Solve the simultaneous equations: 5x + 3y = 41 and 2x + 3y = 20 Do not use trial and error.

  1. 5x + 3y = 41   2) 2x + 3y = 20 Subtract equation 2) from equation 1) to cancel the y's (3y-3y = 0) .  Next we subtract  the x's (5x - 2x = 3x) and then finally we have 41 - 20 = 21. We now have that 3x = 21 ( remember both the y's have cancelled) so we can divide both sides by 3 to find that; x = 7
Answered by David T. Maths tutor

17581 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do we complete the square with questions involving a number stuck with the x-squared?


Solve these simultaneous equations and find the values of x and y. Equation 1: 2x + y = 7 Equation 2: 3x - y = 8


How do you solve algebraic fractions with quadratics?


Expand and simplify (6x + 10)(3x - 2)


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