Simultaneous equations are two equations, each with the same two unknowns and are "simultaneous" because they are solved together.

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In simple terms, the solution to a pair of simultaneous equations is the x and y values of the coordinates of the point at which the graphs cross or intersect. The example below shows this.

For x + y = 4 this gives (0,4) and (4,0) which we plot and then extend a straight line through.

We can find solutions to simultaneous equations algebraically too. There are two common methods. Which one you choose might depend on the values involved or it might just be the method you like the most. We will use the same pair of equations as above.

4x – 2y = 10

x + y = 4

Multiply x + y = 4 by 2 to give 2x + 2y = 8

4x – 2y = 10

2x + 2y = 8

6x = 18

x = 3

We then add the two equations which "eliminates" the 2y (since -2y + 2y = 0)

and leaves 6x = 18 which, after dividing both sides by 6 leaves x = 3

3 + y = 4

y = 1

We can then replace x in one of the equations with the value 3. (in this example, doing this in x + y =4 is simpler than in 4x - 2y = 10)

4x – 2y = 10

(4 x 3) - (2 x 1) = 10

12 - 2 = 10

It is always a good idea to check the values for x and y in the other equation.

4x – 2y = 10

x + y = 4

Rearrange one equation to get make either x or y subject of the equation.

In this case it will be simpler to rearrange x + y = 4 which we can rearrange to y = 4 -x by subtracting x from each side

4x – 2y = 10

4x - 2(4 - x) = 10

4x - 8 + 2x = 10

6x - 8 (+8) = 10 (+8)

6x = 18

x = 3

We then substitute 4 - x for y in the other equation.

3 + y = 4

y = 1

As with the elimination method we then replace x in one of the equations with the value 3. (in this example, doing this in x + y =4 is simpler than in 4x - 2y = 10)

4x – 2y = 10

(4 x 3) - (2 x 1) = 10

12 - 2 = 10

And, as a check, try the values for x and y in the other equation.

There is one example of each the elimination and substitution methods for solving simultaneous equations shown below.

5x - 3y = 16

4x - 5y = 5

With this pair of equations we will need to multiply both to get a common multiple of either x or y. We will multiply the top equation by 5 and the bottom one by 3

25x – 15y = 80

12x -15y = 15

13x = 65

x = 5

We then subtract the two equations which "eliminates" the -15y

and leaves 13x = 65 which, after dividing both sides by 13 leaves x = 5

(5 x 5) - 3y = 16

25 - 3y = 16

-3y = 16 - 25

-3y = -9

y = 3

We then replace x in the top equations with the value 5 to find y.

4x – 5y = 5

(4 x 5) - (5 x 3) = 5

20 - 15 = 5

Finally, check the values for x and y in the other equation. The two equations are shown below graphically.

3x + 4y = 5

x + 5y = 9

Rearrange one equation to get make either x or y subject of the equation.

In this case it will be simpler to rearrange x + 5y = 9 which we can rearrange to x = 9 - 5y by subtracting 5y from each side

3x + 4y = 5

3(9 - 5y) + 4y = 5

27 - 15y + 4y = 5

27 - 11y = 5

-11y = 5 - 27

-11y = -22

y = 2

We then substitute 9 - 5y for x in the other equation and solve for y.

x + 5y = 9

x + (5 x 2) = 9

x + 10 = 9

x = -1

As with the elimination method we then replace y in one of the equations with the value 2.

3x + 4y = 5

(3 x -1) + (4 x 2) = 5

-3 + 8 = 5

And, as a check, try the values for x and y in the other equation. Both equations are shown graphically below.

Sam and Jack have $50 between them and Sam has $5 more than Jack. How much money does each have?

s + j = 50

s - j = 5

s = 27.50 , j = 22.50

This example is quite simple - you might be able to work it out by trial-and-error - but you can use any of the methods above to solve it.

**And finally,...**

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