8.04 Two-step equations

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

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.

Building 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 variable, 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.

Practice question

Question 1

The following operations are performed on $s$s.

	$-$−$7$7		$\times$×$6$6
$s$s		$\editable{}$		$\editable{}$

1. The value in the first blank will be:

   $7-s$7−s

   A

   $-7s$−7s

   B

   $s-7$s−7

   C

   $s+7$s+7

   D

2. The value in the 2nd blank will be:

   $\frac{s-7}{6}$s−76​

   A

   $6+s-7$6+s−7

   B

   $6\left(s-7\right)$6(s−7)

   C

   $6s-7$6s−7

   D

Reveal SolutionWatch video

Breaking down expressions

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

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 variable:

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

1. Add $9$9
2. Multiply by $6$6

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

1. Divide by $6$6
2. 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 variable.

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.

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

The 'subtract $9$9' and 'add $9$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:

When operations do not cancel or simplify, the expression becomes very complicated.

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

Solving equations

Now that we have a way to isolate variables 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.

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.

Working backwards

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 variable 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.

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.

Using algebra to solve equations

When solving more difficult equations, we want to use algebra to isolate the variable 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 variable 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 variable.

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 variable, but how did we know what order to apply them in?

The order of reverse operations

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:

1. Start with addition and subtraction
2. Then multiplication and division
3. Repeat steps $1$1 and $2$2 for expressions inside brackets

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:

1. Add $8$8
2. Divide by $7$7
3. Subtract $6$6

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

Reversing operations to quickly solve equations

Now that we know what operations to apply and in what order, we can solve equations quickly and accurately using algebra.

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.

Practice questions

Question 4

Question 5

Question 6