How do you solve #2x-4y=4# and #x + 4y = 14#?

2 Answers
Jul 18, 2017

Answer:

See a solution process below:

Explanation:

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

#x + 4y = 14#

#x + 4y - color(red)(4y) = 14 - color(red)(4y)#

#x + 0 = 14 - 4y#

#x = 14 - 4y#

Step 2) Substitute #(14 - 4y)# for #x# in the first equation and solve for #y#:

#2x - 4y = 4# becomes:

#2(14 - 4y) - 4y = 4#

#(2 * 14) - (2 * 4y) - 4y = 4#

#28 - 8y - 4y = 4#

#28 + (-8 - 4)y = 4#

#28 + (-12)y = 4#

#28 - 12y = 4#

#-color(red)(28) + 28 - 12y = -color(red)(28) + 4#

#0 - 12y = -24#

#-12y = -24#

#(-12y)/color(red)(-12) = (-24)/color(red)(-12)#

#(color(red)(cancel(color(black)(-12)))y)/cancel(color(red)(-12)) = 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 = 14 - 4y# becomes:

#x = 14 - (4 * 2)#

#x = 14 - 8#

#x = 6#

The Solution is: #x = 6# and #y = 2# or #(6, 2)#

Jul 18, 2017

Answer:

See below

#x=6" "y=2#

Explanation:

To solve #2x−4y=4# and #x+4y=14# , Add the two together

#" "2x−4y=4#
#" "x+4y=14#

#3x=18" "x=6#

substitute this into one of the equations

#x+4y=14" "6+4y=14#

#4y=8" "y=2#