How do you solve the system of equations #12x + y = 38# and #4x + 5y = 22#?

1 Answer
Feb 6, 2017

Se the entire solution process below:

Explanation:

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

#12x + y = 38#

#12x + y - color(12x) = 38 - color(12x)#

#12x - color(12x) + y = 38 - color(12x)#

#0 + y = 38 - 12x#

#y = 38 - 12x#

Step 2) Substitute #38 - 12x# for #y# in the second equation and solve for #x#:

#4x + 5y = 22#

#4x + 5(38 - 12x) = 22#

#4x + 190 - 60x = 22#

#190 - 60x + 4x = 22#

#190 - 56x = 22#

#190 - color(red)(190) - 56x = 22 - color(red)(190)#

#0 - 56x = -168#

#-56x = -168#

#(-56x)/color(red)(-56) = -168/color(red)(-56)#

#(color(red)(cancel(color(black)(-56)))x)/cancel(color(red)(-56)) = 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 = 38 - 12x#

#y = 38 - (12 xx 3)#

#y = 38 - 36#

#y = 2#

The solution is #x = 3# and #y = 2#