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practice some exam Questions – Simultaneous equations

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practice some exam Questions – Simultaneous equations

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1)View SolutionHelpful TutorialsAsymptotes - horizontal and vertical typesIntersection of a straight line and a hyperbolaParts (a) and (b): Part (c): 2)View SolutionPart (i) and (ii): Part (iii): 3)View SolutionHelpful TutorialsSubstitution method for linear and non-linear equations4)View SolutionHelpful TutorialsSubstitution method for linear and non-linear equations

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Substitution method for linear and non-linear equations

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Substitution method for linear and non-linear equations

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In these tutorials you are shown how to solve a simultaneous equation by the substitution method. You are also shown how it relates to the intersection of two graphs and why there are two sets of solutions.

Example 1

Example in the video

Solve the simultaneous equations:

[latex]x + y = 3[/latex]

Example 2

Example in the video

Solve the simultaneous equations:

[latex]2x + y = 2[/latex]

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Substitution method for linear and non-linear equations

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In these tutorials you are shown how to solve a simultaneous equation by the substitution method. You are also shown how it relates to the intersection of two graphs and why there are two sets of solutions.

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Introduction

This is an introduction to solving simultaneous equations by the method of elimination.

x terms are the same

In the next example only the x terms are the same.

y terms the same but with opposite signs

In this example the two equations have the same y terms but with opposite signs.

Care must be taken over the signs as mistakes can easily be made.

y terms are the same and both signs are negative

In this example the y terms are exactly the same and both signs are negative.

Care must be taken over the signs as mistakes can easily be made.

y terms are the same and both signs are positive

In this example the y terms are exactly the same and both signs are positive.

Neither the x or y terms are the same

In this example neither the x or y terms are the same so an adjustment needs to be made.

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Tutorials on this page: Introduction to solving simultaneous equations by the method of elimination x terms are the same y terms the same but with opposite signs y terms are the same and both signs are negative y terms are the same and both signs are positive Neither

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Solving linear systems with 3 variables: no solution

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Solving linear systems with 3 variables: no solution

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Solving linear systems with 3 variables: no solution

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Sal solves a system with three variables that turns out to have no solution.

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Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, "Solve the system of linear equations," they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, or we could just pick some points here. You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get x equals 7/5, which is the same thing as 1 and 2/5. So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations by graphing, they normally give you a little bit cleaner numbers, but we'll try our best and see if we can see where these two lines intersect. So now, let's worry about the second one right over there. So we have 3x minus 2y is equal to 8. So I'll do the same thing. So 3x minus 2y is equal to 8. We'll just look at the x- and the y-intercepts. So first, the y-intercept. When x is equal to 0, this whole thing boils down to 3 times 0 minus 2y is equal to 8. That's just 0. So you have negative 2y is equal to 8. Divide both sides by negative 2, we get y is equal to negative 4. So the y-intercept is 0, negative 4. Right over here, and we mark it 0, negative 4. And then, let's set y equal to 0. So when y is equal to 0, this term right over here just becomes 0. So we get 3x minus 2 times 0, so that's just 0. So 3x is equal to 8. Divide both sides by 3, you get x is equal to 8/3. And 8/3 is the same thing as 2 and 2/3. So it puts it right about there. That's the point, 0, 8/3. Now let me try my best to graph it. Connect these two dots. So let me do my very best. I drew a dotted line there. It goes something like that. And just eyeballing it, it looks like these two lines intersect right over there. I'm hoping that this will give us a clean answer. And this is the point 2, negative 1. So that is the point, 2, negative 1. Right? The x value here is 2. y value is negative 1. Now, that's what we got just by eyeballing it, and clearly these are hand-drawn graphs. Not very precise. Let's verify or let's see if 2, negative 1 does satisfy both of these equations, if it's an x and y value that satisfies both and lies on both the graphs. So if you put 2, negative 1 in this first equation, you get 5 times 2, plus 3 times negative 1, and we're going to see if that is equal to 7. So this is 10 plus negative 3. So that's the same thing as 10 minus 3. Does that equal 7? And yeah, it does. 10 minus 3 is equal to 7. So 2, negative 1 is definitely on that graph or definitely satisfies that equation. And then, let's do it with the other one. So if we do 2, negative 1, you have 3 times 2, minus 2 times negative 1. And we're testing to see if that is equal to 8. So 3 times 2. 3 times 2 is 6. And then 2 times negative 1 is negative 2. But then, we're subtracting that, so it's 6 plus 2. 6 plus 2 is equal to 8. And it definitely does equal 8. So we have the coordinate or we have the point, 2, negative 1 satisfying both equations. So we've solved the system of linear equations by graphing.

