How do you solve the system of equations #4x - y = 0# and #8x + y = 18#?

1 Answer
Dec 28, 2016

#x = 3/2# and #y = 6#

Explanation:

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

#8x + y - color(red)(8x) = 18 - color(red)(8x)#

#8x - color(red)(8x) + y = 18 - color(red)(8x)#

#0 + y = 18 - 8x#

#y = 18 - 8x#

Step 2) Substitute #color(red)(18 - 8x)# for #y# in the first equation and solve for #x#:

#4x - (color(red)(18 - 8x)) = 0#

#4x - color(red)(18 + 8x) = 0#

#4x + 8x - 18 = 0#

#12x - 18 = 0#

#12x - 18 + color(red)(18) = 0 + color(red)(18)#

#12x - 0 = 18#

#12x = 18#

#(12x)/color(red)(12) = 18/color(red)(12)#

#(color(red)(cancel(color(black)(12)))x)/cancel(color(red)(12)) = (6 xx 3)/(6 xx 2)#

#x = (cancel(6) xx 3)/(cancel(6) xx 2)#

#x = 3/2#

Step 3) Substitute #color(red)(3/2)# for #x# in the solution to the second equation in Step 1 to calculate #y#

#y = 18 - (8 xx color(red)(3/2))#

#y = 18 - (color(red)(cancel(color(black)(8)))4 xx color(red)(3)/cancel(color(red)(2)))#

#y = 18 - (4 xx 3)#

#y = 18 - 12#

#y = 6#