1. Simultaneous Equations

2. Simultaneous equations
Equations in two unknowns have an infinite number of solution pairs. For example,
x + y = 3
is true when
x = and y = 2
x = and y = 0
x = –2 and y = and so on … 3 x y
x + y = 3
We can represent the set of solutions on a graph:
Ask pupils to suggest other pair of values that would solve this equation. Praise correct answers involving negative or fractional solutions.
Verify that all of the suggested solutions lie on the line x + y = 3.

3. Simultaneous equations
Another equation in two unknowns will also have an infinite number of solution pairs. For example,
y – x = 1
is true when
x = 1 and y = 2
x = 3 and y = 4
x = –2 and y = –1 and so on … x y 3
y – x = 1
This set of solutions can also be represented in a graph:
Ask pupils to suggest other pair of values that would solve this equation. Praise correct answers involving negative or fractional solutions.

4. Simultaneous equations
There is one pair of values that solves both these equations:
x + y = 3
y – x = 1
We can find the pair of values by drawing the lines x + y = 3 and y – x = 1 on the same graph. x y 3
y – x = 1
The point where the two lines intersect gives us the solution to both equations.
x + y = 3
This is the point (1, 2).
At this point x = 1 and y = 2.

5. Simultaneous equations
x + y = 3
y – x = 1
are called a pair of simultaneous equations.
The values of x and y that solve both equations are x = 1 and y = 2, as we found by drawing graphs.
We can check this solution by substituting these values into the original equations.
Tell pupils that simultaneous means ‘happening at the same time’. When two equations are described as simultaneous it means that we have to find values that solve both equations at the same time.
1 + 2 = 3
2 – 1 = 1
Both the equations are satisfied and so the solution is correct.

6. Simultaneous equations with no solutions
Sometimes pairs of simultaneous equations produce graphs that are parallel.
Parallel lines never meet, and so there is no point of intersection.
When two simultaneous equations produce graphs which are parallel there are no solutions.
How can we tell whether the graphs of two lines are parallel without drawing them?
Two lines are parallel if they have the same gradient.

7. Simultaneous equations with no solutions
We can find the gradient of the line given by a linear equation by rewriting it in the form y = mx + c.
The value of the gradient is given by the value of m.
Show that the simultaneous equations
y – 2x = 3
2y = 4x + 1
have no solutions.
Rearranging these equations in the form y = mx + c gives,
y = 2x + 3
y = 2x + ½
The gradient m is 2 for both equations and so there are no solutions.

8. Simultaneous equations with infinite solutions
Sometimes pairs of simultaneous equations are represented by the same graph. For example,
2x + y = 3
6x + 3y = 9
Notice that each term in the second equation is 3 times the value of the corresponding term in the first equation.
Both equations can be rearranged to give
It is not always as obvious as it is here that two lines coincide. Stress that the only way to make sure is to rearrange both equations so that they are in the form y = mx + c.
y = –2x + 3
When two simultaneous equations can be rearranged to give the same equation they have an infinite number of solutions.

9. Simulataneous Equations The Elimination Method

10. The elimination method
If two equations are true for the same values, we can add or subtract them to give a third equation that is also true for the same values. For example, suppose
3x + y = 9
5x – y = 7
Adding these equations:
The y terms have been eliminated.
3x + y = 9
Point out that the y terms have different signs in front of them. If we add the equations together they will be ‘eliminated’.
5x – y = 7 + 8x
= 16
x = 2
divide both sides by 8:

11. The elimination method
Adding the two equations eliminated the y terms and gave us a single equation in x.
Solving this equation gave us the solution x = 2.
3x + y = 9
5x – y = 7
To find the value of y when x = 2 substitute this value into one of the equations.
Substituting x = 2 into the first equation gives us:
3 × 2 + y = 9
6 + y = 9
y = 3
subtract 6 from both sides:

12. The elimination method
We can check whether x = 2 and y = 3 solves both:
3x + y = 9
5x – y = 7
by substituting them into the second equation.
5 × 2 – 3 = 7
10 – 3 = 7
This is true, so we have confirmed that
x = 2
y = 3
solves both equations.

