Solve algebraically the simultaneous equations x2 +y2 =25 and y – 3x = 13
Firstly, let's begin by rearranging the second equation so that we can get y on its own. Therefore we will get, y = 3x +13. Through substitution, we can do this:x2+ (3x +13)2= 25 . (As y is equal to 3x + 13).If we multiply this out then, we will get x2 + (3x+13)(3x+13) = 25Dealing with (3x+13)(3x+13), we can multiply this out using the FOIL method, which stands for multiplying the First two values from each bracket (3x), then the Outside (furthest) two, (3x and 13), then the inside values (13 and 3x), and then finally the Last remaining values (13). Thus we get 9x2+ 39x + 39x + 169. Tidying this up, we are left with 9x2+ 78x + 169. Thus plugging this into our original equation, we will get: x2 + 9x2+ 78x + 169 = 25.Again, tidying this up for simplicity, this leaves us with (adding the values in correlation to their respective power):10x2 + 78x + 144 = 0.We can this divide this by the common factor 2, again just to make our lives a little easier.5x2+ 39x - 72 =0Let's now then use the quadratic formula. We'll assign 5 as a, 39 as b and 72 as c. Thus, we will get:-39+- root(392-4x5x72)/2x5Leaving us with -39 + 9/10 and -39-9/10, thus x= -3 and, x = -24/5.Plugging both of these solutions into the equation y= 3x+13 ,we will get y= 3(-3) + 13, y= 3(-24/5) + 13.Therefore y = 4 and y = -7/5.Well done! You've just completed a really hard question from the Edexcel GCSE exam paper!
