How do you solve this system of equations by substitution: #x+ 4y = - 14 and y = 2x - 8#?

2 Answers
Mar 26, 2018

Answer:

#(x, y)=(2, -4)#

Explanation:

We're given #y=2x-8.# This means that we can replace all instances of #y# in the first equation with #2x-8.# The goal when given a system of equations in two variables is two get an equation with one variable and solve for that variable, and subsequently back-substitute your solution in to get the other variable.

#x+4(2x-8)=-14#

#x+8x-32=-14#

Solve for #x:#

#9x-32=-14#

#9x=18#

#x=2#

To solve for #y,# simply plug in #x=2# into #y=2x-8.#

#y=2(2)-8=-4#

So, the solution is

#(x, y)=(2, -4)#

Mar 26, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Because the second equation is already solve for #y# we can substitute #(2x - 8)# for #y# in the first equation and solve for #x#:

#x + 4y = -14# becomes:

#x + 4(2x - 8) = -14#

#x + (4 xx 2x) - (4 xx 8) = -14#

#x + 8x - 32 = -14#

#1x + 8x - 32 = -14#

#(1 + 8)x - 32 = -14#

#9x - 32 = -14#

#9x - 32 + color(red)(32) = -14 + color(red)(32)#

#9x - 0 = 18#

#9x = 18#

#(9x)/color(red)(9) = 18/color(red)(9)#

#(color(red)(cancel(color(black)(9)))x)/cancel(color(red)(9)) = 2#

#x = 2#

Step 2) Substitute #2# for #x# in the second equation and calculate #y#:

#y = 2x - 8# becomes:

#y = (2 xx 2) - 8#

#y = 4 - 8#

#y = -4#

The Solution Is:

#x = 2# and #y = -4#

Or

#(2, -4)#