How do you solve the system #y=4x; 2y-x=16#?

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Answer 1 · 2017-01-09T22:02:54

See full solution process below

Step 1) Because the first equation is already solved for #y# we can substitute #4x# for #y# in the second equation and solve for #x#:

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

#8x - x = 16#

#8x - 1x = 16#

#(8 - 1)x = 16#

#7x = 16#

#(7x)/color(red)(7) = 16/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 16/7#

#x = 16/7#

Step 2) We can now substitute #16/7# for #x# in the first equation and calculate #y#:

#y = 4 xx 16/7#

#y = 64/7#

The solution is:

#x = 16/7# and #y = 64/7#