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9.04 Methods for solving two-step equations

Lesson

Introduction

When working with expressions that involve a pronumeral, we can see how they were constructed by observing the order in which operations were applied to that pronumeral.

We have seen this kind of thinking with simple equations already. For example, the expression x+4 can be broken down into 'an unknown number plus four', where the unknown number is x and the operation is '+4'.

In this lesson, we will look at how we can apply this concept to more complicated expressions and then use that to solve equations with those expressions.

Build expressions

Before we start breaking expressions down, we should first understand how they are built up.

Consider the operations 'multiply by 6' and 'add 9'.

If we apply these operations to some pronumeral, we can build an expression:\begin{array}{ccccc} & {\times6} & & {+9} & \\ y & \longrightarrow & 6y &\longrightarrow & 6y+9 \end{array}

Do we get the same expression if we apply the operations in a different order?\begin{array}{ccccc} & {+9} & & {\times6} & \\ y & \longrightarrow & y+9 & \longrightarrow & 6(y+9) \end{array}

No.

Although these two expressions are built from the same operations, they are different because the operations were applied in a different order.

Notice that the 'multiply by 6' operation was only applied to y in the first expression, while it was applied to y+9 in the second.

Examples

Example 1

The following operations are performed on s.\begin{array}{ccccc} & {-7} & & {\times 6} & \\ s & \longrightarrow & ⬚ & \longrightarrow & ⬚ \end{array}

a

The value in the first blank will be:

A
7-s
B
-7s
C
s-7
D
s+7
Worked Solution
Create a strategy

Apply the first operation to s.

Apply the idea

The first operation is to subtract 7 from s. So we will have s-7.

So option C is the correct answer.

b

The value in the second blank will be:

A
\dfrac{s-7}{6}
B
6+s-7
C
6(s-7)
D
6s-7
Worked Solution
Create a strategy

Apply the second operation to the resulting expression in part (a).

Apply the idea

In part (a) we have, s-7. So the second operation is to multiply 6 with s-7.

So we have 6(s-7).

So option C is the correct answer.

Idea summary

When we apply operations to expressions, we need to apply that operation to the whole expression. We can represent this by placing a pair of brackets around the expression before applying an operation to it.

In other words, applying the operation 'multiply by 6' to the expression y+9 gives us 6(y+9). It will not give us y+9\times6.

We could also re-write 6(y+9) as 6y+6\times9=6y+54, by expanding the brackets.

Break down expressions to solve equations

When breaking down expressions, our aim is to apply operations that will turn the expression into an isolated pronumeral.

Returning to the expression 6(y+9), we can see that applying the operations 'divide by 6' and 'subtract 9' will turn the expression back into an isolated pronumeral:\begin{array}{ccccc} & {\div6} & & {-9} & \\ 6(y+9) & \longrightarrow & y+9 & \longrightarrow & y \end{array}

Notice that when we built the expression 6(y+9), we applied the operations:

  1. Add 9

  2. Multiply by 6

And when we broke the expression back down to y, we applied the operations:

  1. Divide by 6

  2. Subtract 9

We can see that, when breaking down an expression, we can reverse the operations used to build the equation (and the order in which they are applied) to cancel out the operations applied to the pronumeral.

The 'divide by 6' and 'multiply by 6' operations cancel each other out.

The image shows the effect of opposite operations. Ask your teacher for more information.
The image shows the effect of opposite operations. Ask your teacher for more information.

The 'subtract 9' and 'add 9' operations cancel each other out.

Notice again that the order in which we apply the operations is important. We can see in the image that the operations that wil cancel out must come one after the other. If this is not the case, we will get something like this:\begin{array}{ccccc} & {+9} & & {\times6} & & {-9} & & {\div6}\\ y & \longrightarrow & y+9 & \longrightarrow & 6(y+9) & \longrightarrow & 6(y+9)-9 & \longrightarrow & \dfrac{6(y+9)-9}{6} \end{array}

If we apply our reverse operations in the wrong order, our expression will get even more complicated.