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Systems of equations with graphing: 5x+3y=7 & 3x-2y=8 (video) | Khan Academy

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practice solving equations with substitution

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practice solving equations with substitution

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Solving systems of equations with substitution

Solving systems of equations with substitution

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Systems of equations with substitution (practice) | Khan Academy

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Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4. So what we want to do, when they say substitution, what we want to do is substitute one of the variables with an expression so that we have an equation and only one variable. And then we can solve for it. Let me show you what I'm talking about. So let me rewrite this first equation. 2y is equal to x plus 7. And we have the second equation over here, that x is equal to y minus 4. So if we're looking for an x and a y that satisfies both constraints, well we could say, well look, at the x and y have to satisfy both constraints, both of these constraints have to be true. So x must be equal to y minus 4. So anywhere in this top equation where we see an x, anywhere we see an x, we say well look, that x by the second constraint has to be equal to y minus 4. So everywhere we see an x, we can substitute it with a y minus 4. So let's do that. So if we substitute y minus 4 for x in this top equation, the top equation becomes 2y is equal to instead of an x, the second constraint tells us that x needs to be equal to y minus 4. So instead of an x, we'll write a y minus 4, and then we have a plus 7. All I did here is I substituted y minus 4 for x. The second constraint tells us that we need to do it. y minus 4 needs to be equal to x or x needs to be equal to y minus 4. The value here is now we have an equation, one equation with one variable. We can just solve for y. So we get 2y is equal to y, and then we have minus 4 plus 7. So y plus 3. We can subtract y from both sides of this equation. The left hand side, 2y minus y is just y. y is equal to-- these cancel out. y is equal to 3. And then we could go back and substitute into either of these equations to solve for x. This is easier right over here, so let's substitute right over here. x needs to be equal to y minus 4. So we could say that x is equal to 3 minus 4 which is equal to negative 1. So the solution to this system is x is equal to negative 1 and y is equal to 3. And you can verify that it works in this top equation right over here. 2 times 3 is 6 which is indeed equal to negative 1 plus 7. Now I want to show you that over here we substituted-- we had an expression that, or we had an equation, that explicitly solved for x. So we were able to substitute the x's. What I want to show you is we could have done it the other way around. We could have solved for y and then substituted for the y's. So let's do that. And we could have substituted from one constraint into the other constraint or vice versa. Either way, we would have gotten the same exact answer. So instead of saying x is equal to y minus 4, in that second equation, if we add 4 to both sides of this equation, we get x plus 4 is equal to y. This and this is the exact same constraint. I just added 4 to both sides of this to get this constraint over here. And now since we've solved this equation explicitly for y, we can use the first constraint, the first equation. And everywhere where we see a y, we can substitute it with x plus 4. So it's 2 times-- instead of 2 times y, we can write 2 times x plus 4. 2 times x plus 4 is equal to x plus 7. We can distribute this 2. So we get 2x plus 8 is equal to x plus 7. We can subtract x from both sides of this equation. And then we can subtract 8 from both sides of this equation, subtract 8. The left hand side, that cancels out. We're just left with an x. On the right hand side, that cancels out, and we are left with a negative 1. And then we can substitute back over here we have y is equal to x plus 4, or so y is equal to negative 1 plus 4 which is equal to 3. So once again, we got the same answer even though this time we substituted for y instead of substituting for x. Hopefully you found that interesting.

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Systems of equations with substitution: 2y=x+7 & x=y-4 (video) | Khan Academy

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Use substitution to solve for x and y. And we have a system of equations here. The first equation is 2y is equal to x plus 7. And the second equation here is x is equal to y minus 4.

