How to solve the simultaneous equations 3x+2y=7 and 5x+y=14

Firstly we rearrange one of these equations so that we have y on one side of the equation on its own. Let's do this with the second equation.

So from 5x+y=14, we can minus 5x from both sides to get:

y=14-5x

Then we can substitute this expression for y into our first equation that is 3x+2y=7

So we have 3x+2(14-5x)=7

Then we expand the bracket: 

3x+(2)(14)+(2)(-5x)=7

By simplifying this equation we get:

3x+28-10x=7

And simplifying further gives:

7x=21

By dividing both sides by 7, we find that x=3.

We substitute this value for x into either of our original equations to find the value of y.

3(3)+2y=7

So 2y=7-9

And therefore y=-1.

Finally we can check our solutions by substituting x=3 and y=-1 into the other original equation.

Therefore the solutions are x=3 and y=-1.

IB
Answered by Imogen B. Maths tutor

9140 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Factorise and solve 4(x)^2-2x=0


Solve for x, y, and z: 5x - 2y = 19 , 3x + 3z = 21 , y + z = 2


Find the coordinates of the minimum point of the curve y=x^2+6x+5.


Show that (x+2)(x+3)(x+4) can be written in the form of ax^3+bx^2+cx+d where a, b, c and d are positive integers.


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:

© 2026 by IXL Learning