... Well, I was subtracting 6/13, now I add 6/13. Those are just going to add to zero. 6/13 minus 6/13 is just zero, so you're left with T. So T is equal to this. If I have 5/13 and I add to that 6/13, well I'm gonna have 11/13. So this is going to be 11/13 is equal to T, or I could write that the other way around. I could write T is equal to 11/13.

# data[].attributes.attachment_link.title
One-step addition & subtraction equations: fractions & decimals (video) | Khan Academy

# data[].attributes.attachment_link.description
Voiceover] Let's give ourselves some practice solving equations. So let's say we had the equation 1/3 plus A is equal to 5/3. What is the A that makes this equation true? If I had 1/3 plus this A, what does A need to be in order for 1/3 plus that to be equal to 5/3?

# data[].attributes.body
One-step equations- addition and subtraction

# data[].attributes.attachment_link.title
One-step addition & subtraction equations (practice) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
One-step equations with fractions and decimals

# data[].attributes.attachment_link.title
One-step addition & subtraction equations (practice) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Two-step equations

# data[].attributes.attachment_link.readable_text
We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2. And the best way to get rid of that 2 is to subtract it. But if we want to subtract it from the right-hand side, we also have to subtract it from the left-hand side, because this is an equation. If this is equal to that, anything we do to that, we also have to do to this. So let's subtract 2 from both sides. So you subtract 2 from the right, subtract 2 from the left, and we get, on the left-hand side, negative 16 minus 2 is negative 18. And then that is equal to x over 4. And then we have positive 2 minus 2, which is just going to be 0, so we don't even have to write that. I could write just a plus 0, but I think that's a little unnecessary. And so we have negative 18 is equal to x over 4. And our whole goal here is to isolate the x, to solve for the x. And the best way we can do that, if we have x over 4 here, if we multiply that by 4, we're just going to have an x. So we can multiply that by 4, but once again, this is an equation. Anything you do to the right-hand side, you have to do to the left-hand side, and vice versa. So if we multiply the right-hand side by 4, we also have to multiply the left-hand side by 4. So we get 4 times negative 18 is equal to x over 4, times 4. The x over 4 times 4, that cancels out. You divide something by 4 and multiply by 4, you're just going to be left with an x. And on the other side, 4 times negative 18. Let's see, that's 40. Well, let's just write it out. So 18 times 4. If we were to multiply 18 times 4, 4 times 8 is 32. 4 times 1 is 4, plus 1 is 72. But this is negative 18 times 4, so it's negative 72. So x is equal to negative 72. And if we want to check it, we can just substitute it back into that original equation. So let's do that. Let's substitute this into the original equation. So the original equation was negative 16 is equal to-- instead of writing x, I'm going to write negative 72-- is equal to negative 72 over 4 plus 2. Let's see if this is actually true. So this right-hand side simplifies to negative 72 divided by 4. We already know that that is negative 18. So this is equal to negative 18 plus 2. This is what the equation becomes. And then the right-hand side, negative 18 plus 2, that's negative 16. So it all comes out true. This right-hand side, when x is equal to negative 72, does indeed equal negative 16.

# data[].attributes.attachment_link.title
Worked example: two-step equations | Algebra (video) | Khan Academy

# data[].attributes.attachment_link.description
We have the equation negative 16 is equal to x over 4, plus 2. And we need to solve for x. So we really just need to isolate the x variable on one side of this equation, and the best way to do that is first to isolate it-- isolate this whole x over 4 term from all of the other terms. So in order to do that, let's get rid of this 2.

