Solve algebraically the simultaneous equations: x^2+y^2 = 25 and y-3x=13

To answer this question, we need to make y the subject of the second equation. We can do this through simple rearrangement:

y-3x=13 so y=13+3x

Now that we have y on its own, this means we can substitute for it in the first equation. We simply put 13+3x in place of y, so we get:

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

If we then expand the brackets on (13+3x)^2, we get (13+3x)(13+3x) = 169+78x+9x^2 , so the equation becomes x^2+9x^2+78x+169 = 25

This will then equal 10x^2+78x+169=25 

We then want to make this equal to zero, so we then can solve it as a standard quadratic equation.

10x^2+78x+144=0

We can simplify this further by dividing each side by 2 to get:

5x^2+39x+72=0

This is quite tricky to factorise and ultimately, we need to do some trial and error with certain numbers that multiply to equal 72. There is 36 and 2, but this clearly won't give us 39x, so this doesn't work. We can try 12 and 6 (126=72), but this will not work as 56 = 30 and plus 12 is 42. 24 and 3 may work - 5*3 = 15 and 15+24 = 39 - this will work and therefore we factorise it to be (5x+24)(x+3)=0. Therefore the two solutions for x are -3 or 5x=-24 so x=-24/5. If x=-3, then y=4 and if x=-24/5, then y=-7/5