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Solving systems of equations by elimination 

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Solving systems of equations by elimination 

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Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations. If you think of it graphically, this would be the intersection of the lines that represent the solution sets to both of these equations. So how can we proceed? We saw in substitution, we like to eliminate one of the variables. We did it through substitution last time. But is there anything we can add or subtract-- let's focus on this yellow, on this top equation right here-- is there anything that we can add or subtract to both sides of this equation? Remember, any time you deal with an equation you have to add or subtract the same thing to both sides. But is there anything that we could add or subtract to both sides of this equation that might eliminate one of the variables? And then we would have one equation in one variable, and we can solve for it. And it's probably not obvious, even though it's sitting right in front of your face. Well, what if we just added this equation to that equation? What I mean by that is, what if we were to add 5x minus 4y to the left-hand side, and add 25.5 to the right-hand side? So if I were to literally add this to the left-hand side, and add that to the right-hand side. And you're probably saying, Sal, hold on, how can you just add two equations like that? And remember, when you're doing any equation, if I have any equation of the form-- well, really, any equation-- Ax plus By is equal to C, if I want to do something to this equation, I just have to add the same thing to both sides of the equation. So I could, for example, I could add D to both sides of the equation. Because D is equal to D, so I won't be changing the equation. You would get Ax plus By, plus D is equal to C plus D. And we've seen that multiple, multiple times. Anything you do to one side of the equation, you have to do to the other side. But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation. On the right-hand side, you're adding 25.5 to the equation. Aren't you adding two different things to both sides of the equation? And my answer would be no. We know that 5x minus 4y is 25.5. This quantity and this quantity are the same. They're both 25.5. This second equation is telling me that explicitly. So I can add this to the left-hand side. I'm essentially adding 25.5 to it. And I could add 25.5 to the right-hand side. So let's do that. If we were to add the left-hand side, 3x plus 5x is 8x. And then what is 4y minus 4y? And this was the whole point. When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. If you just add these two together, they are going to cancel out. They're going to be plus 0y. Or that whole term is just going to go away. And that's going to be equal to 2.5 plus 25.5 is 28. So you divide both sides. So you get 8x is equal to 28. And you divide both sides by 8, and we get x is equal to 28 over 8, or you divide the numerator and the denominator by 4. That's equal to 7 over 2. That's our x value. Now we want to solve for our y value. And we could substitute this back into either of these two equations. Let's use the top one. You could do it with the bottom one as well. So we know that 3 times x, 3 times 7 over 2-- I'm just substituting the x value we figured out into this top equation-- 3 times 7 over 2, plus 4y is equal to 2.5. Let me just write that as 5/2. We're going to stay in the fraction world. So this is going to be 21 over 2 plus 4y is equal to 5/2. Subtract 21 over 2 from both sides. So minus 21 over 2, minus 21 over 2. The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. That's negative 16 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Or we could write that-- let's continue up here-- 4y-- I'm just continuing this train of thought up here-- 4y is equal to negative 8. Divide both sides by 4, and you get y is equal to negative 2. So the solution to this equation is x is equal to 7/2, y is equal to negative 2. This would be the coordinate of their intersection. And you could try it out on both of these equations right here. So let's verify that it also satisfies this bottom equation. 