Solve the simultaneous equations: x^2 + y^2 = 29 and y - x = 3
To solve these equations we need to eliminate one of the variables, so make y the subject of the second equation: y = x + 3. Now y can be substituted into the first equation: x^2 + (x+3)^2 = 29. Expanding (x+3)^2 = x^2 + 6x + 9, so the equation is now x^2 + x^2 + 6x + 9 = 29
Simplifying the equation gives this: 2x^2 + 6x - 20 = 0, which can be simplified further by dividing through by 2: x^2 + 3x - 10 = 0. Factorising the equation gives (x-2)(x+5) = 0, so x = 2 or -5.
Substitute the values of x into y = x+3, to give y = 5 when x = 2 and y = -2 when x = -5.
TG
Related Maths GCSE answers
All answers ▸A football pitch has a length of the xm. Its width is 25m shorter than the length. The area of the pitch is 2200m2. Show that x2 - 25x - 2200 =0 and work out the length of the football pitch.
Solve the equation 3a^2+4a+1=3 for all values of a. Give your answers to 3 significant figures.
Solve the quadratic equation x^2 + 7x + 6 = 0
Solve the equation x^2-9x+20