13. The elimination method
Solve these equations:
3x + 7y = 22
3x + 4y = 10
Subtracting gives:
The x terms have been eliminated.
3x + 7y = 22
3x + 4y = 10 – 3y
= 12
divide both sides by 3:
y = 4
Substituting y = 4 into the first equation gives us,
The x terms have the same number in front of them, but this time the signs are the same. Subtracting the equations will eliminate the x terms.
3x + 7 × 4 = 22
3x + 28 = 22
subtract 28 from both sides:
3x = –6
x = –2
divide both sides by 3:

14. The elimination method
We can check whether x = –2 and y = 4 solves both,
3x + 7y = 22
3x + 4y = 10
by substituting them into the second equation.
3 × –2 + 7 × 4 = 22
– = 22
This is true and so,
x = –2
y = 4
solves both equations.

15. The elimination method
Sometimes we need to multiply one or both of the equations before we can eliminate one of the variables. For example,
4x – y = 29 1
3x + 2y = 19 2
We need to have the same number in front of either the x or the y before adding or subtracting the equations.
Call these equations 1 and 2 .
2 × 1 :
8x – 2y = 58 3
3x + 2y = 19 + :
11x
= 77
x = 7
divide both sides by 11:

16. The elimination method
To find the value of y when x = 7 substitute this value into one of the equations,
4x – y = 29 1
3x + 2y = 19 2
Substituting x = 7 into 1 gives,
4 × 7 – y = 29
28 – y = 29
subtract 28 from both sides:
–y = 1
y = –1
multiply both sides by –1:
Check by substituting x = 7 and y = –1 into 2 ,
3 × × –1 = 9
21 – 2 = 19

17. The elimination method
Solve:
2x – 5y = 25
3x + 4y = 3 1 2
Call these equations 1 and 2 .
3 × 1
6x – 15y = 75 3
2 × 2 –
6x + 8y = 6 4
3 – 4 ,
– 23y = 69
y = –3
divide both sides by –23:
This slide shows a more difficult example where both equations have to be multiplied to make the coefficients of x the same.
Discuss the alternative of multiplying equation 1 by 4 to make the coefficient of y –20. Multiplying the second equation by 5 would make the coefficient of y 20. These equations could then be added together to eliminate the terms containing y.
Substitute y = –3 in 1 ,
2x – 5 × –3 = 25
2x + 15 = 25
2x = 10
subtract 15 from both sides:
x = 5
divide both sides by 2:

18. Simultaneous equations
Substitution method

19. The substitution method
Two simultaneous equations can also be solved by substituting one equation into the other. For example,
y = 2x – 3
y = 2x – 3 1
2x + 3y = 23 2
Call these equations 1 and 2 .
Substitute equation 1 into equation 2 .
2x + 3(2x – 3) = 23
expand the brackets:
2x + 6x – 9 = 23
simplify:
8x – 9 = 23
add 9 to both sides:
8x = 32
x = 4
divide both sides by 8:

20. The substitution method
To find the value of y when x = 4 substitute this value into one of the equations,
y = 2x – 3 1
2x + 3y = 23 2
Substituting x = 4 into 1 gives
y = 2 × 4 – 3
y = 5
Check by substituting x = 4 and y = 5 into 2 ,
2 × × 5 = 23
= 23
This is true and so the solutions are correct.

21. The substitution method
How could the following pair of simultaneous equations be solved using substitution?
3x – y = 9 1
8x + 5y = 1 2
One of the equations needs to be arranged in the form x = … or y = … before it can be substituted into the other equation.
Call these equations 1 and 2 .
Rearrange equation 1 .
3x – y = 9
add y to both sides:
3x = 9 + y
subtract 9 from both sides:
3x – 9 = y
y = 3x – 9

22. The substitution method
3x – y = 9 1
8x + 5y = 1 2
Now substitute y = 3x – 9 into equation 2 .
8x + 5(3x – 9) = 1
expand the brackets:
8x + 15x – 45 = 1
simplify:
23x – 45 = 1
add 45 to both sides:
23x = 46
divide both sides by 23:
x = 2
Substitute x = 2 into equation 1 to find the value of y.
3 × 2 – y = 9
6 – y = 9
–y = 3
y = –3

23. The substitution method
3x – y = 9 1
8x + 5y = 1 2
Check the solutions x = 2 and y = –3 by substituting them into equation 2 .
8 × × –3 = 1
16 – 15 = 1
This is true and so the solutions are correct.
Solve these equations using the elimination method to see if you get the same solutions for x and y.