Solve the simultaneous equations: 3x+5y=3 and 6x+6y=10

(1): 3x+5y=3

(2): 6x+6y=10

Multiply number one by two to give a common factor of 6x in both equations.

(3)=[2*(1)]- 6x+10y=6

Take away equation (2) from equation (3) to leave us with only y's and numbers so that we can solve a value of y.

(3)-(2): 4y=-4

Divide both sides by 4

y=-1

With this value sub back into an original equation wherever there is a y, this does not matter if you choose (1) or (2) as it will give you the same value.

(1): 3x+5y=3

(1): 3x+5(-1)=3

(1): 3x-5=3

(1): 3x=8

(1): x=8/3

(1): x=2.66666666

(1): x=2.67

Final values

y=-1; x=8/3 or 2.67

Answered by Ethan O. Maths tutor

3862 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

A bag contains 4 red counters and some blue counters in the ratio 1:4 respectively. How many counters are there in the bag altogether?


factorise the quadratic: v^2+20v+19


Solve the simultaneous equations: 5x + 3y = 9 and 7x - 2y = 25.


Bhavin, Max and Imran share 6000 rupees in the ratios 2 : 3 : 7. Imran then gives 3/5 of his share of the money to Bhavin. What percentage of the 6000 rupees does Bhavin now have? Give your answer correct to the nearest whole number.


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