How do you solve the system of equations #y = .5x + 20# and #y = .15x#?

1 Answer
Ken C.
Jul 1, 2016

Use substitution to get a solution of #(-400/7,-60/7)#.

Explanation:

We know that #y=y#, so that must also mean, if #y=.5x+20# and #y=.15x#:
#.5x+20=.15x#

This is now a linear equation:
#.5x+20=.15x#
#cancel(.5x)+20-cancel(.5x)=.15x-.5x#
#20=-.35x#
#x=20/-.35=20/-(35/100)#
#=cancel20^4*-100/cancel35^7#
#=-400/7~~-57.14#

Since #y=.15x#:
#y=.15*-400/7#
#=15/cancel100^1*-cancel400^4/7#
#=-60/7~~-8.57#

Thus, the ordered pair that is the solution to this equation is #(-400/7,-60/7)#.

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