How do you solve the following system: #x-4y=2, 3x+4y=-10 #?

1 Answer
Mar 18, 2018

Answer:

See a solution process below:

Explanation:

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

#x - 4y + color(red)(4y) = 2 + color(red)(4y)#

#x - 0 = 2 + 4y#

#x = 2 + 4y#

Step 2) Substitute #(2 + 4y)# for #x# in the second equation and solve for #y#:

#3x + 4y = -10# becomes:

#3(2 + 4y) + 4y = -10#

#(3 xx 2) + (3 xx 4y) + 4y = -10#

#6 + 12y + 4y = -10#

#6 + (12 + 4)y = -10#

#6 + 16y = -10#

#6 - color(red)(6) + 16y = -10 - color(red)(6)#

#0 + 16y = -16#

#16y = -16#

#(16y)/color(red)(16) = -16/color(red)(16)#

#(color(red)(cancel(color(black)(16)))y)/cancel(color(red)(16)) = -1#

#y = -1#

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

#x = 2 + 4y# becomes:

#x = 2 + (4 xx -1)#

#x = 2 + (-4)#

#x = 2 - 4#

#x = -2#

The Solution Is:

#x = -2# and #y = -1#

Or

#(-2, -1)#