Middle Years

Lesson

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 $x+4$`x`+4 can be broken down into 'an unknown number plus four', where the unknown number is $x$`x` and the operation is '$+4$+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.

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

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

If we apply these operations to some pronumeral, we can build an expression:

Do we get the same expression if we apply the operations in a different order?

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$6' operation was only applied to $y$`y` in the first expression, while it was applied to $y+9$`y`+9 in the second.

Applying operations to expressions

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$6' to the expression $y+9$`y`+9 gives us $6\left(y+9\right)$6(`y`+9). It will not give us $y+9\times6$`y`+9×6.

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

The following operations are performed on $s$`s`.

$-$−$7$7 | $\times$×$6$6 | |||

$s$s |
$\editable{}$ | $\editable{}$ |

The value in the first blank will be:

$7-s$7−

`s`A$-7s$−7

`s`B$s-7$

`s`−7C$s+7$

`s`+7DThe value in the 2nd blank will be:

$\frac{s-7}{6}$

`s`−76A$6+s-7$6+

`s`−7B$6\left(s-7\right)$6(

`s`−7)C$6s-7$6

`s`−7D

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

Returning to the expression $6\left(y+9\right)$6(`y`+9), we can see that applying the operations 'divide by $6$6' and 'subtract $9$9' will turn the expression back into an isolated pronumeral:

Notice that when we built the expression $6\left(y+9\right)$6(`y`+9), we applied the operations:

- Add $9$9
- Multiply by $6$6

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

- Divide by $6$6
- Subtract $9$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.

In the images below, we can see that the operation 'divide by $6$6' cancels out the multiplication in the expression, then 'subtract $9$9' cancels out the addition.

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:

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

Consider the expression $\frac{k}{9}-5$`k`9−5, built with the following operations:

What operations should we apply to break down the expression to isolate $k$`k`?

**Think**: What operations were applied to $k$`k` to build this expression and in what order? We want to apply the reverse of these operations to isolate $k$`k`.

**Do**: We can see from the image that the expression was built by applying the operations:

- Divide by $9$9
- Subtract $5$5

The reverse of these operations are 'multiply by $9$9' and 'add $5$5'. We want to apply these to the expression so that each pair of reverse operations will cancel out. This means that we will apply our reverse operations to the expression in the order:

- Add $5$5
- Multiply by $9$9

We can double check that these operations and their order are correct by applying them to the expression:

Since applying these operations isolated $k$`k`, these operations successfully broke down the expression.

**Reflect**: We can break down an expression by reversing the operations we applied when building that expression.

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.

Solve the equation $2\left(r-7\right)=4$2(`r`−7)=4.

**Think**: To solve the equation, we want to isolate $r$`r`. What operations can we apply to the expression $2\left(r-7\right)$2(`r`−7) so that we are left with just $r$`r`?

**Do**: We can see that the two operations applied to $r$`r` to build the expression were 'multiply by $2$2' and 'subtract $7$7'. But what order were they applied in?

Since there is a pair of brackets around $r-7$`r`−7 we know that the multiplication was applied after the subtraction.

To break the equation back down to just $r$`r`, we apply the reverse operations 'divide by $2$2' and 'add $7$7' in the order:

- Divide by $2$2
- Add $7$7

We can see that applying these operations to the left-hand side of the equation will isolate $r$`r`:

We can then solve the equation by applying these operations to the right-hand side of the equation as well:

Since we applied the same operations to both sides of the equation, the results after applying those operations will be equal. This means that the solution to the equation is:

$r=9$`r`=9

**Reflect**: 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.

Consider the equation $\frac{m+11}{2}=10$`m`+112=10

Which pair of operations will make $m$

`m`the subject of the equation?Step $1$1 Step $2$2 $\frac{m+11}{2}$ `m`+112$m+11$ `m`+11$m$ `m`Divide by $2$2, then add $11$11

AMultiply by $11$11, then subtract $2$2

BMultiply by $2$2, then subtract $11$11

CSubtract $11$11, then multiply by $2$2

DApply these operations to the right-hand side of the equation as well.

Step $1$1 Step $2$2 $10$10 $\editable{}$ $\editable{}$ Using your answer from part (b), what value of $m$

`m`will make the equation $\frac{m+11}{2}=10$`m`+112=10 true?

Consider the equation $4\left(s-29\right)=4$4(`s`−29)=4.

Which of the following pairs of operations will make $s$

`s`the subject of the equation?Add $29$29, then divide by $4$4

ADivide by $4$4, then add $29$29

BAdd $4$4, the divide by $29$29

CSubtract $29$29, then multiply by $4$4

DApply these operations to the equation to find the solution.