How to solve y+2x=4 and 3y+2x=5y-2x? Different variables

I totally blanked out and forgot how to solve equations like these with different variables. How do you solve for x and y? Show steps please!

You need to get them into the same format. Normally, it would be ax + by = c

2x + y = 4
4x - 2y = 0

Next, multiply one of the equations by something such that if you added the two equations together, one of the variables would cancel. Let's multiply the first equation by 2 so the y's cancel.

4x + 2y = 8
4x - 2y = 0

Now, add them together. The y's cancel, and you're left with:

8x = 8

x = 1

Now, substitute 1 for x in the very first equation.

y + 2(1) = 4

y + 2 = 4

y = 2

If those answers are correct, you should be able to substitute 1 for x and 2 for y in the second equation.

3(2) + 2(1) = 5(2) - 2(1)

6 + 2 = 10 - 2

8 = 8

They are correct.


ANSWER

x = 1
y = 2

If you transpose y + 2x = 4 to y = 4 - 2x & then use this y value in the second equation you get
3(4 - 2x) + 2x = 5(4 - 2x) - 2x
Expanding this :~ 12 - 6x + 2x = 20 -10x - 2x
Rationalize to 12 - 4x = 20 - 12x
Add 12x to each side & subtract 12 from each side :~ 8x = 8 i.e. x = 1
From y + 2x = 4 if x = 1. :~ y + 2 = 4 Subtract 2 from each side y = 2

1) 3y + 2x = 5y - 2x simplified:
4x - 2y = 0
2) y + 2x = 4
multiply eq(2) by 2, add to eq(1):
4x - 2y = 0
4x + 2y = 8
___________
8x = 8
x = 8/8 = 1
using eq(2):
y + 2(1) = 4
y = 2
check using eq(1):
4(1) - 2(2) = 0
4 - 4 = 0

x = 1, y = 2

- .--

1. y + 2x = 4

2. 3y +2x = 5y - 2x

From 2. 4x = 2y

from 1. y = 4 - 2x

Substitute in 2. 4x = 8 - 4x
8x = 8 so x = 1

from 2. y = 2