How do you solve the system #x^2-y=5# and #-3x+y=-7#?

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Answer 1 · 2016-07-06T18:38:41

#x = 2, y = -1" or " x = 1, y = -4#

Conveniently there is a single #y# term in each equation so they can both be changed to have #y# as the subject.

#y = x^2 - 5 " and "y = 3x -7#

Since #y = y# it follows that

#x^2 - 5 =3x -7#

#x^2 -3x +2 = 0 " factorise the trinomial"#

#(x -2)(x -1) = 0#

#x = 2" or "x = 1#

There are 2 possible answers for x, find the corresponding value for y..

When # x = 2," " y = 3 xx 2 - 7 = -1#
When #x = 1" "y = 3 xx 1 -7 = -4#