How do you solve $y = x^{2} - 4$ $y= x^2-4$ and $y = x - 2$ $y= x-2$ ?

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Answer 1 · 2015-08-10T09:33:00

The answer is $x = - 1$ $x = -1$ ; $x = 2$ $x = 2$

Since

${(y = x^2 -4), (y = x - 2) :}$

we shall have

$x^2 -4 = x - 2$

$x^2 -4 - x + 2 = 0$

It becomes a quadratic equation

$x^2 - x - 2 = 0$

$x^2 - 2x + x - 2 = 0$

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

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

This means that you have

$x + 1 = 0 implies x = -1$

or

$x - 2 = 0 implies x = 2$

If you substitute the two values of $x$ in the original equations, they satisfy them simultaneously.

How do you solve y= x^2-4y=x2−4y= x^2-4 and y= x-2y=x−2y= x-2?