How do you solve the system of equations #-5x - 7y = - 12# and #7x + 6y = - 6#?

1 Answer
Feb 16, 2017

Answer:

See the entire solution process below:

Explanation:

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

#7x + 6y = -6#

#-color(red)(7x) + 7x + 6y = -color(red)(7x) - 6#

#0 + 6y = -color(red)(7x) - 6#

#6y = -color(red)(7x) - 6#

#(6y)/color(red)(6) = (-color(red)(7x) - 6)/color(red)(6)#

#(color(red)(cancel(color(black)(6)))y)/cancel(color(red)(6)) = (-7x)/6 - 6/color(red)(6)#

#y = -7/6x - 1#

Step 2) Substitute #-7/6x - 1# for #y# in the first equation and solve for #x#:

#-5x - 7y = -12# becomes:

#-5x - 7(-7/6x - 1) = -12#

#-5x + (7xx 7/6x) + (7xx 1) = -12#

#-5x + 49/6x + 7 = -12#

#(6/6 xx -5)x + 49/6x + 7 - color(red)(7) = -12 - color(red)(7)#

#-30/6x + 49/6x + 0 = -19#

#19/6x = -19#

#color(red)(6)/color(blue)(19) xx 19/6x = color(red)(6)/color(blue)(19) xx -19#

#cancel(color(red)(6))/cancel(color(blue)(19)) xx color(blue)(cancel(color(black)(19)))/color(red)(cancel(color(black)(6)))x = color(red)(6)/cancel(color(blue)(19)) xx -color(blue)(cancel(color(black)(19)))#

#x = -6#

Step 3) Substitute #-6# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -7/6x - 1# becomes:

#y = (-7/6 xx -6) - 1#

#y = (-7/color(red)(cancel(color(black)(6))) xx -color(red)(cancel(color(black)(6)))) - 1#

#y = 7 - 1#

#y = 6#

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