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## Algebra 1

### Course: Algebra 1 > Unit 6

Lesson 3: Solving systems of equations with elimination- Systems of equations with elimination: King's cupcakes
- Elimination strategies
- Combining equations
- Elimination strategies
- Systems of equations with elimination: x-4y=-18 & -x+3y=11
- Systems of equations with elimination
- Systems of equations with elimination: potato chips
- Systems of equations with elimination (and manipulation)
- Systems of equations with elimination challenge
- Elimination method review (systems of linear equations)

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# Systems of equations with elimination: x-4y=-18 & -x+3y=11

Sal solves the following system of equations by eliminating x: x-4y=-18 and -x+3y=11. Created by Sal Khan.

## Want to join the conversation?

- is this really the easiest way or can there be different ways??(69 votes)
- There are different ways, but this is most likely the simplest. But, the simplest may not always be the best way for you. ;)(43 votes)

- Now, what happens if you can't cancel out one of your variables? For example, 2x - 3y = 3, and 4x + 2y = 2. Then what?(8 votes)
- If you can't cancel one of your variable, you make it so that they can be canceled.

In your example, you would have to multiple 2x - 3y = 3 by -2

So, -2 (2x - 3y = 3)

After you multiply that, you would get -4x + 6y = -6

Now you are able to eliminate with 4x + 2y = 2

Now eliminate "4x" and add those equations together and you get 8y = -4

So, y = -1/2

Now that you know y, you can just plug it into any of the equations and get x

So, 4x + 2(-1/2) = 2

4x - 1 = 2

4x = 3

x = 3/4

So your answer would be (3/4, -1/2).(16 votes)

- This whole thing is a bit KHAN-fusing (im not funny, for real, its hard.) can someone please help?(8 votes)
- Okay, this is a quick review with what the video was talking about in my point of view. I recommend you to write all this in your words on a notebook or sheet of paper.

Say you have two equations:

x-y=1

x+y=3

To do elimination the equations MUST be...

-aligned.

-in the same format.

-atop one another.

You also want ONE of the variables to be cancelled out/eliminated.

x-y=1 => We have a choice. We can either do addition

x+y=3 and have y eliminated, or do subtraction and

have x eliminated. Let's do addition.

x-y=1

+(x+y=3) => y is being cancelled out.

2x =4

x = 2 =>Nice! Now let's solve for y.

2-y=1 => Here, 2 is representing x. This time, we

-(2+y=3) to solve for y, so we cannot do addition

because addition will cancel y out. Let's

do subtraction.

-2y = -2

y = 1 => Huzzah! Joy is brought forth! The

solution is (2,1).

What Sal did was to multiply/divide each equation so they could be eliminated easily. Not all systems of equations can just go on and start eliminating. Sometimes, you need to multiply an equation so one of the values can eliminate the other.

Hope this helped. :)(8 votes)

- Can someone explain to me why this works? Sorry if it's a silly question.(5 votes)
- Elimination works because it eliminates one of the variables, we can solve the equation only if there is 1 variable and not 2 variables. This system of equation was super easy because you could eliminate the x variable right off the bat because x -x =0, so we cancel those two out and were left with just 1 variable in the equation, y, and then we can just solve it.

Basically, you manipulate the equations (multiplying both sides by a number, for instance) to cancel out one of the variables, leaving only 1 left, thus allowing you to solve for the variable left. Then you use that number and plug in back into one of the original equations to find the other variable you are looking for.

1 Eliminate one of the variables (doesn't matter which one)

2 Solve

3 Plug answer back into one of the original equations to find other variable(12 votes)

- Can anyone help me figure out how to do this equation 3x+8y=15

2x−8y=10?

i can't solve this without messing up.

(5 votes)- You're posting under the elimination method, so you'll want to find the least common multiple of either the coefficient of x or y. Then, you subtract the two equations to find the value of one variable. From there you substitute your now known variable to find the other unknown variable. I hope this helped!(7 votes)

- can you explain how to solve a system of equations with elimination? I need major help, I am pulling out my hair right now.(3 votes)
- According to Google, elimination is the complete removal or destruction of something. Therefore, you must multiply an equation by a certain number to eliminate a variable so that you're left with one variable. From there you can find out that one variable and substitute it back into the equation to find your other variable. I hope this helped!(7 votes)

- How many different ways to solve this?(4 votes)
- I think four: graphing, elimination, substitution, and matrices. All four are taught on Khan Academy, though matrices are a little higher level math than Algebra 1.(5 votes)

- I am working on a summer assignment and I have stumbled upon a problem that I do not understand, I am supposed to solve the systems of equations using elimination substitution and graphing. y1 = -3x+5 and y2 = 4x+12 the one and the two coefficient to the y's are tiny and I do not know their purpose nor do I know how to solve this problem please help explain it to me.(0 votes)
- The 1 and 2 subscripts really do not matter that much. This is asking you to find where the two equations intersect.
**Elimination**:

y= -3x+5

y=4x+12.

Let's eliminate y by subtracting equation 2 from equation 1.

=> 0= -7x-7

=> 7= -7x

=>x= -1.

Plug back in => y=8.

Answer: (-1,8).**Substitution**:

y= -3x+5.

y=4x+12.

