Solve the following simultaneous equations. x^2+y^2=25. y-3x=13

The key to solving simultaneous equations is trying to reduce them into one equation. You can see that one of these equations is quadratic, so you should be thinking there will be 2 answers for x and y. You want to simplify one equation as much as possible so that you can use it to eliminate a variable in the second equation. This often means using the linear equation (no quadratics). Using this equation, we can take the -3x to the other side to get y=13+3x. As y is the subject of this equation (on its own on one side), you can now substitute this into the first equation, remembering the whole (13+3x) is squared. So you get x^2+(13+3x)^2=25. If you expand this squared bracket out using the FOIL method, and collected like terms, you get 10x^2+78x+169=25. Take 25 from both sides to get 10x^2+78x+144=0. This looks scary, and the quadratic formula wouldn't help much. However, all coefficients are multiples of 2, so we can divide by 2 to get 5x^2+39x+72=0, which is still daunting but manageable. We need to factorise this. The method i use is to recognise what the coefficients of x need to be in the brackets. As 5 is a prime number, we need the brackets need to be (5x+a)(x+b). I then write out all factors of the final term, 72, and then i go down the list and see if any combinations of (5x+a)(x+b) achieve the result. (not written out for lack of space) When we reach 3 and 24, we see that 324=72, and 53+24=39 - we have our factors! So our brackets are (5x+24)(x+3)=0, and so 5x+24=0, x+3=0, so x=-4.8, x=-3. Now we have these, we can go back into the original question to work out y. We figured out y=13+3x, so subbing our answers of x in, we get that y=-1.4, y= 4.