How do you solve #x^2-2x-35=y, x=y-5#?

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Answer 1 · 2016-11-05T23:32:07

Let's start by isolating #y# in the second equation.

#y = x+ 5#

Substitute into the first.

#x^2 - 2x - 35 = x + 5#

#x^2 - 3x - 40 = 0#

#(x - 8)(x + 5) = 0#

#x= 8 and 5#

#:.y = x +5#

#y = 13 and y = 10#

Hence, the solution set is #{8, 13}# and #{5,10}#.

Hopefully this helps!