solve the simultaneous equations 3x+7y=18 and 7x+9y=8

3x+7y=18 (eqn 1)  -multiply by 7->   21x+49y=126 (eqn 1*)

7x+9y=8  (eqn 2)  -multiply by 3->    21x+27y=24 (eqn 2*)

(eqn 1*)-(eqn 2*)  --> 22y=102 therefor y=51/11

sub in this y value back into (eqn1) -->  3x+7(51/11)=18 --> 3x+357/11=18 --> 3x=-159/11

therefore x=-53/11

RS
Answered by Radhika S. Maths tutor

3470 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do you add fractions? And how do you multiply fractions?


Use the factor theorem to show that (x+2) is a factor of g(x)= 4x^3 - 12x^2 - 15x + 50


Find x when: (2^x)(e^(3x+1))=10. Give your answer in the form (a + ln(b)) / (c + ln(d)) , where a,b,c,d are integers.


Work out the increase in price from £1.50 to £2.00 as a percentage.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning