How do you solve the following system?: # -9x - 3y = 2 , 4x-y=-27 #

1 Answer
Jun 3, 2018

Answer:

The solution to this system of linear equations is #(-83/21,235/21)# or #~~(-3.952,11.19)#

Explanation:

#"Equation 1":# #-9x-3y=2#

#"Equation 2":# #4x-y=-27#

The solution to a system of linear equations is the point or points that they have in common. These systems can be solved by elimination, substitution, or graphing. I'm going to use elimination to determine one coordinate and substitution to find the second coordinate.

Multiply Equation 1 by #-1#. This will reverse the signs.

#-1(-9x-3y)=2xx-1#

#9x+3y=-2#

Multiply Equation 2 by #3#.

#3(4x-y)=27xx3#

#12x-3y=81#

Add Equation 1 and Equation 2.

#color(white)(.)9x+3y=color(white)(.)-2#
#12x-3y=-81#
#---------#
#21xcolor(white)(.......)=color(white)()-83#

Divide both sides by #21#.

#x=-83/21# or #~~-3.952#

Substitute #-83/21# for #x# in Equation 1 and solve for #y#..

#-9(-83/21)-3y=2#

#747/21-3y=2#

Subtract #747/21# from both sides.

#-3y=2-747/21#

Multiply #2# by #21/21# to get an equivalent fraction with #21# as its denominator. Since #21/21=1#, the numbers will change, but the value remains the same.

#-3y=2xx21/21-747/21#

#-3y=42/21-747/21#

#-3y=-705/21#

Divide both sides by #-3#.

#y=-705/21-:3#

Invert and multiply.

#y=-color(red)cancel(color(black)(705))^235/21xx1/color(red)cancel(color(black)(3))^1#

#y=235/21# or #~~11.19#

The solution to this system of linear equations is #(-83/21,235/21)# or #~~(-3.952,11.19)#

graph{(-9x-3y-2)(4x-y+27)=0 [-13.66, 11.65, 2.91, 15.57]}