# data[].attributes.body
Two-step equations with fractions and decimals

# data[].attributes.attachment_link.readable_text
- Let's get some practice solving equations that involve fractions and decimals. And this equation clearly involves fractions. Let's see. We have negative 1/3 = j/4 - 10/3. So I encourage you to pause the video and see if you could solve for j. What j would make this equation true? Alright. Now let's work through this together. So what I like to do, I like to isolate the variable that I'm trying to solve for on one side. Since it's already on the right-hand side, let's try to get all the things that involve j on the right-hand side, and then get rid of everything else on the right-hand side. So I want to get rid of this negative 10/3, and the best way I can think of doing that is by adding 10/3. Now I can't just do that to one side of the equation. Then it wouldn't be equal anymore. If this is equal to that, in order for the quantity to be true, whatever I do to this I have to do to that as well. So I have to add 10/3 to both sides. I have to add 10/3 to both sides of the equation. And so what am I going to get? On the left-hand side, I'm going to have negative 1/3 + 10/3, which is 9/3. 9/3. And then that's going to be equal to. And on the right-hand side, the negative 10/3 and the positive 10/3, those cancel out to get a zero, and I'm just left with j/4. It's equal to j/4. Now you might recognize 9/3, that's the same thing as nine divided by three. So this is just going to be three. So that simplifies a little bit. Three, let me just rewrite it so you don't get confused. Three is equal to j/4. Now, to solve for j, I could just multiply both sides by four. 'Cause if I divide something by four and then multiply by four, I'm just going to be left with that something. If I start with j and I divide by four, and then I multiply, and then I multiply by four, so I'm just going to multiply by four, then I'm just going to be left with j on the right-hand side. But I can't just multiply the right-hand side by four. I have to do it with the left-hand side as well. So I multiply the left-hand side by four as well. And what I will be left with, four times three is twelve. And then j divided by four times four, well that's just going to be j. So we get j is equal to 12. And the neat thing about equations is you can verify that you indeed got the right answer. You can substitute 12 for j here, and verify that negative 1/3 is equal to 12/4 - 10/3. Does this actually work out? Well 12/4 is the same thing as three, and if I wanted to write that as thirds, this is the same thing as 9/3. And 9/3 - 10/3 is indeed equal to negative 1/3. So we feel very good about that. Let's do another example. So I have n/5 + 0.6 = 2. So let's isolate this term that involves n on the left-hand side. So let's get rid of this 0.6. So let's subtract 0.6 from the left-hand side. But I can't just do it from the left. I have to do it from both sides if I want the equality to hold true. So, subtract 0.6. Now on the left-hand side, I'm just going to be left with n/5, and on the right-hand side, 2 - 0.6, that;'s going to be 1.4. And if you don't want to do this in your head, you could work this out separately. It's going to be 2.0 - 0.6. You could say, "Oh, this is 20/10-6/10" which is going to be 14/10, which is that there. Or if you want to do it a little bit kind of the traditional method, you say, "Oh, I'm trying to subtract six from zero, let me re-group." That's going to be a 10. I'm going to take from the ones place. If I take a one from the one's place, and that's going to be equal to 10/10. 10/10 - 6/10 is 4/10. And then, bring down one one minus zero ones is just one. So it's 1.4. And now, to solve for n. Well on the left have n being divided by five. If I just want n here, I can just multiply by five. So, if I multiply by five, five times n divided by five is going to be just n. But I can't just multiply the left-hand side by five. I have to multiply the right-hand side by five as well. And so what is that going to get us? We are going to get n = 1.4 x 5. 1.4 x 5. Now you might be able to do this in your head, 'cause this is one and 2/5. So this thing should all be equal to seven, but I'll just do it this way as well. Five times four is 20. Re-group the two. One times five is five plus two is seven. And when I look at all the numbers that I'm multiplying, I have one digit to the right of the decimal point. So my answer will have one digit to the right of the decimal point. So it's 7.0, or just 7. n = 7. And you can verify this works, 'cause seven divided by five is going to be equal to 1.4, plus 0.6 is equal to two. Let's do one more example. This is too much fun. Alright. 0.5 times the whole quantity (r + 2.75) = 3. Now there's a bunch of ways that you could tackle this. A lot of times when you see something like this, your temptation might be, "Let's distribute the 0.5." But that makes it a little bit hairy, 'cause 0.5 times 2.75. You could calculate that, and you will get the right answer if you do it correctly. But a simpler thing might be, well, let's just divide both sides by 0.5. That way I'm going to get more whole numbers involved. So if I divide. Remember, whatever I do to the left-hand side I have to do to the right-hand side. And the way my brain thought about it was, well, if I divide by 0.5 on the left-hand side I can get rid of this. And if I divide by 0.5 on the right-hand side I'm still going to get an integer. Three divided by 0.5 is six. It's the same thing as three divided by a half. How many halves fit into three? Six halves fit into three. So this going to be six right over here. So these cancel out. And then this is going to be equal to six. So the whole thing is simplified now to r + 2.75 is going to be equal to is equal to six. And now to just isolate the r on the left-hand side, I can substract the 2.75 from the left. But, like we've seen multiple times, I can't just do it from, I can't just do it from the left. I'm having, my brain is malfunctioning. 2.75. I can't just do it from the left. I have to do it from the right as well. 2.75. So this simplifies to r, this simplifies to r, Is equal to -- what's six minus 2.75? Well if you want to do it in your head, six minus two would be four, and then if you take 0.75 from that it would be 3.25. If you don't feel comfortable doing it in your head, we could just write it out. 6.00 - 2.75, be careful to align the decimals and then I've got to do some re-grouping. Let's see, I have zero hundreds, trying to subtract five hundreds. That's not going to work out. So I try to re-group from here, but I have nothing here. So let me re-group from here. One one. Let me take away one of these ones. I'm going to have five ones. And then that's going to be equivalent, so I have five ones here, and that one one I took away is going to be ten tenths. And then I can take one of those tenths away. So I'm going to have nine tenths. And that's going to be ten hundredths. And now I could subtract. Ten hundreths minus five hundredths is five hundredths. Nine tenths minus seven tenths is two tenths. My decimal is going to be there. Five minus two is three. 3.25. And you can verify that. 3.25 + 2.75 is six, times 0.5 is indeed equal to three. So we feel, once again, really good about this.

