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What does it mean for an equation to have ONE SOLUTION, NO SOLUTION, OR INFINITELY MANY SOLUTIONS? | Educreations

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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.

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Adding and subtracting negative numbers

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Adding & subtracting negative numbers (practice) | Khan Academy

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Multiplying and Dividing Integers

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Equivalent expressions

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Practice determining whether or not two algebraic expressions are equivalent by manipulating the expressions. These problems require you to combine like terms and apply the distributive property.

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Combining like terms with negative coefficients & distribution

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Combining like terms with negative coefficients & distribution

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Practice combining like terms using the distributive property. These problems involve adding and subtracting negative numbers.

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Combining like terms with rational coefficients

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Combining like terms with rational coefficients

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Practice combining like terms for expressions that have rational coefficients. This requires adding and subtracting fractions.

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Why is it important to know what operation to use in different situations?

Why is it important to compare measures using the same type of unit? (Ex. Inches to inches rather than inches to feet)

How are linear equations and inequalities solved by using the properties of equality?

How are linear models used to represent real world situations?

How can writing equations that model situations help us to solve problems?

How do you use patterns to understand mathematics and model situations?

How can we communicate and generalized algebraic relationship?

How can I convert from one unit of measure to another?

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Students will strengthen foundational arithmetic skills and academic vocabulary.

Students will master order of operations and evaluating expressions.

Students will use unit conversions appropriately and choose a level of accuracy appropriate to limitations on measurement.

Students will master solving basic one and two step equations and inequalities and graph their solutions.

Students will solve multi-step equations and construct a viable argument using the equality properties of
numbers to justify their solution.

Students will solve multi-step inequalities and interpret their solutions graphically.

Students will solve absolute value equations, compound inequalities and absolute value inequalities.

Students will solve linear models involving equations and inequalities as well as with absolute value and
compound inequalities.

Students will rearrange formulas to solve for one of the variables.

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PRIORITY:
NQ.1​ Use units as a way to understand problems and to guide the solution of multi-step problems;
choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.
A.SSE.1​ Interpret expressions that represent a quantity in terms of its context.
a. Interpret parts of an expression, such as terms, factors, and coefficients
b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For
example, interpret P(1 + r)n
as the product of P and a factor not depending on P.
A.CED.1​ Create equations and inequalities in one variable including ones with absolute value ​and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. CA
A.CED.2​ Create equations in two or more variables to represent relationships between quantities;
graph equations on coordinate axes with labels and scales.
A.CED.4​ Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V = IR to highlight resistance R.

SUPPORTING:
NQ.2​ Define appropriate quantities for the purpose of descriptive modeling.
NQ.3​ Choose a level of accuracy appropriate to limitations on measurement when reporting
quantities
A.CED.3​ Represent constraints by equations or inequalities, and by systems of equations and/or
inequalities, and interpret solutions as viable or non-viable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of
different foods.
A.REI.1 ​Explain each step in solving a simple equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method.
A.REI.3: ​Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
A.REI.3.1 (CA) ​Solve one-variable equations and inequalities involving absolute value, graphing the
solutions and interpreting them in context.

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Testing solutions to equations

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Testing solutions to equations (practice) | Khan Academy

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Practice plugging in values to equations and checking to see if the equations hold true.

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Mixing Fertilizer

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Solving multi-step equations

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To solve an equation we find the value of the variable that makes the equation true. For more complicated, fancier equations, this process can take several steps.

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Solving equations with distributive property

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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.

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Equations with parentheses (video) | Khan Academy

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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.

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Equations with parentheses

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Linear equations with parentheses

Linear equations with parentheses

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Equations with parentheses (practice) | Khan Academy

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Multi-step Inequalities

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Sal solves several multi-step linear inequalities.

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Equations with parentheses: decimals & fractions

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Equations with parentheses: decimals & fractions

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Solve linear equations that contain expressions with parentheses and the coefficients are rational numbers.

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Multi-step Inequalities

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Inequalities with variables on both sides (with parentheses)

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Sal solves the inequality 5x+7>3(x+1), draws the solution on a number line and checks a few values to verify the solution.

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Multi-step Inequalities

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Inequalities with variables on both sides

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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.

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Multi-step linear inequalities

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Solve linear inequalities that take multiple steps to solve.

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Intro to absolute value inequal

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Sal introduces the concept of inequalities that contain absolute value expressions, and solves a bunch of examples.

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Compound Inequalities

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Compound inequalities: OR | Algebra (video) | Khan Academy

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Sal solves the compound inequality 5z+7<27 OR -3z≤18.

