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

1 Answer
Mar 7, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Because the first equation is already solved for #y# we can substitute #-2x + 6# for #y# in the second equation and solve for #x#:

#2y + x = -5# becomes:

#2(-2x + 6) + x = -5#

#(2 xx -2x) + (2 xx 6) + x = -5#

#-4x + 12 + x = -5#

#-4x + x + 12 = -5#

#-3x + 12 = -5#

#-3x + 12 - color(red)(12) = -5 - color(red)(12)#

#-3x + 0 = -17#

#-3x = -17#

#(-3x)/color(red)(-3) = (-17)/color(red)(-3)#

#(color(red)(cancel(color(black)(-3)))x)/cancel(color(red)(-3)) = 17/3#

#x = 17/3#

Step 2) Subsitute #17/3# for #x# in the first equation and calculate #y#:

#y = -2x + 6# becomes:

#y = -34/3 + (3/3 xx 6)#

#y = -34/3 + 18/3#

#y = -16/3#

The solution is: #x = 17/3# and #y = -16/3# or #(17/3, -16/3)#