How do you solve the system #2x-3y=-24# and #x+6y=18# using substitution?

2 Answers
Jul 15, 2017

Answer:

See a solution process below:

Explanation:

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

#x + 6y = 18#

#x + 6y - color(red)(6y) = 18 - color(red)(6y)#

#x + 0 = 18 - 6y#

#x = 18 - 6y#

Step 2) Substitute #(18 - 6y)# for #x# in the first equation and solve for #y#:

#2x - 3y = -24# becomes:

#2(18 - 6y) - 3y = -24#

#(2 xx 18) - (2 xx 6y) - 3y = -24#

#36 - 12y - 3y = -24#

#36 + (-12 - 3)y = -24#

#36 + (-15y) = -24#

#36 - 15y = -24#

#-color(red)(36) + 36 - 15y = -color(red)(36) - 24#

#0 - 15y = -60#

#-15y = -60#

#(-15y)/color(red)(-15) = (-60)/color(red)(-15)#

#(color(red)(cancel(color(black)(-15)))y)/cancel(color(red)(-15)) = 4#

#y = 4#

Step 3) Substitute #4# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = 18 - 6y# becomes:

#x = 18 - (6 * 4)#

#x = 18 - 24#

#x = -6#

The solution is: #x = -6# and #y = 4# or #(-6, 4)#

Jul 15, 2017

Answer:

#color(blue)(y=4,x=-6#

Explanation:

#:.2x-3y=-24----(1)#

#x+6y=18----(2)#
:.
in (2) # x=18-6y#

substitute #color(blue)(x=18-6y# in (1)

#:.2(color(blue)(18-6y))-3y=-24#

#:.36-12y-3y=-24#

#:.-15y=-24-36#

multiply both sides by#-1#

#:.15y=24+36#

#:.15y=60#

#:.color(blue)(y=4#

substitute #color(blue)(y=4 # in (2)

#:.x+6(color(blue)4)=18#

#:.x+24=18#

#:.x=18-24#

#:.color(blue)(x=-6#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(check:#

substitute #color(blue)(y=4# and#color(blue)(x=-6# in (2)

#:.(color(blue)(-6))+6(color(blue)(4))=18))#

#:.-6+24=18#

#:.color(blue)(18=18#