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Compound Inequalities

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Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this "and" is here. So we have 3y plus 7 is less than 2y. So let's isolate the y's on the left-hand side. So let's get rid of this 2y on the right-hand side, and we can do that by subtracting 2y from both sides. So we're going to subtract 2y from both sides. The left-hand side, we have 3y minus 2y, which is just y, plus 7 is less than 2y minus 2y. And there's nothing else there. That's just going to be 0. And then we can get rid of this 7 here by subtracting 7 from both sides. So let's subtract 7 from both sides. Left-hand side, y plus 7 minus 7. Those cancel out. We just have y is less than 0 minus 7, which is negative 7. So that's one of the constraints. That's this constraint right over here. Now let's work on this constraint. We have 4y plus 8 is greater than negative 48. So let's get rid of the 8 from the left-hand side. So we can subtract 8 from both sides. The left-hand side, we're just left with a 4y because these guys cancel out. 4y is greater than negative 48 minus 8. So we're going to go another 8 negative. So 48 plus 8 would be a 56, so this is going to be negative 56. And now to isolate the y, we can divide both sides by positive 4, and we don't have to swap the inequality since we're dividing by a positive number. So it's divide both sides by 4 over here. So we get y is greater than-- what is 56/4, or negative 56/4? Let's see. 40 is 10 times 4, and then we have another 16 to worry about. So it's 14 times 4. So y is greater than negative 14. Is that right? 4 times 10 is 40, 4 times 4 is 16. Yep, 56. So y is greater than negative 14 and-- let's remember, we have this "and" here-- and y is less than negative 7. So we have to meet both of these constraints over here. So let's draw them on the number line. So I have my number line over here. And let's say negative 14 is over here. So you have negative 13, 12, 11, 10, 9, 8, 7-- that's negative 7-- and then negative 6, 5, 4, 3, 2, 1. This would be 0, and then you could keep going up more positive. And so we're looking for all of the y's that are less than negative 7. So let's look at this, less than negative 7. So not including negative 7, so we'll do an open circle around negative 7, and less than negative 7. And if that was the only constraint, we would keep going to the left. But we have this other constraint-- and y has to be greater than negative 14. So you make a circle around negative 14, and everything that's greater than that. And if you didn't have this other constraint, you would keep going. But the y's that satisfy both of them are all of the y's in between. These are the y's that are both less than negative 7 and greater than negative 14. And we can verify that things here work. So let's try some values out. So a value that would work, well, let me just do negative 10 is right here, 8, 9, this is negative 10. That should work. So let's try it out. So we'd have 3 times negative 10 plus 7 should be less than 2 times negative 10. So this is negative 30 plus 7 is negative 23, which is indeed less than negative 20. So that works. And negative 10 has to work for this one as well. So you have 4 times negative 10, which is negative 40, plus 8. Negative 40 plus 8 should be greater than negative 48. Well, negative 40 plus 8 is negative 32. We're going 8 in the positive direction, so we're getting less negative. And negative 32 is greater than negative 48. It's less negative. So this works. So negative 10 works. Now, let's just verify some things that shouldn't work. So 0 should not work. It's not in the solution set. So let's try it out. We've got 3 times is 0 plus 7. That would be 7. And 7 is not less than 0. So it would violate this condition right over here if we put a 0 over here. If you put a negative 15 over here, it should violate this condition right over here because it wasn't in this guy's solution set. Anyway, hopefully you found that useful.

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Compound inequalities: AND | Algebra (video) | Khan Academy

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Solve for y. We have 3y plus 7 is less than 2y and 4y plus 8 is greater than negative 48. So we have to find all the y's that meet both of these constraints. So let's just solve for y in each of the constraints and just remember that this "and" is here.

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Compound inequalities

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Compound inequalities | Algebra (practice) | Khan Academy

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Rate conversion

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Sal shows how we can treat units of measurement algebraically, and use these tools in order to convert between different units of the same quantity.

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Rate conversion

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Convert a given rate to an equivalent rate in different units. For example, convert 10 km/h to 36 m/s.

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Same rate with different units

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Sal shows how we can describe the rate 50 km per hour in a variety of different units, using dimensional analysis.

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Appropriate units

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Examples for determining the appropriate units to use to measure a quantity of interest.

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Appropriate units

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Pick the appropriate units to use when modeling real-world phenomena.

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Solving for a variable

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Video transcript

Solve P equals 2l plus 2w for l. So this right here, this is just the formula for the perimeter of a rectangle. Perimeter is equal to 2 times the length plus 2 times the width. But they just want us to solve this equation right here, solve for l. So let's do that. So we have P is equal to 2 times l plus 2 times w. So we need to solve for l. So let's isolate all of the l terms on one side. And the best way, we could just do that by leaving it here on the right and then getting rid of this 2w. And the best way to get rid of this 2w is to subtract it from the right. But if you're going to subtract it from the right, you also have to subtract it from the left if you want this equality to hold. So you have to also subtract it from the left. And so the left-hand side becomes P minus 2w. And the right-hand side, you get-- this 2w minus 2w cancels out. You just have a 2l. And then if you want to solve for l, you just have to divide both sides of this equation by 2. You just divide both sides of this equation by 2. And we have isolated our l. We get l is equal to P minus 2w over 2. Or if we wanted to write it the other way, you could write l is equal to P minus 2w over 2. And we are done.

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Solving an equation for a variable (video) | Khan Academy

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Length
Weight
Capacity
Justify
Isolate
Rearrange
Isolate

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Power
Base
Exponent
Expression
Coefficients
Constants
Like Terms
Distributive Property
Evaluate
Substitute
Variable
Units
Measurement
Conversion
Proportion
Ratio
Evaluate
Substitute
Equations
Inequalities
Term
Integers
Power
Base
Exponent
Expression
Coefficients
Constants
Equations
Inequalities
Compound Inequalities
Absolute Value
Linear Models

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Absolute Value

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Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum. For K-12 kids, teachers and parents.

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One, None or Infinite Solutions

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Solving two-step equations powerpoint

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Proportions

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Equations with Infinite and No Solutions

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Integer worksheet

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Adding\Subtracting Integers

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Understanding how positive and negative numbers interact

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The Distributive Property

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Solving multi-step equations lesson

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Number of solutions to a linear equation

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Number of solutions to equations (practice) | Khan Academy

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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?

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Solving Inequalities

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Equations worksheet

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module 1

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Two step equations

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Conversions- part 1

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Conversions- part 2

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Absolute Value Inequalities

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Absolute Value Inequalities (goes along with kuta software worksheet).mov

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Absolute Value Inequalities