Solve the following simultaneous equations: x^2 + y^2 = 29 and y - x =3

This question is slightly trickier than normal simultaneous equations, because we have values to the power of 2. What we can do is make either y or x the subject of the 2nd equation (y-x = 3). For this example, I will choose to make y the subject, which then gives us y = 3 + x. We can use this equation, and substitute it into the value of "y" in the 1st equation, as follows --> x^2 + (3+x)^2 = 29 We can then expand the brackets and simplify: x^2 + x^2 + 6x + 9 = 29 2x^2 + 6x -20 = 0 (it is important to make the equation equal to 0 so that we can solve the equation to find the values of x) x^2 + 3x - 10 = 0 (we can divide the whole equation by 2, as this is a common factor) (x + 5) (x - 2) = 0 (we can factorise the equation to give us 2 brackets; we have found two numbers which multiply to give -10 and add to give +3) .˙. x = -5 and x = 2 (each bracket is made equal to 0 and solved separately, we have two values for x because this is a quadratic equation)Each value is then substituted back into the rearranged 2nd equation (y = 3 + x) which gives us y = -2 and y = 5

Answered by Trushna D. Maths tutor

3205 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Complete the square on this equation: 2x^2 + 20x + 15 = 0.


Solve the simultaneous equations: 3x + 2y =4 4x + 5y =17


I know the formula, but I don't understand it.


How would I find the nth term of this sequence? 15, 18, 21, 24, ...


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences