How do you solve the following system: #5x+y=-7, 6x + 7y = -9 #?

1 Answer
Jul 2, 2017

See a solution process below:

Explanation:

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

#5x + y = -7#

#-color(red)(5x) + 5x + y = -color(red)(5x) - 7#

#0 + y = -5x - 7#

#y = -5x - 7#

Step 2) Substitute #(-5x - 7)# for #y# in the second equation and solve for #x#:

#6x + 7y = -9# becomes:

#6x + 7(-5x - 7) = -9#

#6x + (7 * -5x) - (7 * 7) = -9#

#6x + (-35x) - 49 = -9#

#6x - 35x - 49 = -9#

#(6 - 35)x - 49 = -9#

#-29x - 49 = -9#

#-29x - 49 + color(red)(49) = -9 + color(red)(49)#

#-29x - 0 = 40#

#-29x = 40#

#(-29x)/color(red)(-29) = 40/color(red)(-29)#

#(color(red)(cancel(color(black)(-29)))x)/cancel(color(red)(-29)) = -40/29#

#x = -40/29#

Step 3) Substitute #-40/29# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = -5x - 7# becomes:

#y = (-5 xx -40/29) - 7#

#y = 200/29 - 7#

#y = 200/29 - (29/29 xx 7)#

#y = 200/29 - 203/29#

#y = -3/29#

The solution is: #x = -40/29# and #y = -3/29# or #(-40/29, -3/29)#