(Q20 Non-Calculator paper, Higher Tier) Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y – 3x = 13
We have a quadratic equation (unknowns are raised to the power of 2) and a liner equation (power 1). We will use the linear equation to express one of the variables, e.g. y, in terms of the other, i.e. x. Then, we will substitute the obtained expression for y in the quadratic equation and solve for x, using the quadratic formula. Having found x, we will use its value to find y.Execute: From y - 3x = 13 => y = 13 + 3xFrom x2 + y2 = 25 => x2 + (13+3x)2= 25. Expanding the brackets we get: 10x2 + 78x + 144 = 0. Using the quadratic formula: x= {-b +/- sqrt (b2-4ac)}/(2a) we get: x = -3 and x = -4.8. Plugging those values in y = 13 + 3x, we get: y = 4 and y = -1.4 in this order. Answer: x = -3, y = 4 & x = -4.8, y = -1.4. Note: lack of formatting options make the formulae look much more complicated than they are. Using whiteboard will resolve this issue.
