While we can solve simple equations by reversing a single operation or just by observing a solution, more complicated equations require a more methodical method – algebra.
Before we start using algebra to solve equations we should first understand why it works. As we saw when solving simple equations, algebra can be used to isolate the pronumeral in an equation by reversing the operation that had been applied to it.
We can use this same idea to solve harder equations by breaking them down into a series of one-step equations.
Consider the equation $4x+7=31$4x+7=31.
To solve this equation, we want to isolate $x$x.
We can work backwards to isolate $x$x by considering the term that it is a part of, $4x$4x.
Since $31$31 is equal to $7$7 more than $4x$4x, we know that $4x$4x must be $7$7 less than $31$31.
Writing this out as a mathematical sentence, we get $4x=31-7$4x=31−7 which we can evaluate to get:
$4x=24$4x=24
This is now a one-step equation which we can solve by isolating $x$x.
What number can we multiply by $4$4 to get $24$24?
We can remember from our four times tables that $4\times6=24$4×6=24, so the solution must be:
$x=6$x=6
We can also double check our answer by substituting this solution back into our starting equation to see if makes the equation true.
Replacing $x$x with $6$6 in the equation gives us:
$4\times6+7=31$4×6+7=31
If we evaluate both sides of the equation, we can see that they both have a value of $31$31, so the equation is true and we have confirmed that $x=6$x=6 is the solution.
When working backwards, we think of each step as a ‘fill in the blank’ type question.
While this method isn’t as applicable as using algebra, since it requires strong arithmetic to observe the value for each blank, it can be a faster method if our mental calculation skills are very good.
When solving more difficult equations, we want to use algebra to isolate the pronumeral by reversing the operations applied to it.
This is similar to working backwards, where we reversed the operations one at a time to isolate the pronumeral except, instead of observing the value for simpler expressions, we are directly reversing the operations.
Consider the equation we solved above, $4x+7=31$4x+7=31.
To solve this equation with algebra, we notice that the expression on the left-hand side of the equation has been built by applying the operations ‘multiply by $4$4’ and ‘add $7$7’ to the pronumeral.
To reverse these operations, we can apply the reverse operations ‘subtract $7$7’ and ‘divide by $4$4’ to both sides of the equation.
By applying these operations, we isolate $x$x and find the solution:
We knew that reversing the operations that were used to build the expression would isolate that pronumeral, but how did we know what order to apply them in?
One way to know what order to apply the reverse operations in is to consider the order in which the operations were applied to build the expression. This approach is explained in more detail when looking at breaking down expressions when using non-algebraic methods for solving two-step equations.
However, a faster way to find the solution is to apply the reverse operations according to the order of reverse operations.
What does this mean?
If we take another look at the equation $4x+7=31$4x+7=31, we can see that we should reverse the addition first since that will simplify the left-hand side to a single term, while reversing the multiplication first would result in expressions containing fractions which does not simplify the expression.
It is for this reason that we usually apply the reverse operations according to the steps:
We can see this in action with a more complicated equation, $7\left(y+6\right)-8=55$7(y+6)−8=55.
We can see that the operations used to build the expression on the left-hand side were ‘multiply by $7$7’, ‘add $6$6’ and ‘subtract $8$8’.
This means that the reverse operations we want to apply to the equation will be ‘divide by $7$7’, ‘subtract $6$6’ and ‘add $8$8’.
We can then apply these in the order of reverse operations:
Notice that we apply the reverse operation ‘subtract $6$6’ last since $y+6$y+6 is contained within a pair of brackets.
Applying these operations isolates $y$y on the left-hand side of the equation to give us the solution:
$y=3$y=3
When applying reverse operations to isolate a pronumeral, we apply them according to the order of reverse operations:
Notice that this order is the reverse of the usual order of operations.
Now that we know what operations to apply and in what order, we can solve equations quickly and accurately using algebra.
Solve the equation $\frac{p}{9}+4=11$p9+4=11.
Think: We can see from the equation that the expression on the left-hand side was built by applying the operations ‘divide by $9$9’ and ‘add $4$4’ to the pronumeral $p$p. We want to reverse these operations to isolate $p$p and find the solution.
Do: We know that the reverse operations we want to apply will be ‘multiply by $9$9’ and ‘subtract $4$4’. Using the order of reverse operations, we know that we want to apply them in the order:
Doing this gives us:
So the solution to the equation is:
$p=63$p=63
Solve the equation $18=3\left(m-7\right)$18=3(m−7).
Think: We can see from the equation that the expression on the right-hand side was built by applying the operations ‘multiply by $3$3’ and ‘subtract $7$7’ to the pronumeral $m$m. We want to reverse these operations to isolate $m$m and find the solution.
Do: We know that the reverse operations we want to apply will be ‘divide by $3$3’ and ‘add $7$7’. Using the order of reverse operations, we know that we want to apply them in the order:
Noting that the subtraction we want to reverse is inside a pair of brackets.
Doing this gives us:
So the solution to the equation is:
$m=13$m=13
Reflect: When solving an equation using algebra, we first identify which operations were used to build the expression containing the pronumeral, then we reverse those operations and apply them according to the order of reverse operations.
Notice that, in the second worked example, we wrote the solution as $m=13$m=13 even though the final step in our working out gave us $13=m$13=m. We can do this because the two equations are equivalent.
When writing an equation, swapping the left and right-hand sides of the equation does not change the solution.
Consider the equation $5\left(n-7\right)=35$5(n−7)=35.
If we want to make $n$n the subject of the equation, what is the first step we should take?
Add $7$7 to both sides to find the value of $5n$5n.
Add $7$7 to only the left-hand side of the equation.
Divide only the left-hand side of the equation by $5$5.
Divide both sides by $5$5 to find the value of $n-7$n−7.
What is the second step we should take to make $n$n the subject?
Multiply only the left-hand side of the equation by $5$5.
Multiply both sides by $5$5 to find the value of $5\left(n-7\right)$5(n−7).
Add $7$7 to both sides to find the value of $n$n.
Add $7$7 to only the left-hand side of the equation.
Apply the steps found in parts (a) and (b) to find the solution to the equation.
Enter each line of working as an equation.
Solve the equation $\frac{u+7}{2}=5$u+72=5.
Enter each line of working as an equation.
Solve the equation $6=\frac{t}{5}-3$6=t5−3.
