maths help....

x^2+y^2=25
y=x-7

Put y into the first equation:

x^2+(x-7)^2=25

x^2+(x-7)(x-7)=25

x^2+x^2-14x+49=25

Take away 25 from both sides

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

2x^2 - 14x + 24 = 0

Factorise:

(2x - 8)(x - 3) = 0

-8 x -6 is +24, - 8 + -6 = 14 (don't forget, you are timesing -3 by 2) and 2x^2

Solution 1: 0 x (x - 3) = 0

2x - 8 = 0
x - 4 = 0
x = 4

Solution 2: (2x - 8) x 0 = 0

x - 3 = 0
x = 3

So x has two solutions because its x squared. x is 3 or 4

Let's put that into the original y equation:

y=x-7

3 - 7 = -4

-4 - 7 = -11

Therefore y is either -4 or -11. This is because the original equation had y squared, so it has 2 solutions.

X = 3 or 4
Y = -4 or -11

For the record, you're wrong Dunno quite what you've done