How do you solve the system $Y=x+7$ , $x^2+y^2=25$ ?

Question

4360 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-01T17:21:29

Substitute $y = x+7$ into $x^2+y^2 = 25$ to get a quadratic in $x$ , hence solutions:

$(x, y) = (-3, 4) or (-4, 3)$

Substitute $y = x+7$ into $x^2+y^2 = 25$ ...

$25= x^2+y^2$

$=x^2+(x+7)^2$

$=x^2+x^2+14x+49$

$=2x^2+14x+49$

Subtract $25$ from both ends to get:

$2x^2+14x+24 = 0$

Divide through by $2$ to get:

$x^2+7x+12 = 0$

We can factor this by finding a pair of factors of $12$ whose sum is $7$ . The pair $3, 4$ works.

Hence:

$0 = x^2+7x+12 = (x+3)(x+4)$

This has roots $x=-3$ and $x=-4$ , corresponding to $y = 4$ and $y=3$ respectively.

How do you solve the system Y=x+7Y=x+7, x^2+y^2=25x^2+y^2=25?