Question 28584
I'm trying to solve for addition for the following problem 
2x-4y=7
4x+2y=9 
I began to solve multiplying by 2 for the 2nd equation to get rid of the variable y 
2x-4y=7
8x+4y=18
10x=25
25/10 = 2.5 or x = 2.5 
However, something is wrong, I can't seem to solve for y which makes me think I started out solving this problem of addition the wrong way. 
2(2.5)-4y=7
-4y=7
y=3 
2(2.5)-4(3)=
5-12=-7 

You may edit the question. Maybe convert formulae to the same formula notation {{{x^2-1}}} as in your solutions.
I'm trying to solve for addition for the following problem

2x-4y=7
4x+2y=9

I began to solve multiplying by 2 for the 2nd equation to get rid of the variable y

2x-4y=7
8x+4y=18
10x=25
25/10 = 2.5 or x = 2.5

However, something is wrong, I can't seem to solve for y which makes me think I started out solving this problem of addition the wrong way.

2(2.5)-4y=7
TILL HERE YOU ARE RIGHT!
YOU ARE ALMOST THERE.DON'T WORRY!
(Are you hungry? you have swallowed 2X(2.5). That is the culprit)
Let me delete the wrong portions and continue
5-4y =7
5-7 = 4y
-2=4y
y= -2/4 = -1/2
Therefore  x=5/2 and y= (-1/2)
Verification: consider 4x+2y =9 the other equation.
Putting x=5/2 and y= (-1/2),we have 
LHS= 4x+2y =4X(5/2)+2X(-1/2) = 10-1=9 which is the right LHS.
Therefore our values are correct