How do you solve a simultaneous equation by 'substitution'?
Firstly, demonstrate with an example: Solve: 3x + y = -9 and x2 + 2x - 3 = yRearrange the first equation to get "y" by itself by moving parts of the equation to the other side e.g. y = - 9 - 3xSubstitute your new value for "y" in terms of "x" into the second equation e.g. x2 + 2x -3 = -9 - 3xMove all the terms onto the same side of the equation to make it equal 0 e.g. x2 + 5x + 6 = 0Factorise your quadratic e.g. (x + 2)(x + 3) = 0 (you can check this by expanding the brackets back out using the 'FOIL' methodFrom this we know x = -2 and x = -3We then plug these values back into our equation we made earlier: y = -9 - 3xWhen x = -2, y = -3 and when x = -3, y = 0
SH
Related Maths GCSE answers
All answers ▸Factorise 2b^2 + 6b
A circle with diameter 6cm is cut from a square with side length 7cm. What is the remaining area of the square? You may assume pi = 3 for this question.
15 machines work at the same rate, 15 machines can complete an order in 8 hours, however 3 of the machines break down after 6 hours. The other machines continue until the order is complete. In total how many hours does EACH machine work? (3 mark question)
Find the point of intersection between the lines 2y=-4x+4 and 3y=10x-3