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# Systems of equations with substitution: potato chips

To solve a system of equations using substitution...

1. Isolate one of the variables in one of the equations, e.g. rewrite 2x+y=3 as y=3-2x.
2. You can now express the isolated variable using the other one. *Substitute* that expression into the second equation, e.g. rewrite x+2y=5 as x+2(3-2x)=5.
3. Now you have an equation with one variable! Solve it, and use what you got to find the other variable.

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Created by Sal Khan.

## Want to join the conversation?

• What does he mean by the last video?
• Does Sal draw these pictures? if he does, I'm very impressed.
• I believe he does
• So, which method is faster/easier/recommended: elimination or substitution?
• It really depends on the situation. Sometimes it's more convenient to use one over the other.
• I don't know why my teacher put me into this class because i really don't know any of this. Every time i look at this question i get over whelmed because i dont get the problems he is doing.
😵😠🤯
• Let's use this example:
{7x+10y=36
{2x-y=-9
First, label the equations. 7x+10y=36 would be Equation 1, and 2x-y=-9 would be Equation 2. Next, multiply Equation 2 by an integer so that either the x values (2x and 7x) or the y values (-y and 10y) are equal. You could multiply Equation 2 by -10, for example, so the y values would be equal, since -y • -10 = 10y. Label the new equation, -20x+10y=90, Equation 3. Next, subtract Equation 3 from Equation 1. You should get 27x = -54. Divide both sides of the equation by 27, and you should get x=-2, so you've solved for x. Now, substitute x=-2 into Equation 2. You should get (-4)-y=(-9) (it is okay if you don't have the parentheses, I just put them there to emphasize the fact that -4 and -9 are negative). Add 4 to both sides of the equation to get - y=-5. Divide both sides of the equation by -1 to get y=5. You are done! (x,y)=(-2,5)!

My explanation is a little complicated, you might have to reread this. Hope it helps!
• the value for m = -4w+ 11 @ essentially the equation for the line which you could then use to find the answer to any combination/ variable to the problem
• This does not make any sense to me at all, could you explain this in easier terms?
• I can try, though I'm not sure if this is easier or not. A system of equations is where you have more than one equation with the same variables and you need to find out what values of the variables will work for all the equations.
Here is an example: 2x+3y=12; 5x+y=17
Substitution is one way to solve it.
First, we can rearrange one of the equations in order to isolate one of the variables:
5x+y=17
y=17-5x
We now have a way to express y in terms of x, so we can put it into the other equation instead of y in order to solve for x:
2x+3y=12
2x+3(17-5x)=12
2x+51-15x=12
51-13x=12
-13x=-39
x=3
Now we know what value x needs to be to satisfy both equations, so we can use that in place of x to solve for y.
5x+y=17
5(3)+y=17
15+y=17
y=2
We now know the values for both x and y, and in order to be quite sure, we can check them by putting them both into the first equation again:
2(3)+3(2)?=12
6+6?=12
12=12
This system of equations could also be solved in several other ways, such as elimination or graphing. In graphing, the point at which all equations in the system intersect is the solution.
• Why does and how does this substitution work?
• because "m" is always equal to the same thing. "M' will always equal -4w + 11, no matter which equation it's in
• Im still confused on everything..
• Would you like my help? Reply if you would like my help!
• How do we solve y intercept?
• I believe the elimination method is the easiest, but when would the substitution method be more useful?
• When one has messy equations that cannot be ''scaled'' easily in such a way that allows you to make an elimination. For instance is you have these 2 equations:

ln(3x + 2y) + 3*log(x/3) = 5

(x - 2)²/9 + (y + 7 )²/16 = 17x

It seems difficult to try making one of the sides of one of them equal to one side of the other equation. Painfully, substitution is the best choice to combine them.

Another part where substitution is useful is when you start to deal with composition of functions (videos on the Algebra II playlist talk about it), which is an important aspect when you start to learn calculus.