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

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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 $y$ in the first equation and substitute it in the second one and, then, find $x$ :

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

Using Bhaskara:

$(1+-sqrt(1^2-4(1)(-2)))/2*1$
$(1+-3)/2$
$x_1=2$ and $x_2=-1$

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

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

So, your answers are:

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

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