How do you solve the following system: #5x + 8y = -2, 6x+2y=-4#?

1 Answer
Aug 31, 2017

Answer:

See a solution process below:

Explanation:

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

#6x + 2y = -4#

#-color(red)(6x) + 6x + 2y = -color(red)(6x) - 4#

#0 + 2y = -6x - 4#

#2y = -6x - 4#

#(2y)/color(red)(2) = (-6x - 4)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = (-6x)/color(red)(2) - 4/color(red)(2)#

#y = -3x - 2#

Step 2) Substitute #(-3x - 2)# for #y# in the first equation and solve for #x#:

#5x + 8y = -2# becomes:

#5x + 8(-3x - 2) = -2#

#5x + (8 * -3x) - (8 * 2) = -2#

#5x - 24x - 16 = -2#

#(5 - 24)x - 16 = -2#

#-19x - 16 = -2#

#-19x - 16 + color(red)(16) = -2 + color(red)(16)#

#-19x - 0 = 14#

#-19x = 14#

#(-19x)/color(red)(-19) = 14/color(red)(-19)#

#(color(red)(cancel(color(black)(-19)))x)/cancel(color(red)(-19)) = -14/19#

#x = -14/19#

Step 3) Substitute #-14/19# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -3x - 2# becomes:

#y = (-3 xx -14/19) - 2#

#y = 42/19 - 2#

#y = 42/19 - (19/19 * 2)#

#y = 42/19 - 38/19#

#y = 4/19#

The Solution Is: #x = -14/19# and #y = 4/19# or #(-14/19, 4/19)#