Simultaneous Linear Equations (I): In One or Two Variables – JSS3 Mathematics Lesson Note (Second Term)

Simultaneous linear equations are two or more linear equations that are solved at the same time to find the values of the unknowns that satisfy all of them.

Behavioural objectives

By the end of the lesson, students should be able to:

1. explain what simultaneous linear equations are;
2. name the two algebraic methods of solving simultaneous equations;
3. list the steps of the substitution method;
4. solve simultaneous linear equations by the substitution method and check their answers.

What are simultaneous linear equations?

Simultaneous linear equations are two linear equations in two unknowns that share the same solution.

• Linear equation: an equation in which the unknowns have power 1, such as 3x + y = 4.
• Unknown (variable): a letter that stands for a number we do not yet know, such as x, y, m or n.
• One unknown: a linear equation in one unknown, such as 2x + 1 = 7, has one solution (x = 3).
• Two unknowns: one equation such as x + y = 5 has many solutions, so a second equation is needed to fix x and y.
• Simultaneous: the two equations are true together, so the same x and y must fit both.

Example of a pair of simultaneous equations:

4x + y = 8 (1)
3x – y = 6 (2)

What is the solution of simultaneous equations?

The solution is the pair of values of the unknowns that makes both equations true.

• Form of answer: a value for each unknown, such as x = 1 and y = 1.
• Test of a solution: substitute both values into each original equation; both must balance.

What are the methods of solving simultaneous equations?

Simultaneous equations can be solved graphically or algebraically, and there are two algebraic methods: substitution and elimination.

• Graphical method: draw both lines on the same axes and read the point where they cross.
• Substitution method: make one unknown the subject of one equation, then put it into the other equation.
• Elimination method: add or subtract the equations to remove one unknown.

This lesson covers the substitution method.

How do you use the substitution method?

In the substitution method, you make one variable the subject of one equation, then substitute it into the other equation.

1. Label the equations (1) and (2).
2. Re-arrange one equation so that one variable is the subject of the formula.
3. Substitute this expression into the other equation.
4. Solve the resulting equation to obtain the value of one variable.
5. Find the other variable by substituting your answer into the re-arranged equation.
6. Check the solution by substituting both answers into the original equations.

Which variable should you make the subject?

• Choose a variable with coefficient 1: in 3x + y = 4, y has coefficient 1, so y = 4 – 3x has no fractions.
• Already a subject: if an equation is given as y = 5x + 2, substitute it straight away.
• No coefficient 1: any variable can be used, but the working may contain fractions, as in 4m – 3n = 0.

Worked examples

Example 1

Solve the following simultaneous equations by the substitution method.

(i) y = 5x + 2
x + 2y = 15

(ii) 2x + 3y = 5
3x + y = 4

(iii) 4m – 3n = 0
m + 2n = 3

(iv) x + 6y = -2
3x + 2y = 10

Solution to (ii)

1. Label the equations.
   2x + 3y = 5 (1)
   3x + y = 4 (2)
2. Make y the subject of equation (2).
   y = 4 – 3x (3)
3. Substitute y = 4 – 3x into equation (1).
   2x + 3(4 – 3x) = 5
4. Open the brackets and solve for x.
   2x + 12 – 9x = 5
   12 – 7x = 5
   12 – 5 = 7x
   7 = 7x
   x = 1
5. Substitute x = 1 into equation (3).
   y = 4 – 3(1)
   y = 4 – 3
   y = 1

Hence x = 1, y = 1.

Check:

• Equation (1): 2(1) + 3(1) = 2 + 3 = 5
• Equation (2): 3(1) + 1 = 3 + 1 = 4

Solution to (iii)

1. Label the equations.
   4m – 3n = 0 (1)
   m + 2n = 3 (2)
2. Make m the subject of equation (2).
   m = 3 – 2n (3)
3. Substitute m = 3 – 2n into equation (1).
   4(3 – 2n) – 3n = 0
4. Open the brackets and solve for n.
   12 – 8n – 3n = 0
   12 – 11n = 0
   12 = 11n
   n = 12/11
5. Substitute n = 12/11 into equation (3).
   m = 3 – 2(12/11)
   m = 3 – 24/11
   m = 33/11 – 24/11
   m = 9/11

Hence m = 9/11, n = 12/11.

Check:

• Equation (1): 4(9/11) – 3(12/11) = 36/11 – 36/11 = 0
• Equation (2): 9/11 + 2(12/11) = 9/11 + 24/11 = 33/11 = 3

Example 2

Solve x + y = 7 and x – y = 1 by substitution.

1. Label the equations.
   x + y = 7 (1)
   x – y = 1 (2)
2. Make x the subject of equation (1).
   x = 7 – y (3)
3. Substitute into equation (2).
   (7 – y) – y = 1
4. Solve for y.
   7 – 2y = 1
   6 = 2y
   y = 3
5. Substitute y = 3 into equation (3).
   x = 7 – 3 = 4

Hence x = 4, y = 3.

Check: 4 + 3 = 7 and 4 – 3 = 1.

What mistakes should you avoid?

• Sign errors: when you move a term across the equals sign, change its sign.
• Brackets: multiply every term inside the bracket, as in 3(4 – 3x) = 12 – 9x.
• Wrong equation: never substitute back into the same equation you re-arranged; use the other one.
• Stopping early: find both unknowns, then check in both original equations.

Evaluation

Solve the following simultaneously using the substitution method.

1. x + 6y = -2
   3x + 2y = 10
2. -2 = 5x – y
   15 = x + 2y
3. 4x + 7y = 20
   3x + y = -2

Assignment

Solve the simultaneous equations in Exercise 16.3, page 149, No 11 to 1, by the substitution method.