How do you solve the system #5x+5y= -10# and #-4x+2y= -10#?

1 Answer
Mar 21, 2018

See a solution process below:

Explanation:

Step 1) Solve the first equation for #x#:

#5x + 5y = -10#

#(5x + 5y)/color(red)(5) = -10/color(red)(5)#

#(5x)/color(red)(5) + (5y)/color(red)(5) = -2#

#x + y = -2#

#x + y - color(red)(y) = -2 - color(red)(y)#

#x + 0 = -2 - y#

#x = -2 - y#

Step 2) Substitute #(-2 - y)# for #x# in the second equation and solve for #y#:

#-4x + 2y = -10# becomes:

#-4(-2 - y) + 2y = -10#

#(-4 xx -2) + (-4 xx -y) + 2y = -10#

#8 + 4y + 2y = -10#

#8 + (4 + 2)y = -10#

#8 + 6y = -10#

#8 - color(red)(8) + 6y = -10 - color(red)(8)#

#0 + 6y = -18#

#6y = -18#

#(6y)/color(red)(6) = -18/color(red)(6)#

#(color(red)(cancel(color(black)(6)))y)/cancel(color(red)(6)) = -3#

#y = -3#

Step 3) Substitute #-3# for #y# in the solution to the first equation at the end of Step 1 and calculate #x#:

#x = -2 - y# becomes:

#x = -2 - (-3)#

#x = -2 + 3#

#x = 1#

The Solution Is:

#x = 1# and #y = -3#

Or

#(1, -3)#