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

Question

6651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-21T16:57:38

There is no solution.

We can use the fact that $color(red)y=color(blue)(x-7$ to substitute this expression of $y$ into the other equation.

$x^2+color(red)y^2=7" "=>" "x^2+color(blue)((x-7))^2=7$

Expanding the resultant equation and then solving it:

$x^2+(x^2-14x+49)=7$

$2x^2-14x+42=0$

Dividing through by $2$ :

$x^2-7x+21=0$

Examining this, we see that the discriminant $b^2-4ac=49-4(21)=-35$ . Since the discriminant is negative, this system has no solutions.

We can graph the two equations given originally:

graph{(x^2+y^2-7)(y-x+7)=0 [-19.64, 20.92, -13.17, 7.1]}

The two graphs never intersect, so this confirms our conclusion of no solution.

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