I'm assuming that those are two simultaneous equations:

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

y - 3x = 13 ------------ b

From the second equation (b), we can rearrange the equation to get an equation for y which requires us to add 3x to both sides of the equal sign.

We then get:

y = 13 + 3x --------- c

This is now in a suitable form to allow us to substitute it back into the first original simultaneous equation (a) so that we have an equation just in terms of x so that we can solve for x:

x^2 + (13 + 3x)^2 = 25

The expansion of (13 + 3x)^2 = 9x^2 + 78x + 169

(If you're unsure about how I arrived at this, you can get in touch with me and I'll be more than happy to explain)

We then have:

x^2 + 9x^2 + 78x + 169 = 25

This is in the form of a quadratic equation, so we can shift all the terms to one side of the equation, equating the equation to 0:

10x^2 + 78x + 144 = 0

There are many ways to solve a quadratic equation. The most straight forward way when you have such big numbers would probably be to use the quadratic formula, but you can also try to find suitable factors. I'll use the quadratic formula which is:

x = (-b +- sqrt(b^2 - 4ac))/2a

This can be seen more clearly on the website (https://www.bbc.com/bitesize/guides/z3hb97h/revision/2)

From our quadratic equation, a = 10 , b = 78 , c = 144

Hence;

x = (-(78) +- sqrt ( 78^2 - 4*10*144))/2*10

x = (-78 +- sqrt (6084 - 5760))/20

x = (-78 +- sqrt (324))/20

x = (-78 +- 18)/20

So therefore:

x = (-78+18)/20 OR x = (-78-18)/20

This gives us values for x as:

x = -3 OR x = -4.8

This is not the right place to stop because we still need to find the corresponding values of y from equation c.

y = 13 + 3x

When x = -3 AND When x = -4.8

y = 13 + 3(-3) y = 13 + 3(-4.8)

y = 13 - 9 y = 13 - 14.4

y = 4 y = -1.4

I hope this helped and if you have any questions, you can get in touch with me!