Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y – 3x = 13
(1) x2 + y2 = 25 (2) y - 3x = 13 To solve simultaneous equations algebraically we want to rearrange one of the equations to be able to substitute this in to the other equation. In this example, we have squared x and y terms, which makes this equation the more complex one. We therefore decide to rearrange equation (2), and as the y term here has a coefficient of 1 it is easiest to rearrange for y. By adding 3x to both sides of equation (2):y = 13 + 3x. We then substitute this in to equation (1): (13 + 3x)2 + x2 = 25. And write this as : (13 + 3x)(13 + 3x) + x2 = 25. By expanding the brackets we get: 169 + 39x + 39x + 9x2 + x2 = 25. By collecting like terms and simplifying: 169 + 78x +10x2 = 25. Making this equal to 0: 10x2 + 78x + 144 = 0. And divide every term by 2: 5x2 + 39x + 72 = 0. We then just solve this equation like a normal quadratic equation as follows: Split up the x terms: 5x2 + 15x + 24x + 72 = 0. Factorise: 5x(x + 3) + 24(x + 3) = 0. Put in to brackets: (5x + 24)(x + 3) = 0. Find 2 equations with x values equal to 0: 5x + 24 = 0 and x + 3 = 0. Taking 5x + 24 = 0 : We get 5x = -24 and so x = -24/5. Taking x + 3 =0 we get that x = -3. We then substitute these values back in to get a y value. For this, we can use equation (2), and use where we found that : y = 13 + 3x. If we first substitute that x = -24/5, we get: y = 13 + 3(-24/5) so y = 13 - 72/5 so y = (65 - 72)/5 and therefore y = -7/5. And now we substitute x = -3 so we get: y = 13 + 3(-3) so y = 13 - 9y = 4. So our final answers are:x = -24/5, y = -7/5, x = -3, y = 4.
