How do you solve the system of equations #6x + 3y = 0# and #- 5x + 2y = 27#?

1 Answer
Aug 4, 2017

See a solution process below:

Explanation:

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

#6x + 3y = 0#

#-color(red)(6x) + 6x + 3y = -color(red)(6x) + 0#

#0 + 3y = -6x#

#3y = -6x#

#(3y)/color(red)(3) = (-6x)/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = -2x#

#y = -2x#

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

#-5x + 2y = 27# becomes:

#-5x + 2(-2x) = 27#

#-5x + (-4x) = 27#

#-5x - 4x = 27#

#(-5 - 4)x = 27#

#-9x = 27#

#(-9x)/color(red)(-9) = 27/color(red)(-9)#

#(color(red)(cancel(color(black)(-9)))x)/cancel(color(red)(-9)) = -3#

#x = -3#

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

#y = -2x# becomes:

#y = (-2 xx -3)#

#y = 6#

The Solution Is: #x = -3# and #y = 6# and #(-3, 6)#