Substitute y= -3x+5 into the second equation.

=> -3x+5=4x+12

=> 5=7x+12

=> 7x= -7

=> x= -1.

Plug back in => y=8.

Answer: (-1,8).

Graphing the equations will show an intersection at (-1,8). So, all three ways give the same answer. In the future, you can use the one that you like.(12 votes)

- I get how to do it, but I don't understand why this works.(5 votes)
- this seems very simple, but I am having trouble with this:

Solve the system of equations:

-10y+9x=-9

10y+5x=-5

How would I solve for the x and y pair?(3 votes)- Since you have -10 y in first and 10y in second, add equations together. -10y + 10y = 0, 9x + 5x=14x, and -9 = 5--14, so you get 14x = -14, divide by 14 to get x=-1. Substitute this into either equation to get y which should be 0.(5 votes)

## Video transcript

- [Instructor] So we have a system of two linear equations here. This first equation, X minus
four Y is equal to negative 18, and the second equation,
negative X plus three Y is equal to 11. Now what we're gonna do
is find an X and Y pair that satisfies both of these equations. That's what solving the
system actually means. As you might already have seen, there's a bunch of X and Y pairs that satisfy this first equation. In fact, if you were to graph them, they would form a line, and there's a bunch of other X and Y pairs that satisfy this other equation, the second equation, and if you were to graph them, it would form a line. And so if you find the X and
Y pair that satisfy both, that would be the
intersection of the lines, so let's do that. So actually, I'm just gonna rewrite the first equation over here, so I'm gonna write X minus four Y is equal to negative 18. So, we've already seen in algebra that as long as we do the same thing to both sides of the equation, we can maintain our equality. So what if we were to add, and our goal here is to
eliminate one of the variables so we have one equation with one unknown, so what if we were to add this negative X plus three Y
to the left hand side here? So negative X plus three Y, well, that looks pretty good because an X and a negative
X are going to cancel out, and we are going to be left with negative four Y, plus three Y. Well, that's just going to be negative Y. So by adding the left hand side of this bottom equation
to the left hand side of the top equation, we were able to cancel out the Xs. We had X, and we had a negative X. That was very nice for us. So what do we do on the right hand side? We've already said that we have to add the same thing to both
sides of an equation. We might be tempted to just say, well, if I
have to add the same thing to both sides, well, maybe I have to add a negative X plus three Y to that side. But that's not going to help us much. We're gonna have negative
18 minus X plus three Y. We would have introduced an X on the right hand side of the equation, but what if we could add something that's equivalent to
negative X plus three Y that does not introduce the X variable? Well, we know that the number 11 is equivalent to negative X plus three Y. How do we know that? Well, that second equation tells us that. So once again, all I'm doing
is I'm adding the same thing to both sides of that top equation. On the left, I'm expressing
it as negative X plus three Y, but the second equation tells us that negative X plus three Y is going to be equal to 11. It's introducing that second constraint, and so let's add 11 to
the right hand side, which is, once again, I
know I keep repeating it, it's the same thing as
negative X plus three Y. So negative 18 plus 11 is negative seven, and since we added the
same thing to both sides, the equality still holds, and we get negative Y is
equal to negative seven, or divide both sides by negative one or multiply both sides by negative one. So multiply both sides by negative one. We get Y is equal to seven, so we have the Y coordinate of the X, Y pair that
satisfies both of these. Now how do we find the X? Well, we can just substitute
this Y equals seven to either one of these. When Y equals seven, we
should get the same X regardless of which equation we use. So let's use the top equation. So we know that X minus four times, instead of writing Y, I'm gonna write four times seven 'cause we're gonna figure out what is X when Y is seven? That is going to be equal to negative 18, and so let's see, negative
four or four times seven. That is 28. So let's see, I could just solve for X. I could add 28 to both sides, so add 28 to both sides. On the left hand side, negative 28, positive
28, those cancel out. I'm just left with an X, and on the right hand side, I get negative 18 plus 28 is 10. So there you have it. I have the X, Y pair that satisfies both. X equals 10. Y equals seven. I could write it here. So I could write it as coordinates. I could write it as 10 comma seven, and notice, what I just did here, I encourage you. Substitute Y equals seven here, and you would also get X equals 10. Either way, you would
have come to X equals 10, and to visualize what is going on here, let's visualize it really fast. Let me draw some coordinate axes. Whoops, I meant to draw a
straighter line than that. Alright, there you go, so let's say that is our Y axis, and that is, whoops, that is our X axis, and then, let's see, the top equation is gonna
look something like this. It's gonna look something like this, and then that bottom equation is gonna look something like, lemme draw a little bit nicer than that. It's going to look something like this, something like that. Lemme draw that bottom one here so you see the point of intersection, and so the point of
intersection right over here, that is an X, Y pair that satisfies both of these equations, and that we just saw, it happens when X is equal to 10, and Y is equal to seven. Once again, this white line, that's all the X and Y pairs that satisfy the top equation. This orange line, that's
all the X and Y pairs that satisfy the orange equation, and where they intersect, that point is on both lines. It satisfies both equations, and once again, take X equals 10, Y equals seven. Substitute it back into
either one of these, and you will see that it holds.