How do you solve $x^{2} - 2 x - 35 = y, x = y - 5$ $x^2-2x-35=y, x=y-5$ ?

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Answer 1 · 2016-11-05T23:32:07

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

$y = x + 5$ $y = x+ 5$

Substitute into the first.

$x^{2} - 2 x - 35 = x + 5$ $x^2 - 2x - 35 = x + 5$

$x^{2} - 3 x - 40 = 0$ $x^2 - 3x - 40 = 0$

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

$x = 8 and 5$ $x= 8 and 5$

$:.y = x +5$

$y = 13 and y = 10$

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

Hopefully this helps!

How do you solve x^2-2x-35=y, x=y-5x2−2x−35=y,x=y−5x^2-2x-35=y, x=y-5?