How do you solve the system of equations #3x-2y=7# and #x+3y=-5#?

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Answer 1 · 2017-02-02T14:58:29

See the entire solution process below:

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

#x + 3y = -5#

#x + 3y - color(red)(3y) = -5 - color(red)(3y)#

#x + 0 = -5 - 3y#

#x = -5 - 3y#

Step 2) Substitute #-5 - 3y# for #x# in the first equation and solve for #y#:

#3(-5 - 3y) - 2y = 7#

#-15 - 9y - 2y = 7#

#-15 - 11y = 7#

#color(red)(15) - 15 - 11y = color(red)(15) + 7#

#0 - 11y = 22#

#-11y = 22#

#(-11y)/color(red)(-11) = 22/color(red)(-11)#

#(color(red)(cancel(color(black)(-11)))y)/cancel(color(red)(-11)) = -2#

#y = -2#

Step 3) Substitute #-2# for #y# in the solution to the second equation at the end of Step 1 and calculate #x#:

#x = -5 - (3 xx -2)#

#x = -5 - (-6)#

#x = -5 + 6#

#x = 1#

The solution is #x = 1# and #y = -2#