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

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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.