How do I solve simultaneous equations?

When given two simultaneous equations, the first step is to make the factor of one letter equal in both equations. For example, in the example below, we will try to make both equations contain '2b':
6a + b = 16
5a - 2b = 19

This can be done by multiplying the entire first equation by 2. To do this, we have to multiply both sides of the equation by 2, as seen below.

2(6a + b) = 2 x 16

Now we must expand the brackets.

12a + 2b = 32.

The two equations are now

12a + 2b = 32
5a - 2b = 19

Now that we have made the factor before 'b' equal in both equations, we must eliminate 'b' from both equations by minusing or adding one equation from another to cancel out the 'b's.

(5a - 2b = 19)

+(12a + 2b = 32)

= (17a + 0 = 51)

17a = 51

At this stage, we have isolated 'a'. If we know that 17a is equal to 51, we can find 1a by dividing 51 by 17.

51/17 = a = 3

This value of 'a' can now be substituted into either of the beginning equations to find the value for b.

6a + b = 16

(6 x 3) + b = 16

18 + b = 16

Now rearrange the equation to find a value for 'b' on its own. b = 16 - 18 = -2 So the final solved values are:

a = 3
b = -2

Answered by Lois B. Maths tutor

2650 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How do I factorise 12y-18


Differentiate dy/dx ((2x^3)+(x^2)-(4x)+7)


f(x)= 5 - x and g(x)= 3x + 7. Simplify f(2x) + g(x-1) (2018 past paper question)


Simplify fully {x^2 -3x -4}/{x^2 + 4x +3}


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