How do you solve #5x - y = -9# and #y + 2x = 2#?

1 Answer
Jun 6, 2018

Answer:

The solution is #(-11/3,28/3)# or #~~(-3.67,9.33)#.

Explanation:

Solve the linear system:

#"Equation 1":# #5x-y=-9#

#"Equation 2":# #y+2x=2#

I am going to solve the system by elimination and substitution.

Rearrange Equation 2 to standard form: #Ax+By=C#

#2x-y=2#

Multiply Equation 2 by #-1#. This will reverse the signs, but the graph will remain the same.

#-1(2x-y)=2xx-1#

#-2x+y=-2#

Add Equation 1 and Equation 2.

#color(white)(..)5x-y=-9#
#-2x+y=-2#
#-------#
#color(white)(..)3xcolor(white)(.....)=-11#

Divide both sides by #3#.

#x=-11/3# or #~~-3.67#

Substitute #-11/3# for #x# in Equation 2 and solve for #y#.

#2(-11/3)+y=2#

#-22/3+y=2#

Multiply both sides by #3#.

#-22+3y=6#

Add #22# to both sides.

#3y=6+22#

Simplify #6+22# to #28#.

#3y=28#

Divide both sides by #3#.

#y=28/3# or #~~9.33#

graph{(5x-y+9)(-2x+y+2)=0 [-12.795, 9.705, -15.865, -4.615]}