SOLUTION: Use any method to solve Equation 1 and Equation 2. 1) 8x + 64= 8y 2) -4x +4y = 32 Answer: ( , ) Solve for x and y

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: Use any method to solve Equation 1 and Equation 2. 1) 8x + 64= 8y 2) -4x +4y = 32 Answer: ( , ) Solve for x and y       Log On


   

Question 1191737: Use any method to solve Equation 1 and Equation 2.
1) 8x + 64= 8y
2) -4x +4y = 32
Answer: ( , )
Solve for x and y


Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
8x+%2B+64=+8y......... 1)
-4x+%2B4y+=+32....... 2)....both sides multiply by 2
_______________________________
8x+%2B+64=+8y......... 1), isolate x
8x=+8y-64.........simplify, both sides divide by 8
x=+y-8........1a)

-4x+%2B4y+=+32....... 2), substitute x
-4%28y-8%29%2B4y+=+32
-4y%2B32%2B4y+=+32
32=+32

=> solution to the system is x=+y-8





Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


To start with, put both equations in the form Ax+By=C and simplify.

8x + 64 = 8y
8x - 8y = -64
x - y = -8

-4x + 4y = 32
x - y = -8

The two equations are equivalent; any solution to one equation is a solution to the other. There is no unique (x,y) solution.

If you want a solution in (x,y) form, then solve the equation for y in terms of x and write the solution parametrically:

x - y = -8
y = x+8

Solution: (x,x+8) where x is any number