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

16893 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

25* 3/2


Simplify: 2x + 6y + 2y - x


Solve the simultaneous equations: 12x - 4y = 12 (1) and 3x + 2y = 12 (2)


express the number 84 as a product of its prime factors


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 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences