How do solve the following linear system?: # 4x-2y=2 , -2x+5y=-3 #?

1 Answer
Apr 24, 2017

See the entire solution process below:

Explanation:

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

#4x - 2y = 2#

#4x - color(red)(4x) - 2y = 2 - color(red)(4x)#

#0 - 2y = 2 - 4x#

#-2y = 2 - 4x#

#(-2y)/color(red)(-2) = (2 - 4x)/color(red)(-2)#

#(color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = 2/color(red)(-2) - (4x)/color(red)(-2)#

#y = -1 + 2x#

Step 2) Substitute #-1 + 2x# for #y# in the second equation and solve for #x#:

#-2x + 5y = -3# becomes:

#-2x + 5(-1 + 2x) = -3#

#-2x + (5 * -1) + (5 * 2x) = -3#

#-2x - 5 + 10x = -3#

#10x - 2x - 5 = -3#

#(10 - 2)x - 5 = -3#

#8x - 5 = -3#

#8x - 5 + color(red)(5) = -3 + color(red)(5)#

#8x - 0 = 2#

#8x = 2#

#(8x)/color(red)(8) = 2/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = 1/4#

#x = 1/4#

Step 3) Subsitute #1/4# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -1 + 2x# becomes:

#y = -1 + (2 * 1/4)#

#y = -1 + 1/2#

#y = -1/2#

The solution is: #x = 1/4# and #y = -1/2# or #(1/4, -1/2)#