How do you solve the following system: # -8x+2y=4 , 4x+y=10 #?

1 Answer
Jan 24, 2017

Answer:

See the entire solution process below:

Explanation:

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

#4x + y = 10#

#4x - 4x + y = 10 - 4x#

#0 + y = 10 - 4x#

#y = 10 - 4x#

Step 2) Substitute #color(red)(10 - 4x)# for #y# in the first equation and solve for #x#:

#-8x + 2(color(red)(10 - 4x)) = 4#

#-8x + 20 - 8x = 4#

#-16x + 20 = 4#

#-16x + 20 - color(red)(20) = 4 - color(red)(20)#

#-16x + 0 = -16#

#-16x = -16#

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

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

#x = 1#

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

#y = 10 - (4 xx color(red)(1))#

#y = 10 - 4#

#y = 6#

The solution is #x = 1# and #y = 6# or (1, 6)