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}
The value in the first blank will be:
The value in the second blank will be:
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:
Add 9
Multiply by 6
And when we broke the expression back down to y, we applied the operations:
Divide by 6
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.
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
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}
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}
Using your answer from part (b), what value of m will make the equation \dfrac{m+11}{2}=10 true?
Example 3
Consider the equation 4(s-29)=4
Which of the following pairs of operations will make s the subject of the equation?
Apply these operations to the equation to find the solution.
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.