How do you solve the system of equations #10x + 2y = 16# and #- 3x - 2y = - 9#?

1 Answer
Aug 3, 2017

See a solution process below:

Explanation:

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

#10x + 2y = 16#

#-color(red)(10x) + 10x + 2y = -color(red)(10x) + 16#

#0 + 2y = -10x + 16#

#2y = -10x + 16#

#(2y)/color(red)(2) = (-10x + 16)/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = (-10x)/color(red)(2) + 16/color(red)(2)#

#y = -5x + 8#

Step 2) Substitute #(-5x + 8)# for #y# in the second equation and solve for #x#:

#-3x - 2y = -9# becomes:

#-3x - 2(-5x + 8) = -9#

#-3x + (-2 xx -5x) + (-2 xx 8) = -9#

#-3x + 10x + (-16) = -9#

#-3x + 10x - 16 = -9#

#(-3 + 10)x - 16 = -9#

#7x - 16 = -9#

#7x - 16 + color(red)(16) = -9 + color(red)(16)#

#7x - 0 = 7#

#7x = 7#

#(7x)/color(red)(7) = 7/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 1#

#x = 1#

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

#y = -5x + 8# becomes:

#y = (-5 xx 1) + 8#

#y = -5 + 8#

#y = 3#

The Solution Is: #x = 1# and #y = 3# or #(1, 3)#