How do you solve the system of equations #-5x - 2y = 9# and #10x + y = - 3#?

1 Answer
May 27, 2017

See a solution process below:

Explanation:

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

#10x + y = -3#

#-color(red)(10x) + 10x + y = -color(red)(10x) - 3#

#0 + y = -10x - 3#

#y = -10x - 3#

Step 2) Substitute #(-10x - 3)# for #y# in the first equation and solve for #x#:

#-5x - 2y = 9# becomes:

#-5x - 2(-10x - 3) = 9#

#-5x + (-2 xx -10x) + (-2 xx -3) = 9#

#-5x + 20x + 6 = 9#

#(-5 + 20)x + 6 = 9#

#15x + 6 = 9#

#15x + 6 - color(red)(6) = 9 - color(red)(6)#

#15x + 0 = 3#

#15x = 3#

#(15x)/color(red)(15) = 3/color(red)(15)#

#(color(red)(cancel(color(black)(15)))x)/cancel(color(red)(15)) = 3/color(red)(3 xx 5)#

#x = color(red)(cancel(color(black)(3)))/color(red)(color(black)(cancel(color(red)(3))) xx 5)#

#x = 1/5#

Step 3) Substitute #1/5# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -10x - 3# becomes:

#y = (-10 xx 1/5) - 3#

#y = -2 - 3#

#y = -5#

The solution is: #x = 1/5# and #y = -5# or #(1/5, -5)#