Algebra
posted by G .
Solving for systems of equation using the elimiation method.
3x2y=x6
3(x+2y)=3

3x  2y = x  6
3(x + 2y) = 3
Lets rearrange the second equation
3(x + 2y) = 3
x + 2y = 1
Now you have:
3x  2y = x  6
x + 2y = 1
You want to eliminate one variable and the easiest one to do so is y. So add the two equations together like so..
3x  2y + x + 2y = x  6 + 1
Simplify..
4x = x  5
3x = 5
x = 5/3
Plug back into one of the original equation and solve for y..
3(5/3)  2y = 5/3  6
5  2y = 5/3  6
2y = 5/3 + 1
y = 4/3
So x is 5/3 and y is 4/3. 
thank you for your response.
I should have included the answers the book gives.
It gives X=7 Y=4 
That's not possible. x and y are the same for both those equations. Although x = 7 and y = 4 does satisfy the second equation it does not for the first one.
3x  2y = x  6
3(7)  2(4) = 7  6
21  8 = 7  6
29 = 13 ??? This is not possible so those two values cannot be x and y. Are you sure you typed the equations right? A way for the x = 7 and y = 4 to be true is if the first equation is 3x + 2y = x  6. So it may be a typo by the book *shrug*.
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