How do you find the solution of the system of equations $y = x^{2} - 4$ $y= x^2-4$ and $y = x - 2$ $y= x-2$ ?

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How do you find the solution of the system of equations $y = x^{2} - 4$ $y= x^2-4$ and $y = x - 2$ $y= x-2$ ?

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Answer 1 · 2015-05-17T13:27:42

You can isolate one variable in one equation and substitute its value in the other one. Let's do it.

Let's get the value of $y$ $y$ in the first equation and substitute it in the second one and, then, find $x$ $x$ :

$x^{2} - 4 = x - 2$ $x^2-4=x-2$
$x^{2} - x - 2 = 0$ $x^2-x-2=0$

Using Bhaskara:

$\frac{1 \pm \sqrt{1^{2} - 4 (1) (- 2)}}{2} \cdot 1$ $(1+-sqrt(1^2-4(1)(-2)))/2*1$
$\frac{1 \pm 3}{2}$ $(1+-3)/2$
$x_{1} = 2$ $x_1=2$ and $x_{2} = - 1$ $x_2=-1$

Now we have these values for $x$ $x$ , let's find the values of $y$ $y$ - using the first equation:

$y_{1} = {(2)}^{2} - 4 = 0$ $y_1=(2)^2-4=0$
$y_{2} = {(- 1)}^{2} - 4 = - 3$ $y_2=(-1)^2-4=-3$

So, your answers are:

$y = 0$ $y=0$ when $x = 2$ $x=2$
$y = - 3$ $y=-3$ when $x = - 1$ $x=-1$

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How do you find the solution of the system of equations $y = x^{2} - 4$ $y= x^2-4$ and $y = x - 2$ $y= x-2$ ?

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How do you find the solution of the system of equations y=x2−4y= x^2-4 and y=x−2y= x-2?

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How do you find the solution of the system of equations $y = x^{2} - 4$ $y= x^2-4$ and $y = x - 2$ $y= x-2$ ?