Question

Hi. Could somebody explain the substitution method to me? Use this equation as an example.
3y - 2x = 11
y + 2x = 9

Answer 1

With the substitution method, you rearrange one of the equations to give you one variable in terms of the others. Then you take the other equation(s) and replace that variable with the expression for it in terms of the others. In this system of equations, the second equation can be easily rearranged to give y as a function of x — just subtract 2x from both sides. Now, take your new equation y = 9-2x, and wherever you see y in the first equation, substitute (9-2x). Solve the result for x. Then, take your value of x, and put it into y = 9-2x to get the value of y.

Finally, and this is perhaps the most important step, take your newly found values of x and y, plug them into all of the original equations and make sure that the result is true. If you make a mistake, you may end up with values that make one equation work, but not the other, so you need to test all of them.

Try it, I'll check your answer, or answer questions if you don't understand...

Answer 2

y = 9 - 2x
The first equation looks like this:
3 times 9 - 2x - 2x = 11
Did I go wrong at this point?

Answer 3

Okay, here's how you do it:
\[3y-2x=11\]\[y=9-2x\]\[3(9-2x) - 2x = 11\]Can you expand that?

Answer 4

Remember that by the distributive property of multiplication, \[a(b-c) = ab - ac\]

Answer 5

Okay. That's totally where I went wrong where I tried answering the question before.
3(9 - 2x) - 2x = 11
27 - 6x - 2x = 11
I am going to subtract 2x from 6x.
27 - 4x = 11
Anything I did wrong in this step?

Answer 6

No, \(-6x -2x = -8x\)

Answer 7

\(6x - 2x =4x\) but that's not what you have here...

Answer 8

It's maybe easier to think of it as
-6x
-2x
-----
-8x

(an addition problem)

Answer 9

Oh...right. Okay.
27 - 8x = 11
Now, I'll subtract 27 from both sides.
-8x = -16
I think I can determine the answer now.

Answer 10

Thank you. :)

Answer 11

As a bit of personal style, instead of subtracting the 27 from both sides, and ending up with -8x = -16, I would have added 8x to both sides, then subtracted 11, giving me a result with mostly (or all) positive signs, which for most people is a slightly safer road to follow...

Answer 12

Probably half the mistakes I see on OpenStudy involve getting a sign wrong somewhere...if there's an easy way to make that less likely, I'm going to take it!

Answer 13

So you found the value of x. What is the value of y?

Answer 14

Let's see:
If x is 2, I'll plug it into the second equation.
y = 9 - 2(2)
y = 9 - 4
y = 5

Answer 15

Okay, now take your values and plug them into the other equation: is the result true?

Answer 16

Here's why you need to do that. x=1, y=7 is a solution to y = 9-2x, right? But what happens if you plug it into 3y-2x=11?

Answer 17

I'll plug the results into the original equations.
3(5) - 2(2) = 11
15 - 4 = 11

5 + 2(2) = 9
5 + 4 = 9

Answer 18

It is true.

Answer 19

That won't be true, in response to your question.

Answer 20

Exactly! So, you need to check your solutions in all of the equations to be sure you have a valid solution for the entire system.

Answer 21

Thank you. :) Are you a teacher?

Answer 22

There's another common way to solve simple sets of equations like this called elimination. I don't know what the context is of your question about how to solve — if you're doing this in a math class, then you'll likely go on to do elimination. If you were told to do this in some other class, you might not. I'll explain elimination to you if you want...

No, I'm not a teacher, just a (mostly) patient explainer...

Answer 23

(I get impatient with people who just want answers, and don't want to learn how to find their own correct answers)

Answer 24

I learned about elimination, and frankly, I prefer it over this mess.

Answer 25

Elimination on this one would certainly be easier :-)

Answer 26

You're certainly good at this. Have you ever had somebody lash out at you because you were actually trying to teach them something and they just wanted an answer?

Answer 27

I'm firmly of the belief that most people can understand and do this stuff, if they don't sabotage themselves by insisting that they cannot ("I'm dumb at math"). It takes finding the right explanation (one size does NOT fit all), and a willingness to practice what you've learned. So much of it is just following a recipe with no deep insight required...just doing enough of them to recognize which recipe!

Answer 28

I don't like to be wrong, so coupled with that belief, I try pretty hard to give clear explanations :-)