# data[].attributes.attachment_link.title
Two-step equations with decimals and fractions (video) | Khan Academy

# data[].attributes.attachment_link.description
Let's get some practice solving equations that involve fractions and decimals. And this equation clearly involves fractions. Let's see. We have negative 1/3 = j/4 - 10/3. So I encourage you to pause the video and see if you could solve for j. What j would make this equation true?

# data[].attributes.body
Solving two-step equations powerpoint

# data[].attributes.body
Two-step equations

# data[].attributes.attachment_link.title
Two-step equations | Algebra (practice) | Khan Academy

# data[].attributes.attachment_link.description
Practice solving equations that take two steps to solve. For example, solve -16 = x/4 + 2.

# data[].attributes.body
Two-step equations with fractions and decimals

# data[].attributes.attachment_link.title
Two-step equations with decimals and fractions (practice) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Two-step equations- find the mistake

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# data[].attributes.attachment_link.title
Find the mistake: two-step equations (practice) | Khan Academy

# data[].attributes.attachment_link.description
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

# data[].attributes.body
Two-step equation word problems

# data[].attributes.attachment_link.title
Two-step equations word problems (practice) | Khan Academy

# data[].attributes.attachment_link.description
Practice writing equations to model and solve real-world situations.

# data[].attributes.body
Algebra Connect Four game (can be played with a partner)

# data[].attributes.attachment_link.readable_text
Interactivate: Algebra Four

Algebra Four: Students play a generalized version of connect four, gaining the chance to place a piece on the board by solving an algebraic equation. Parameters: Level of difficulty of equations to solve and type of problem. Algebra Four is one of the Interactivate assessment games.

On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. Student development of numerical models and simulations integrated with core curriculum provides an opportunity to gain practical experience in computational science. Additionally, the National Computational Science Institute (NCSI) provides nation-wide workshops portraying resources and instructional ideas to middle school, high school, and undergraduate instructors for use in the classroom. Resources and materials offered to these instructors are available free of charge from Shodor's website and are largely developed by Shodor student interns. Shodor's academic program efficiently guides participants from excitement to experience to expertise through computational explorations, research opportunities, and service.

We're sorry!

We have detected JavaScript as being disabled in your browser.

The links below provide instructions for enabling JavaScript dependent on your browser.

Mozilla Firefox Users
Internet Explorer Users
Safari Users

After enabling JavaScript, refresh the page.
You may also try using the help feature of your browser.

# data[].attributes.attachment_link.description
Interactivate: Algebra Four On a mission to transform learning through computational thinking, Shodor is dedicated to the reform and improvement of mathematics and science education through student enrichment, faculty enhancement, and interactive curriculum development at all levels. Student development of numerical models and simulations integrated with core curriculum provides an opportunity to gain practical experience in computational science.

