Solve algebraically the simultaneous equations, x^2 + y^2 = 25 and y – 3x = 13

First you need to pick a variable to solve for( irrelevant which one is picked) so I will choose to solve y first as in this question it is easier (as y has no coefficient). So make y the subject of the simpler equation(without the power) to get y = 13 + 3x, then sub this into the first equation to eliminate the y variable to get x^2 + (13 + 3x)^2 = 25. Then simplify and group terms (also putting all terms on the same side) to get quadratic in x: 10x^2 + 78x + 144 = 0. Then can simplify by factoring out 2 to get 5x^2 + 39x + 72 = 0. Then use quadratic formula or factorising into (5x + 2)(x + 3) get solutions for x being -3 and -24/5. Then can sub these into the original equation to find solutions for y being 4 and -7/5 respectively. So solutions are x = -3, y = 4 and x = -24/5 and y is -7/5.

BU

Related Maths GCSE answers

All answers ▸

You are given a square which you are told has a total area of 100 squared centimetres. You are also told that one side of the square has dimension 4(3x + 2), and the other has dimension 8x - y. What are the values of x and y?


Solve the simultaneous equations; 3x+2y=11 and 2x-2y=14


How to solve the following for x: (2x+3)/(x-4) - (2x-8)(2x+1) = 1


Using the quadratic formula factorise 2x^2 + 7x + 11 =0