How do you solve the simultaneous equations 3x + 4y = 5 and 2x – 3y = 9

We label each equation

3x + 4y = 5 (1)

2x – 3y = 9 (2)

We now want to get rid of one of the variables (x or y). Lets get rid of x:

We need (1) and (2) to have the same number of x's so we multiple (1) by 2 and (2) by 3 so they both have 6 x's

6x + 8y = 10 (1)*2

6x - 9y = 27 (2)*3

Now to get rid of the x's we take one of the equations from the other. It is easier to do (2)*3 -(1)*2

  6x - 9y = 27 (2)*3

- 6x + 8y = 10 (1)*2

                                   

     -17y = 17 

If we devide this equation by -17 we get

y = -1

We can plug this value of y into (1) to get

3x -4 = 5

add 4 to both sides of the equation

3x = 9

Divide by 3

x = 3

Now to check put x and y into (2)

6 - (-3) = 9 

As this is true, we have the solution

x = 3, y = -1

SS
Answered by Sahiti S. Maths tutor

41380 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Hey I am having difficulties solving simultaneous equations? Please may you help me!


Prove that the difference between the squares of any two consecutive integers is equal to the sum of these two integers.


I am getting stuck on how to solve Simultaneous Equations, can you explain how to do this?


A right-angled triangle has one angle size 60 degrees, and hypotenuse of length 32cm. Calculate the length of the side opposite the 60 degree angle, to 3sf.


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