# data[].attributes.body
Proportions

# data[].attributes.body
Multi-step equations (variables on both sides)

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Linear equations with variables on both sides

Linear equations with variables on both sides

# data[].attributes.attachment_link.title
Equations with variables on both sides (practice) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.body
Multi-step equations with distribution

# data[].attributes.attachment_link.readable_text
If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Linear equations with parentheses

Linear equations with parentheses

# data[].attributes.attachment_link.title
Equations with parentheses (practice) | Khan Academy

# data[].attributes.attachment_link.description
If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

# data[].attributes.body
Intro to solving equations with variables on both sides

# data[].attributes.attachment_link.readable_text
Let's try to solve a more involved equation. So, let's say that we have 2x plus 3, 2x plus 3 is equal to is equal to 5x minus 2. So this might look a little daunting at first. We have x's on both sides of the equation. We're adding and subtracting numbers. How do you solve it? And we'll do it in a couple of different ways. The the important thing to remember is, we just want to isolate an x. Once you've isolated an x, you have x equals something. Or x equals something. You're done, you've solved the equation. You can actually go back and check whether that works, So what we're going to do is just do a bunch of operations on both sides of the equation, to eventually isolate the x. But while we do those, I actually want to visualize what's occurring. Because I don't want you just say, oh what are the rules or the steps of solving equations. And I forgot whether this is a allowed or that isn't allowed. If you visualize what's happening, it'll actually be common sense what's allowed. So let's visualize it. So we have 2x right here on the left-hand side. So that's literally, that's x plus x. And then you have plus 3. Plus 3, I'll do it like this. So that's equal to plus 1, plus 1, plus 1. That's the same thing as 3. I could've drawn 3 circles here as well. Let do the same color. Plus 3. And then that is equal to 5 x's. Do that in blue. That is equal to 5 x's. So, 1, 2, 3, 5, 6. And I want to make it clear. You would never actually have to do it this way when you're solving the problem. You would just have to do the algebraic steps. But I'm doing this for you so you can actually visualize what this equation is saying. the left-hand side is these two orange x's plus 3. The right-hand side is 5x minus 2. So minus 2, we could write as -- so let me do this in a different color, I'll do it in pink. So, minus 2, I'll do as minus 1 and minus 1. Now, we want to isolate the x's on the same side of the equation. So, how could we do that? Well, there's two ways of doing it. We could subtract these two x's from both sides of the equation. And that would be pretty reasonable. Because then you'd have 5 x's minus the 2 x's. You'd have a positive number of x's on the right-hand side. Or, you could actually subtract 5x from both sides. And that's what's neat about algebra. As long as you do legitimate operations, you will eventually get the right answer. So let's just start off subtracting 2x from both sides of the equation. And what I mean there, I mean we're going to remove 2 x's from the left-hand side. And if we were to move 2 x's on the left-hand side, we have to remove 2 x's the right-hand side. Just like that. So what does that give us? We're subtracting 2 x's. 2 x's from the left. And we're also going to subtract 2 x's from the right. Now, what does our left-hand side simplify to? We have 2x plus 3 minus 2x. The 2 x's cancel out. So you're just left with -- you're just left with the 3. And you see that over here. We took 2 of these x's away. We're just left with the plus 1, plus 1, plus 1. And then on the right-hand side, 5x minus 2x. We have it over here. We have 5 x's minus 2 x's. You only have 1, 2, 3, x's left over. 3 is equal to 3x. And then you have your minus 2 there. You have your minus 2. So, normally if you were to do the problem, you would just have to write what we have here on the left-hand side. So what can we do next? Remember, we want to isolate the x's. Well, we have all of our x's on the right-hand side right here. If we could get rid of this negative 2, off of the right-hand side, then the x's will be alone. They'll be