Solve the simultaneous equations algebraically.

x^2 + y^2 = 29y - x = 3Rearrange the second equation such that one variable, either x or y, is the subject. I will rearrange to make x the subject. x = y - 3. Substitute the new equation into equation 1 in place of x.(y-3)^2 + y^2 = 29.Rearrange to make the equation equal 0 and simplify. y^2 - 6y + 9 + y^2 = 29. 2y^2 - 6y - 20 = 0. All variables have a factor of 2 and so we can simplify further by dividing by 2.y^2 - 3y - 10 = 0. Factorise. (y-5)(y+2)=0. Therefore, y=5 when x=? and y=-2 when x=?. To find out what x is in each case of y we can substitute the x values back into one of the original equations and see what values of y we receive. In this case I will substitute back into equation 2. When y=5, 5 - x = 3. Therefore, x=2. --> When y=5, x=2.When y=-2, -2 - x= 3. Therefore, x = -5. --> When y=-2, x=-5.

JA
Answered by Jane A. Maths tutor

2448 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Find the possible values of x for x^2 = 36-5x.


Sarah’s collection contains dresses, skirts and blouses. If the ratio of dresses to skirts is 7 to 4 and the ratio of skirts to blouses is 7 to 2, what is the ratio of dresses to blouses?


If you are given the function f(x) = 10-3x and g(x) = (x-3)/2 , find the value of the composite function gf(4) ?


Rationalise the denominator of the following fraction: 1/(√2 + 1)


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences