1.02 Solving linear equations

Lesson

An equation is a mathematical statement of equality. It uses an equals sign to show that the expression on the left-hand side of the statement is equal to the expression on the right-hand side. The equals sign means that the equation is balanced, like a set of scales.

The diagram above represents the equation $3x=x+4$3x=x+4. Only one value of $x$x will make this equation true and keep the scales balanced.

If we subtracted an $x$x from only the right-hand side of the equation, the scales would become unbalanced. To keep the scales balanced we would need to subtract an $x$x from the left-hand side as well.

Keep the equation balanced

Whenever we add, subtract, multiply, divide or apply any other operation to one side of an equation, we must do exactly the same to the other side.

To solve an equation, our aim is to find the value of the variable (in this case, $x$x). To do so, we rearrange the equation to get the variable by itself on one side of the equation, and its value on the other side. This is also called isolating the variable. In the example above, the solution to the equation $3x=x+4$3x=x+4 would be written as $x=2$x=2.

For any equation, we can check the solution by substituting it back into the original equation. If the solution is correct, the left-hand side (LHS) of the equation will be equal to the right-hand side (RHS).

Worked example

Example 1

Check if $x=2$x=2 is a solution to the equation $3x=x+4$3x=x+4:

Solution

Substitute $x=2$x=2 into each side of the equation:

 LHS $=$= $3x$3x $=$= $3\times2$3×2 $=$= $6$6

 RHS $=$= $x+4$x+4 $=$= $2+4$2+4 $=$= $6$6

Since the LHS$=$=RHS, $x=2$x=2 is a solution to the equation. It is also the only possible solution.

As we shall see in the following examples, equations can be solved by a series of steps.

One step equations

The simplest equations can be solved in one step. In the examples below, we think about the opposite (or inverse) operation that can be performed to both sides of the equation in order to find the solution.

This same thinking can be applied to each step of every equation that we solve.

Worked example

example 2

Solve $x-1.39=8.67$x1.39=8.67.

Think: In this equation $1.39$1.39 is subtracted from $x$x, so we do the opposite and add $1.39$1.39 to both sides.

Do:

 $x-1.39$x−1.39 $=$= $8.67$8.67 $x-1.39+1.39$x−1.39+1.39 $=$= $8.67+1.39$8.67+1.39 (Add $1.39$1.39 to both sides) $x$x $=$= $10.06$10.06

Practice questions

Question 1

Find the value of $t$t when $10t=60$10t=60.

Question 2

Solve: $-5=\frac{x}{5}$5=x5

Two step equations

With equations that are solved using more than one step, we need to think carefully about the order in which we perform the operations to both sides.

Often there are several ways a multi-step equation can be solved. To master the art of equation solving, it is good practice to try different ways of solving the same problem. Regardless of our approach, we should always get the same solution. If we don't, it's a good indicator that we have made an error somewhere in our working.

Worked example

example 3

Solve $-5\left(y-1\right)=25$5(y1)=25

Think: In this equation, the brackets tell us that the quantity $y-1$y1 has been multiplied by $-5$5, so our first step will be to divide both sides by $-5$5.

Do:

 $-5\left(y-1\right)$−5(y−1) $=$= $25$25 $\frac{-5\left(y-1\right)}{-5}$−5(y−1)−5​ $=$= $\frac{25}{-5}$25−5​ (Divide both sides by $-5$−5) $y-1$y−1 $=$= $-5$−5 $y-1+1$y−1+1 $=$= $-5+1$−5+1 (Add $1$1 to both sides) $y$y $=$= $-4$−4

Reflect: An alternative approach to solving this question would be to expand the brackets first, taking care to multiply the negative numbers correctly.

Practice questions

Question 3

Find the solution for the following equation: $\frac{x+9}{7}=4$x+97=4

Question 4

Solve the following equation: $\frac{x}{2}-3=1$x23=1

Equations with three or more steps

Equations involving three or more steps may include variables on both sides of the equation. As well, there may be one or more sets of brackets that need to be expanded, and the equation could include one or more fractions.

Just as we have done previously, we continue performing the same inverse operation to both sides, one operation at a time. As always, our aim is to work our way through a series of steps to isolate the variable and find the solution.

Worked example

example 4

Solve $5x+9=x+21$5x+9=x+21

Think: In this equation there are variables on both sides. Our first step is to use an inverse operation to remove the variable from one side of the equation. We do this by subtracting $x$x from both sides. We can then subtract $9$9 from both sides to get the variables on one side and the numbers on the other side.

Do:

 $5x+9$5x+9 $=$= $x+21$x+21 $5x+9-x$5x+9−x $=$= $x+21-x$x+21−x (Subtract $x$x from both sides) $4x+9$4x+9 $=$= $21$21 $4x+9-9$4x+9−9 $=$= $21-9$21−9 (Subtract $9$9 from both sides) $4x$4x $=$= $12$12 $\frac{4x}{4}$4x4​ $=$= $\frac{12}{4}$124​ (Divide both sides by $4$4) $x$x $=$= $3$3

Practice questions

Question 5

Solve the following equation: $x-2\left(x+3\right)=-1$x2(x+3)=1

Question 7

Solve the following equation: $\frac{5x}{3}-3=\frac{3x}{8}$5x33=3x8

Linear vs non-linear equations

In this chapter we cover linear equations. In year 12, non-linear equations will be covered. The difference between them can be summarised as follows:

Linear equations:

• The only power of any variable appearing in the equation is $1$1
• The graph of the equation will be a straight line  Examples: $y=3x-2$y=3x−2 $x=4$x=4 $4.5x+2.3y=0$4.5x+2.3y=0 $p=\frac{7q}{3}-10$p=7q3​−10 $S=V_o-Dn$S=Vo​−Dn

Non-linear equations:

• The power of at least one variable in the equation is not $1$1
• The graph of the equation will be a curve  Examples: $y=x^2+1$y=x2+1 $a^2+b^2=c^2$a2+b2=c2 $A=P\left(1+R\right)^n$A=P(1+R)n $y=2.5p^3-\frac{9}{2}q$y=2.5p3−92​q $D=\sqrt{x}$D=√x

Outcomes

MS11-1

uses algebraic and graphical techniques to compare alternative solutions to contextual problems