isolated. So how can we get rid of this negative 2, if we visualize it over here. This negative 1, this negative 1. Well, we could add 2 to both sides of this equation. Think about what happens there. So, if we add 2, so I'm going to do it like this. Plus 1, plus 1. So you could literally see. We're adding 2. And then we're going to add 2 to the left-hand side. 1 plus, 1 plus. What happens? Let me do it over here as well. So we're going to add 2. We're going to add 2. So what happens to the left-hand side? 3 plus 2 is going to be equal to 5. And that is going to be equal to 3x minus 2 plus 2. These guys cancel out. And you're just left with 3x. And we see it over here. We have the left-hand side is 1 plus 1 plus 1 plus 1 plus 1. We have 5 1's, or 5. And the right-hand side, we have the 3 x's, right over there. And then we have the negative 1, negative 1. Plus 1, plus 1, negative 1, these cancel out. They get us to 0. They cancel out. So we're just left with 5 is equal to 3x. So we have 1, 2, 3, 4, 5 is equal to 3x. Let me clear everything that we've removed, so it looks a little bit cleaner. These are all of the things that we've removed. Let me clear that out. And then let me clear that out, like that. Edit. Clear. So now we are just left with 1, 2, 3, 4, 5. Actually, let me move this over. So I could just move this over right over here. We now have 1, 2, 3, 4, 5. These are the two that we added here, is equal to 3x. These guys canceled out. That's why we have nothing there. Now, to solve this, we just divide both sides of this equation by 3. And this is going to be a little hard to visualize over here. But if we divide over here both sides by 3, what do we get? We divide the left by 3. We divide the right by 3. The whole reason why we divided by 3 is because the x was being multiplied by 3. 3 is the coefficient on the x. Fancy word, it literally just means the number multiplying the variable. The number we're solving, the variable we're solving for. So these 3's cancel out. The right-hand side of the equation is just x. The left-hand side is 5/3. So 5/3, we could say is is equal to 5/3. And this is different than everything we've seen so far. I now have the x on the right-hand side, the value on the left-hand side. That's completely fine. This is the exact same thing as saying 5/3 is is equal to x is the same thing as saying x's equal to 5/3. Completely equivalent. Completely equivalent. We sometimes get more used to this one, but this is completely the same thing. Now, if we wanted to write this as a mixed number, if we want to write this as a mixed number, 3 goes into 5 one time with remainder 2. So it's going to be 1 2/3. So it's going to be 1 2/3. So we could also write that x is equal to 1 2/3. And I'll leave it up for you to actually substitute back into this original equation. And see that it works out. Now, to visualize it over here, you know, how did he get 1 2/3, let's think about it. Instead of doing 1, I'm going to do circles. I am going to do circles. Actually, even better, I'm going to do squares. So I'm going to have 5 squares on the left-hand side. I'll do it in this same yellow color right here. So I have 1, 2, 3, 4, 5. And that is going to be equal to the 3 x's. x plus x plus x. Now, we're dividing both sides of the equation by 3. We're dividing both sides of the equation by 3. Actually, that's where we did it up here, we divided both sides by 3. So how do you do that right-hand side's pretty straightforward. You want to divide these 3 x's into 3 groups. That's 1, 2, 3 groups. 1, 2, 3. Now, how do you divide 5 into 3? And they have to be thought through even groups. And the answer tells us. Each group is going to be 1 2/3. So, 1 2/3. So it's going to be 2/3 of this, the next one. And then we're going to have 1 2/3. So this is 1/3. We're going to need another. Another 1, so this is 1 1/3. We're going to need 1 more 1/3, so this is going to be right here. And then we're left with 2/3 and 1. So we've broken it up into 3 groups. This right here. Let me make it clear. Let me make it clear, this right here is 1 2/3. 1 2/3. And then this right here, this 1/3. That's another 1/3, so that's 2/3, and then that's 1 right there. So that's 1 2/3. And then finally this is 2/3 and this is 1, so this is 1 2/3. So when you divide both sides by 3 you get 1 2/3. Each section, each bucket, is 1 2/3 on the left-hand side. On the left-hand side, or 5/3. And on the right-hand side we just have an x. So it still works. A little bit harder to visualize with fractions.

