Question

I need to solve the following system of inequalities by elimination:
5x + 4y = 12
3x - 3y = 18
I know I need to 'add down' to complete this problem with elimination, but I'm not sure where to start and how to continue through. Do I start by multiplying by 3 so that I can cancel out the 15x when adding down? If someone could walk me through, I'd really appreciate it

Answer 1

https://www.khanacademy.org/math/algebra/systems-of-eq-and-ineq/fast-systems-of-equations/v/solving-systems-of-equations-by-elimination

Answer 2

really hope it helps:)

Answer 3

The problem I'm having is that I can't immediately cancel one of the variables out, though. neither equation has a variable the same as the other, so I'm not sure if how I'm carrying it out is how I should do it. In his example, the variables are the same and can immediately cancel out.

Answer 4

Okay, this isn't as hard as you think :-)

Pick a variable, any variable...

Answer 5

okay, how about x

Answer 6

Well, let's say you picked \(x\): multiply the first equation by the coefficient of \(x\) in the second equation. Multiply the second equation by the coefficient of \(x\) in the first equation. What do you get?

Answer 7

it comes out to
15x + 4y = 12
15x - 3y = 18

Answer 8

\[5x + 4y = 12\]\[
3x - 3y = 18
\]

Multiply the first one by \(3\) and the second by \(5\) and we get:

\[15x + 12y = 36\]\[15x -15y = 90\]Right?

Answer 9

You have to multiply each term to preserve the equations.

\[5x+4y=12\rightarrow3*5x + 3*4y = 3*12\rightarrow15x+12y=36\]

Answer 10

right, so, then after that, 15x cancels out and i have
+ 12y = 36
- 15y = 90
how do i solve down from here? does it become
-3y = 126

Answer 11

now you can combine the two equations by a number of equivalent means.

if the coefficients are equal but opposite in sign, you can just add down the columns:

\[2x + 3y = 5\]\[2x - 3y = 2\]-------------\[4x+0y=7\]for example

Answer 12

so does it become 0x - 3y = 126?

Answer 13

but we have the coefficients equal and of the same sign. so we can either subtract the second equation from the first (instead of adding), which I find slightly more likely to produce errors, or we can multiply one of the equations by -1 before adding (safer, in my experience), or there's another way which I'll describe later if you want.

\[2x + 3y = 5\]\[2x+4y = 3\]subtracting we have \[2x-2x + 3y-4y = 5-3\]\[0x-y = -2\]
obviously, if you multiply one equation by \(-1\) before adding, you get the same:
\[2x+(-1)2x + 3y + (-1)4y = 5+(-1)3\]\[2x -2x + 3y -4y = 5 -3 \]\[0x-y=-2\]

The third way, which you're hinting at, is to rearrange the two equations that we have:
\[2x+3y=5\]\[2x-4y=3\]becomes\[2x=5-3y\]\[2x=3-4y\]now we notice that we have the same thing on the left side of each of those equations, so we can simply set the right sides equal to each other:\[5-3y=2x\]\[3-4y=2x\]so\[5-3y=2x=3-4y\]or\[5-3y=3-4y\]and you can easily solve that for \(y\).

(note that I'm just making up equations now, so as not to do your problem for y

Answer 14

so we had \[15x+12y=36\]\[
15x−15y=90\]

if we combine them by multiplying the first one by \(-1\) and adding, we get:
\[(-1)*15x+(-1)*12y=(-1)*36\]\[15x-15y=90\]------------------\[-15x+15x-12y-15y=-36+90\]\[0x-27y=54\]\[-27y=54\]\[y=\]

Answer 15

I don't know why this buggy system put that crap in the middle of my equation :-(

Answer 16

I trust you can find the value of \(y\) from that equation! Now to find the value of \(x\), put the known value of \(y\) into either of the original equations and solve for the value of \(x\).

Answer 17

so the value of y is -2, and now I just plug that in as the y value in the equations to find x! thank you so much!

Answer 18

exactly!

to your solution (which you always should do), plug the values you find into ALL of the equations in the system and make sure they give you correct statements. It is possible to get a "solution" which will work for some but not all of the equations. There's a technical term for such solutions: wrong :-)

Answer 19

it all checks out! (4, -2) works when plugged back into both :) thank you so much for all your help!