How do you solve the following system: #2x-4y=6 , y + 4x = 16 #?

1 Answer
May 17, 2018

Answer:

See a solution process below:

Explanation:

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

#y + 4x = 16#

#y + 4x - color(red)(4x) = 16 - color(red)(4x)#

#y + 0 = 16 - 4x#

#y = 16 - 4x#

Step 2) Substitute #(16 - 4x)# for #y# in the first equation and solve for #x#:

#2x - 4y = 6# becomes:

#2x - 4(16 - 4x) = 6#

#2x - (4 * 16) + (4 * 4x) = 6#

#2x - 64 + 16x = 6#

#2x + 16x - 64 = 6#

#(2 + 16)x - 64 + color(red)(64) = 6 + color(red)(64)#

#18x - 0 = 70#

#18x = 70#

#(18x)/color(red)(18) = 70/color(red)(18)#

#x = 35/9#

Step 3) Substitute #35/9# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = 16 - 4x# becomes:

#y = 16 - (4 xx 35/9)#

#y = 144/9 - 140/9#

#y = (144 - 140)/9#

#y = 4/9#

The Solution Is:

#x = 35/9# and #y = 4/9#

Or

#(35/9, 4/9)#