# data[].attributes.attachment_link.title
Intro to equations with variables on both sides (video) | Khan Academy

# data[].attributes.attachment_link.description
Let's try to solve a more involved equation. So, let's say that we have 2x plus 3, 2x plus 3 is equal to is equal to 5x minus 2. So this might look a little daunting at first. We have x's on both sides of the equation. We're adding and subtracting numbers.

# data[].attributes.body
Multi-step equations with Distribution

# data[].attributes.attachment_link.readable_text
We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses. And the best way to get rid of these parentheses is to kind of multiply them out. This has a negative 1-- you just see a minus here, but it's just really the same thing as having a negative 1-- times this quantity. And here you have a 3 times this quantity. So let's multiply it out using the distributive property. So the left-hand side of our equation, we have our negative 9. And then we want to multiply the negative 1 times each of these terms. So negative 1 times 9x is negative 9x, and then negative 1 times negative 6 is plus 6, or positive 6. And then that is going to be equal to-- let's distribute the 3-- 3 times 4x is 12x. And then 3 times 6 is 18. Now what we want to do, let's combine our constant terms, if we can. We have a negative 9 and a 6 here, on this side, we've combined all of our like terms. We can't combine a 12x and an 18, so let's combine this. So let's combine the negative 9 and the 6, our two constant terms on the left-hand side of the equation. So we're going to have this negative 9x. So we're going to have negative 9x plus-- let's see, we have a negative 9 and then plus 6-- so negative 9 plus 6 is negative 3. So we're going to have a negative 9x, and then we have a negative 3, so minus 3 right here. That's the negative 9 plus the 6, and that is equal to 12x plus 18. Now, we want to group all the x terms on one side of the equation, and all of the constant terms-- the negative 3 and the positive 18 on the other side-- I like to always have my x terms on the left-hand side, if I can. You don't have to have them on the left, so let's do that. So if I want all my x terms on the left, I have to get rid of this 12x from the right. And the best way to do that is to subtract 12x from both sides of the equation. So let me subtract 12x from the right, and subtract 12x from the left. Now, on the left-hand side, I have negative 9x minus 12x. So negative 9 minus 12, that's negative 21. Negative 21x minus 3 is equal to-- 12x minus 12x, well, that's just nothing. That's 0. So I could write a 0 here, but I don't have to write anything. That was the whole point of subtracting the 12x from the left-hand side. And that is going to be equal to-- so on the right-hand side, we just are left with an 18. We are just left with that 18 here. These guys canceled out. Now, let's get rid of this negative 3 from the left-hand side. So on the left-hand side, we only have x terms, and on the right-hand side, we only have constant terms. So the best way to cancel out a negative 3 is to add 3. So it cancels out to 0. So we're going to add 3 to the left, let's add 3 to the right. And we get-- the left-hand side of the equation, we have the negative 21x, no other x term to add or subtract here, so we have negative 21x. The negative 3 and the plus 3, or the positive 3, cancel out-- that was the whole point-- equals-- what's 18 plus 3? 18 plus 3 is 21. So now we have negative 21x is equal to 21. And we want to solve for x. So if you have something times x, and you just want it to be an x, let's divide by that something. And in this case, that something is negative 21. So let's divide both sides of this equation by negative 21. Divide both sides by negative 21. The left-hand side, negative 21 divided by negative 21, you're just left with an x. That was the whole point behind dividing by negative 21. And we get x is equal to-- what's 21 divided by negative 21? Well, that's just negative 1. Right? You have the positive version divided by the negative version of itself, so it's just negative 1. So that is our answer. Now let's verify that this actually works for that original equation. So let's substitute negative 1 into that original equation. So we have negative 9-- I'll do it over here; I'll do it in a different color than we've been using-- we have negative 9 minus-- that 1 wasn't there originally, it's there implicitly-- minus 9 times negative 1. 9 times-- I'll put negative 1 in parentheses-- minus 6 is equal to-- well, actually, let me just solve for the left-hand side when I substitute a negative 1 there. So the left-hand side becomes negative 9, minus 9 times negative 1 is negative 9, minus 6. And so this is negative 9 minus-- in parentheses-- negative 9 minus 6 is negative 15. So this is equal to negative 15. And so we get negative 9-- let me make sure I did that-- negative 9 minus 6, yep, negative 15. So negative 9 minus negative 15, that's the same thing as negative 9 plus 15, which is 6. So that's what we get on the left-hand side of the equation when we substitute x is equal to negative 1. We get that it equals 6. So let's see what happens when we substitute negative 1 to the right-hand side of the equation. I'll do it in green. We get 3 times 4 times negative 1 plus 6. So that is 3 times negative 4 plus 6. Negative 4 plus 6 is 2. So it's 3 times 2, which is also 6. So when x is equal to negative 1, you substitute here, the left-hand side becomes 6, and the right-hand side becomes 6. So this definitely works out.

# data[].attributes.attachment_link.title
Equations with parentheses (video) | Khan Academy

# data[].attributes.attachment_link.description
We have the equation negative 9 minus this whole expression, 9x minus 6-- this whole thing is being subtracted from negative 9-- is equal to 3 times this whole expression, 4x plus 6. Now, a good place to start is to just get rid of these parentheses.

# data[].attributes.body
Number of solutions to a linear equation

# data[].attributes.attachment_link.title
Number of solutions to equations (practice) | Khan Academy

# data[].attributes.attachment_link.description
Practice telling whether an equation has one, zero, or infinite solutions. For example, how many solutions does the equation 8(3x+10)=28x-14-4x have?

