Solve the simultaneous equations x^2+ y^2 = 29 and y–x = 3
For this question we will be using the substitution method in order to solve this question. Firstly we will take the less complex equation and rearrange to make either x or y the subject of the formula. We will make y the subject of the formula in this case. So y-x = 3 becomes y = x+3. We will now substitute this equation for y into our other equation. So x^2+y^2 = 29 will now become x^2+(x+3)^2 = 29. Next we will expand and simplify this equation. x^2+(x+3)^2 = 29 now becomes x^2+3x-10=0. This is a quadratic equation, this can be solved using the quadratic formula giving answers of x = -5 and x = 2. These two values for x can now be substituted back into the eqaution y-x = 3 to give y = -2 (for x = -5) and y = 5 (for x=2). These are the solutions to the equation.
Related Maths GCSE answers
All answers ▸Express 56 as the product of its prime factors
There are 10 boys and 20 girls in a class. The mean mark in a test for all the class is 60. The mean mark for the girls is 54. Work out the mean mark for the boys.
Jon and Nik share money in the ratio 5 : 2 Jon gets £150 more than Nik. How much money do they share altogether?
Show that 6sin(60◦) + 5tan(60◦) can be written in the form √k where k is an integer.