NSW Year 8 - 2020 Edition
5.03 Distributing to solve
Lesson

Algebra is useful when solving equations because it allows for multiple different methods. Depending on the question and the numbers involved, we can choose which is the most appropriate approach, and hopefully save some time.

One of these methods is distributing to solve, and it can save a lot of work when used appropriately.

### Distributing in an equation

Using the distributive law in an equation with algebraic terms follows the same rules that we use to distribute a numeric expression.

The distributive law

The distributive law allows us to expand multiplication applied to an expression inside a pair of brackets according to the rule:

$A\left(B+C\right)=AB+AC$A(B+C)=AB+AC

Using the distributive law when solving an equation can help us reduce the equation to only the four basic operations, since we can use it to expand brackets.

#### Worked Example

Solve the equation: $7\left(r-2\right)+5=37$7(r2)+5=37

Think: We can start by expanding the brackets using the distributive law, then reverse the operations as usual.

Do: If we follow these steps, we get:

 $7\left(r-2\right)+5$7(r−2)+5 $=$= $37$37 $7\times r+7\times\left(-2\right)+5$7×r+7×(−2)+5 $=$= $37$37 Use the distributive law to expand the brackets $7r-14+5$7r−14+5 $=$= $37$37 Simplify the products $7r-9$7r−9 $=$= $37$37 Collect the constant terms $7r$7r $=$= $46$46 Reverse the subtraction $r$r $=$= $\frac{46}{7}$467​ Reverse the multiplication

As we can see, by expanding the brackets first, the order in which we apply reverse operations becomes more obvious.

### Why are we distributing?

Good question.

Let's have a look at what happens if we solve the same equation $7\left(r-2\right)+5=37$7(r2)+5=37 using only reverse operations.
If we do this, we get the working out:

 $7\left(r-2\right)+5$7(r−2)+5 $=$= $37$37 $7\left(r-2\right)$7(r−2) $=$= $32$32 Reverse the addition $r-2$r−2 $=$= $\frac{32}{7}$327​ Reverse the multiplication $r$r $=$= $\frac{32}{7}+2$327​+2 Reverse the subtraction $r$r $=$= $\frac{32}{7}+\frac{14}{7}$327​+147​ Convert $2$2 to a fraction with denominator $7$7 $r$r $=$= $\frac{46}{7}$467​ Perform the addition

We find that it takes more steps to solve the equation using only reverse operations and it involves a lot more fractions.

As such, while both methods for solving the equation work just fine, using the distributive law can save us some time.

#### Practice questions

##### Question 1

Solve this equation for $x$x by first expanding the brackets:
$-6\left(4x+5\right)=-143$6(4x+5)=143

Enter each line of working as an equation.

##### Question 2

Solve this equation for $x$x by first expanding the brackets and collecting like terms:
$4\left(4x-6\right)-3x+8=36$4(4x6)3x+8=36

Enter each line of working as an equation.

### Multiple pairs of brackets

The other reason to use the distributive law when solving equations is that there are some problems we simply can't solve using just reverse operations. Consider the equation:

$3\left(p-2\right)+4\left(2p+8\right)=81$3(p2)+4(2p+8)=81

Because we can't reverse the multiplications of $3$3 and $4$4 at the same time, it's actually impossible to solve this equation using only reverse operations. But we can solve it using the distributive law.

If we use the distributive law to expand all the brackets first, we will only have multiplication, addition and subtraction left. We can then gather any like terms and apply reverse operations, like so:

 $3\left(p-2\right)+4\left(2p+8\right)$3(p−2)+4(2p+8) $=$= $81$81 $3\times p+3\times\left(-2\right)+4\times2p+4\times8$3×p+3×(−2)+4×2p+4×8 $=$= $81$81 Use the distributive law to expand the brackets $3p-6+8p+32$3p−6+8p+32 $=$= $81$81 Simplify the products $11p+26$11p+26 $=$= $81$81 Gather any like terms together $11p$11p $=$= $55$55 Reverse the addition $p$p $=$= $5$5 Reverse the multiplication

#### Practice question

##### Question 3

Solve this equation for $x$x by first expanding the brackets:
$-3\left(2x-6\right)+5\left(4x-5\right)=21$3(2x6)+5(4x5)=21

Enter each line of working as an equation.

### Outcomes

#### MA4-10NA

uses algebraic techniques to solve simple linear and quadratic equations