# data[].attributes.body
One, None or Infinite Solutions

# data[].attributes.attachment_link.title
What does it mean for an equation to have ONE SOLUTION, NO SOLUTION, OR INFINITELY MANY SOLUTIONS? | Educreations

# data[].attributes.attachment_link.description
Educreations is a community where anyone can teach what they know and learn what they don't. Our software turns any iPad or web browser into a recordable, interactive whiteboard, making it easy for teachers and experts to create engaging video lessons and share them on the web.

# data[].attributes.body
Number of Solutions

# data[].attributes.attachment_link.readable_text
Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions. You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number. Let's say x is equal to-- if I want to say the abstract-- x is equal to a. Or if we actually were to solve it, we'd get something like x equals 5 or 10 or negative pi-- whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. And if you were to just keep simplifying it, and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no. No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5-- and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for. Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, we could subtract 2 from both sides. If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. And then you would get zero equals zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side, this is going to go away and you're just going to be left with a 2 there. So all I did is I added 7x. I added 7x to both sides of that equation. And now we've got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides, just to get rid of this constant term. So we're going to get negative 7x on the left hand side. On the right hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides. To subtract 2x from both sides, you're going to get-- so subtracting 2x, you're going to get negative 9x is equal to negative 1. Now you can divide both sides by negative 9. And you are left with x is equal to 1/9. So we're in this scenario right over here. We very explicitly were able to find an x, x equals 1/9, that satisfies this equation. So this right over here has exactly one solution.

# data[].attributes.attachment_link.title
Number of solutions to equations | Algebra (video) | Khan Academy

# data[].attributes.attachment_link.description
Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions.

# data[].attributes.body
Number of Solutions Example

# data[].attributes.attachment_link.readable_text
Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's from both sides of the equation because we have a 24x there, and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14.

# data[].attributes.attachment_link.title
Worked example: number of solutions to equations (video) | Khan Academy

# data[].attributes.attachment_link.description
Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides.

# data[].attributes.body
Equations with No Solutions

# data[].attributes.body
Creating an equation with infinitely many solutions

# data[].attributes.attachment_link.readable_text
We're asked to use the drop-down to form a linear equation with infinitely many solutions. So an equation with infinitely many solutions essentially has the same thing on both sides, no matter what x you pick. So first, my brain just wants to simplify this left-hand side a little bit and then think about how I can engineer the right-hand side so it's going to be the same as the left no matter what x I pick. So right over here, if I distribute the 4 over x minus 2, I get 4x minus 8. And then I'm adding x to that. And that's, of course, going to be equal to 5x plus blank. And I get to pick what my blank is. And so 4x plus x is 5x. And of course, we still have our minus 8. And that's going to be equal to 5x plus blank. So what could we make that blank so this is true for any x we pick? Well, over here we have 5 times an x minus 8. Well, if we make this a minus 8, or if we subtract 8 here, or if we make this a negative 8, this is going to be true for any x. So if we make this a negative 8, this is going to be true for any x you pick. You give me any x, you multiply it by 5 and subtract 8, that's, of course, going to be that same x multiplied by 5 and subtracting 8. And if you were to try to somehow solve this equation, subtract 5x from both sides, you would get negative 8 is equal to negative 8, which is absolutely true for absolutely any x that you pick. So let's go-- let me actually fill this in on the exercise. So I want to make 5-- it's going to be 5x plus negative 8.

# data[].attributes.attachment_link.title
Creating an equation with infinitely many solutions (video) | Khan Academy

# data[].attributes.attachment_link.description
We're asked to use the drop-down to form a linear equation with infinitely many solutions. So an equation with infinitely many solutions essentially has the same thing on both sides, no matter what x you pick.

# data[].attributes.body
Equations with Infinite and No Solutions

# data[].attributes.body
Adding\Subtracting Integers

# data[].attributes.attachment_link.title
Understanding how positive and negative numbers interact

# data[].attributes.body
Solution to an equation

# data[].attributes.attachment_link.title
What is a Solution to an Equation? | Virtual Nerd

# data[].attributes.attachment_link.description
Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In this non-linear system, users are free to take whatever path through the material best serves their needs. These unique features make Virtual Nerd a viable alternative to private tutoring.

# data[].attributes.body
Solving multi-step equations lesson

# data[].attributes.body
This can only be completed in Mrs. Campos' class

# data[].attributes.body
Multi-step equations practice on white boards

# data[].attributes.body
More multi-step equation practice on white boards

# data[].attributes.body
One-step equation practice on white boards

# data[].attributes.body
Two-step equation practice (on white boards)

# data[].attributes.body
Final Exam Review