The solution to simultaneous equations is where the values of x and y satisfy both equations. In this case, we need to find values of x and y that satisfy the equations x² + y² =25 and y - 3x = 13. If we notice that x² + y² =25 is actually an equation of a circle with centre (0,0) and radius of 5 and y – 3x = 13 is a straight line that has an intercept of 13 and a gradient of 3. We can make a rough sketch of these and see that there should be two solutions to this problem, meaning the answer should include two values of y and two values of x.Firstly, I would label x² + y² =25 as equation 1 and y - 3x = 13 as equation 2. Secondly, I would rearrange equation 2 to get y = 3x + 13, which I shall label as equation 3. Then I would substitute the value of y from equation 3 into equation 1 to get x² + (3x + 13)² = 25. This is done so that we can use substitution to get an equation where there is only one type of unknown either x or y, in this case it is the x. After the substitute I will expand (3x + 13)² to get x² + 9x² + 39x + 39x + 169 = 25. Then I would simplify the equation (this is where you would add up all like terms so all the Xs go together and all the X² go together and the numbers as well) so x² + 9x² to get 10 x² and 39x + 39x to get 72x which results in 10x² + 78x + 169 = 25. Then you can simplify the equation fully by subtracting 25 from both sides to get 10x² + 78x + 169 -25 = 0, which results in 10x² + 78x + 144 = 0. Then I would divide both sides of the equation by 2 to get 5x² + 39x + 72 = 0, this will help make the problem easier to solve as we will now be dealing with much simpler numbers. Afterwards, I would factorise the equation (this is a process in which we find expressions you can multiply together to get the equation), this can be done by finding out the product of the coefficient (the constant that comes before variables like x or x²) of x² and 72, which is 360. Then I would find the pair of factors of 360 which add up to give the coefficient of x. The factors that match are 15 and 24. So then I would split 39x into 15x and 24x to get 5x² + 15x + 24x + 72 = 0. We can see that both 15 and 72 are the third multiple of 5 and 24 respectively, from this we can say that x + 3 is a factor of this quadratic equation. Resulting in 5x(x + 3) + 24(x+3) = 0 which results in (x + 3)(5x + 24)=0 .For a product to result in 0 one of the brackets should be equal to zero. So, if we assume x+3 = 0 then we get x = -3 and if we assume 5x + 24 = 0 then we get x = -(24/5). Once we have two solutions for x we can substitute them into equation 3 to get the corresponding solutions for the value of y.So, if x = -3 then y = 3(-3) + 13, y = 13 - 9, y = 4if x = -(24/5) then y = 3(-(24/5)) + 13, y = -(7/5)So, the solutions are:x = -3, y = 4x = -(24/5), y = -(7/5)

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