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

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Answer 1 · 2016-07-06T18:38:41

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

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

$y = x^{2} - 5 and y = 3 x - 7$ $y = x^2 - 5 " and "y = 3x -7$

Since $y = y$ $y = y$ it follows that

$x^{2} - 5 = 3 x - 7$ $x^2 - 5 =3x -7$

$x^{2} - 3 x + 2 = 0 factorise the trinomial$ $x^2 -3x +2 = 0 " factorise the trinomial"$

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

$x = 2 or x = 1$ $x = 2" or "x = 1$

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

When $x = 2, y = 3 \times 2 - 7 = - 1$ $x = 2," " y = 3 xx 2 - 7 = -1$
When $x = 1 y = 3 \times 1 - 7 = - 4$ $x = 1" "y = 3 xx 1 -7 = -4$

How do you solve the system x^2-y=5x2−y=5x^2-y=5 and -3x+y=-7−3x+y=−7-3x+y=-7?