How do you solve #y=x^2 # and #y= 0.5x + 5# using substitution?

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Answer 1 · 2018-08-02T14:59:13

Please see below.

Just substitute value of #y# from equation #y=0.5x+5# in #y=x^2#

and we get #0.5x+5=x^2# and multiplying by #2#, we get

#x+10=2x^2# or #2x^2-x-10=0#

i.e. #2x^2-5x+4x-10=0#

or #x(2x-5)+2(2x-5)=0#

or #(x+2)(2x-5)=0#

i.e. either #x+2=0# i.e. #x=-2#

or #2x-5=0# i.e. #x=5/2=2.5#

Putting these values of #x# in #y=0.5x+5#, we get

#y=4# or #y=6.25#