Now that we have a way to isolate pronumerals in an expression, we can apply this method to solve equations that involve more than one step. We can do this since, as long as we apply an operation to both sides of the equation, expressions on either side of the equals sign will be equal in value.

Examples

Example 2

Consider the equation \dfrac{m+11}{2}=10

a

Which pair of operations will make m the subject of the equation?\begin{array}{ccccc} & \text{Step } 1 & & \text{Step } 2 & \\ \dfrac{m+11}{2} & \longrightarrow & m+11 & \longrightarrow & m \end{array}

A
Divide by 2, then add 11.
B
Multiply by 11, then subtract 2.
C
Multiply by 2, then subtract 11.
D
Subtract 11, then multiply by 2.
Worked Solution
Apply the idea

Step 1 is the operation that can cancel out the division of 2 which is multiply by 2:\begin{array}{ccccc} & {\times2} & & \text{Step } 2 & \\ \dfrac{m+11}{2} & \longrightarrow & m+11 & \longrightarrow & m \end{array}

Step 2 is the operation that can cancel out the addition of 11 which is subtraction of 11:\begin{array}{ccccc} & {\times2} & & {-11} & \\ \dfrac{m+11}{2} & \to & m+11 & \to & m \end{array}

So option C is the correct answer.

b

Apply these operations to the right-hand side of the equation as well.\begin{array}{ccccc} & \text{Step } 1 & & \text{Step } 2 & \\ 10 & \longrightarrow & ⬚ & \longrightarrow & ⬚ \end{array}

Worked Solution
Create a strategy

Apply the operations: multiply by 2, then subtract 11.

Apply the idea

\begin{array}{ccccc} & \times2 & & -9 & \\ 10 & \longrightarrow & 20 & \longrightarrow & 11 \end{array}

c

Using your answer from part (b), what value of m will make the equation \dfrac{m+11}{2}=10 true?

Worked Solution
Create a strategy

Combine the results in parts (a) and (b).

Apply the idea

In parts (a) and (b), we applied the same pair of operations to both the left-hand side and the right-hand side. We have:\begin{array}{ccccc} & {\times2} & & {-11} & \\ \dfrac{m+11}{2} & \longrightarrow & m+11 &\longrightarrow & m \\ \\ 10 & \longrightarrow & 20 & \longrightarrow & 11 \end{array}

If we apply the same operations to two equal expressions, then the results of those operations also equal.

So the value of m that will make the equation \dfrac{m+11}{2}=10 true is m=11.

Example 3

Consider the equation 4(s-29)=4

a

Which of the following pairs of operations will make s the subject of the equation?

A
Add 29, then divide by 4.
B
Divide by 4, then add 29.
C
Add 4, then divide by 29.
D
Subtract 29, then multiply by 4.
Worked Solution
Create a strategy

Find the reverse operations in building the expression on the left-hand side of the equation in the reverse order.

Apply the idea

The expression on the left-hand side of the equation is 4(s-29). To build this expression the following operations happened:

  1. 29 was subtracted

  2. Multiply by 4

So the reverse operations in the reverse order are:

  1. Divide by 4

  2. Add 29

So option B is the correct answer.

b

Apply these operations to the equation to find the solution.

Worked Solution
Create a strategy

Apply the operations: divide by 4, then add 29 to both sides of the equation.

Apply the idea
\displaystyle 4(s-29)\displaystyle =\displaystyle 4Write the equation
\displaystyle \dfrac{4(s-29)}{4}\displaystyle =\displaystyle \dfrac{4}{4}Divide both sides by 4
\displaystyle s-29\displaystyle =\displaystyle 1Evaluate
\displaystyle s-29+29\displaystyle =\displaystyle 1+29Add 29 to both sides
\displaystyle s\displaystyle =\displaystyle 30Evaluate
Idea summary

We can break down an expression to just the pronumeral by reversing the operations that were applied to build it. We can use this to solve equations by applying the reverse operations to both sides of the equation.

Outcomes

MA4-10NA

uses algebraic techniques to solve simple linear and quadratic equations

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