Question

What is the value of the x variable in the solution to the following system of equations?

4x + 2y = 12
x – y = 3

Answer 1

Ok! First of all solve for x in the second equation :

\(x-y =3\)
\(x = 3 +y\)

Now put this value in the first equation .

Answer 2

So I solve for y. I plug into the other equation, and I simplify, solving for x. My equation could look like: x=a for some number a. If it looks like this, I have one solution for x, and it is simply a. It could look something that is always true, but not very interesting like 3=3 or 0=0 or -1=-1 These kinds of statements are always true, which means that it doesn't matter what x is. There will be a solution for any x. It could look like a statement that is never true like 1=0 or -2 = 2 or 10 = 20 This is always false, which means that it doesn't matter what x is, there will be no solutions.

Answer 3

In the first equation we have : \(4x+2y=12\)
But x = 3 +y

So \(4(3+y)+2y=12\) must satisfy the condition :

Solve for y and put the value of it in x = 3 +y . Can you do it now? @tiffpink

Answer 4

I dont get it still im confused

Answer 5

See, we have second equation as : x - y= 3

right?

Answer 6

I thought I had to multiply in the second equation so that I can eliminate

Answer 7

You can also do like that.

Answer 8

Multiply the second equation by 4 . What do you get as new second equation?

Answer 9

4x- 4y = 12

Answer 10

Great work... Now we have two equations :
\[4x+2y=12\]
\[4x-4y=12\]

Subtract both these equations :
\[(4x+2y)-(4x-4y) = 0\]
\[4x+2y-4x+4y= 0 \]

Can you solve this now for "y" ?

Answer 11

x = 2 , y = 2

Answer 12

See : \(\cancel{4x} + 2y -\cancel{4x} + 4y=0\)
\(6y = 0 \)
\(y = 0\)

And as x = y + 3

Thus x = ? , can you tell me now?

Answer 13

Sorry, but I am more confused. I just never did it that way.