Solve the simultaneous equations y = x^2 - 6x and 2y + x - 6 = 0
Rearrange the second equation in terms of y: meaning that the equation is of the form y = ....-this will give y = 3 - x/2You may now substitute the y in the left hand equation with what y in the right hand equation equals-this will give 3 - x/2 = x2 - 6xNow rearrange to give the form ax2 + bx + c = 0 where a, b and c are rational numbers-this will give 2x2 -11x - 6 = 0You can now solve the quadratic either using the quadratic formula or via factorisation. I prefer factorisation:multiply the coefficients of x2 and units (a and c) together: this gives 12find a pair of numbers which can be added to make the coefficient of x (b): these are -12 and 1express the quadratic with the coefficients of x separated in this way: 2x2 -12x + x - 6 = 0factorise each half of the equation - you should be able to find a common factor: 2x(x - 6) + 1(x - 6) = 0you now know you have 2x + 1 lots of (x - 6) - so this quadratic can be factorised as (2x + 1)(x - 6) = 0To solve the factorised quadratic you find the x value that makes each bracket equal zero - so x = 6 and x = -1/2These are simultaneous equations, so you also need to find the y values for these x values: We found earlier that y = 3 - x/2, therefore y = 0 and y = 13/4. Make sure to indicate these values in pairs, such as coordinates: (6, 0) and (-1/2, 13/4)There are two solutions to this question as the graph of y = x2 - 6x is a curve, and 2y + x - 6 = 0 is a straight line. The solutions will be the places where the lines meet, meaning the lines will cross over twice at the coordinates we've found.