5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus negative 8. That's equivalent to-- let's see, this is 17.5 plus 8. And that indeed does equal 25.5. So this satisfies both equations. Now let's see if we can use our newly found skills to tackle a word problem, our newly found skills in elimination. So here it says, Nadia and Peter visit the candy store. Nadia buys 3 candy bars and 4 Fruit Roll-Ups for $2.84. Peter also buys 3 candy bars, but can only afford 1 additional Fruit Roll-Up. His purchase costs $1.79. What is the cost of each candy bar and each Fruit Roll-Up? So let's define some variables. Let's just use x and y. Let's let x equal cost of candy bar-- I was going to do a c and a f for Fruit Roll-Up, but I'll just stick with x and y-- cost of candy bar. And let y equal the cost of a Fruit Roll-Up. All right. So what does this first statement tell us? Nadia buys 3 candy bars, so the cost of 3 candy bars is going to be 3x. And 4 Fruit Roll-Ups. Plus 4 times y, the cost of a Fruit Roll-Up. This is how much Nadia spends. 3 candy bars, 4 Fruit Roll-Ups. And it's going to cost $2.84. That's what this first statement tells us. It translates into that equation. The second statement. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. So plus 1 additional Fruit Roll-Up. His purchase cost is equal to $1.79. What is the cost of each candy bar and each Fruit Roll-Up? And we're going to solve this using elimination. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. So how can we do this? Remember, with elimination, you're going to add-- let's focus on this top equation right here. Is there something we could add to both sides of this equation that'll help us eliminate one of the variables? Or let me put it this way, is there something we could add or subtract to both sides of this equation that will help us eliminate one of the variables? Well, like in the problem we did a little bit earlier in the video, what if we were to subtract this equation, or what if we were to subtract 3x plus y from 3x plus 4y on the left-hand side, and subtract $1.79 from the right-hand side? And remember, by doing that, I would be subtracting the same thing from both sides of the equation. This is $1.79. How do I know? Because it says this is equal to $1.79. So if we did that we would be subtracting the same thing from both sides of the equation. So let's subtract 3x plus y from the left-hand side of the equation. And let me just do this over on the right. If I subtract 3x plus y, that is the same thing as negative 3x minus y, if you just distribute the negative sign. So let's subtract it. So you get negative 3x minus y-- maybe I should make it very clear this is not a plus sign; you could imagine I'm multiplying the second equation by negative 1-- is equal to negative $1.79. I'm just taking the second equation. You could imagine I'm multiplying it by negative 1, and now I'm going to add the left-hand side to the left-hand side of this equation, and the right-hand side to the right-hand side of that equation. And what do we get? When you add 3x plus 4y, minus 3x, minus y, the 3x's cancel out. 3x minus 3x is 0x. I won't even write it down. You get 4x minus-- sorry, 4y minus y. That is 3y. And that is going to be equal to $2.84 minus $1.79. What is that? That's $1.05. So 3y is equal to $1.05. Divide both sides by 3. y is equal to-- what's $1.05 divided by 3? So 3 goes into $1.05. It goes into 1 zero times. 0 times 3 is 0. 1 minus 0 is 1. Bring down a 0. 3 goes into 10 three times. 3 times 3 is 9. Subtract. 10 minus 9 is 1. Bring down the 5. 3 goes into 15 five times. 5 times 3 is 15. Subtract. We have no remainder. So y is equal to $0.35. So the cost of a Fruit Roll-Up is $0.35. Now we can substitute back into either of these equations to figure out the cost of a candy bar. So let's use this bottom equation right here. Which was originally, if you remember before I multiplied it by negative 1, it was 3x plus y is equal to $1.79. So that means that 3x plus the cost of a Fruit Roll-Up, 0.35 is equal to $1.79. If we subtract 0.35 from both sides, what do we get? The left-hand side-- you're just left with the 3x; these cancel out-- is equal to-- let's see, this is $1.79 minus $0.35. That's $1.44. And 3 goes into $1.44, I think it goes-- well, 3 goes into $1.44, it goes into 1 zero times. 1 times 3 is 0. Bring down the 1. Subtract. Bring down the 4. 3 goes into 14 four times. 4 times 3 is 12. I'm making this messy. 14 minus 12 is 2. Bring down the 4. 3 goes into 24 eight times. 8 times 3 is 24. No remainder. So x is equal to 0.48. So there you have it. We figured out, using elimination, that the cost of a candy bar is equal to $0.48, and that the cost of a Fruit Roll-Up is equal to $0.35.

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Solving systems of equations by elimination (video) | Khan Academy

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Let's explore a few more methods for solving systems of equations. Let's say I have the equation, 3x plus 4y is equal to 2.5. And I have another equation, 5x minus 4y is equal to 25.5. And we want to find an x and y value that satisfies both of these equations.