How do you solve #4x + y = 8# and #y = 4x# using substitution?

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Answer 1 · 2016-07-20T21:41:51

#x = 1, y = 4#

The y values in both equations are the same.
MAke y the subject in each equation.

#y = -4x+8" and " y = 4x#

Now, # " "y " " = subst for each "y#

#:. -4x +8 = 4x#

#8 = 8x#

#x=1#

If #x= 1#, then #y = 4 xx1 = 4#