Just like we saw with addition and subtraction, multiplication and division are also inverse operations. For example, multiplying a number by two is the opposite of dividing it by two.

Looking at a balance scale model again, we can see how we can multiply or divide by the same number on both sides of an equation to keep it balanced.

**Multiplication property of equality**: Multiplying each side of an equation by the same number produces an equivalent equation. Example:

\begin{aligned}&\text{If } &x &= 3 \\ &\text{Then } &x\cdot{2} &= 3\cdot{2}\end{aligned}

We can visualize this with a scale.

The scale tells us that x=3. We can double the values on both sides and the scale will still be balanced.

We multiplied the tiles on both sides of the scale by 2 and it is still balanced. We could have also multiplied both sides of the original scale by 3 to create 3 equal groups on each side of the balance and it would still be balanced.

If we treat each side of the scale like a group that aligns with the other side, we can keep applying the multiplication property of equality.

**Division property of equality**: Dividing each side of an equation by the same number produces an equivalent equation. Example:

\begin{aligned}&\text{If } &4x &= 8 \\ &\text{Then } &\dfrac{4x}{4} &= \dfrac{8}{4}\end{aligned}

We can visualize this with a scale.

Notice that the groups on the left and right align where every 2x aligns with 4 unit tiles. We can divide both sides by 2 (removing half of the tiles) and the scale will still be balanced.

We can also divide both sides of the original scale by 4, since we can see there are 4 groupings where every x tile is equal to 2 unit tiles.

By dividing both sides of the scale by the same amount, we ensure that the left and right side of the scale are still the same size and weight to keep the scale balanced.

When we use these properties with specific numbers to eliminate them on one side of the equation, we are applying the multiplicative inverse and identity properties.

For example, with the equation 8x=30 we want to isolate x which requires getting rid of the coefficient 8. To do this we can use the inverse operation of dividing by 8.

\dfrac{8x}{8}=\dfrac{30}{8}

This can also be written as multiplication: \dfrac{1}{8} \cdot 8x=\dfrac{1}{8}\cdot 30

This allows us to see the multiplicative inverse in action: 1\cdot x=\dfrac{1}{8}\cdot 30

Now applying the multiplicative identity: x=\dfrac{1}{8}\cdot 30

Then we can simplify the right side of the equation: x=\dfrac{30}{8}=\dfrac{15}{4}

We don't always write out all of these steps, but it is still important to know what is happening algebraically.

Scale 1 is balanced.

Scale 1:

Scale 2:

Which of the following options could go in place of the question mark to balance scale 2?

A

B

C

D

Worked Solution

Solve 3x=18

Worked Solution

Solve: \dfrac{x}{8}=6

Worked Solution

5 is a solution to the equation 8x=40.

a

Verify using substitution.

Worked Solution

b

Verify using a model.

Worked Solution

At a beach fruit stand, fresh squeezed juice is sold at \$ 6 per quart. You and your friends spent a total of \$ 84 on quarts of juice. Write an equation that shows the relationship between the total cost and the number of quarts of juice purchased.

Worked Solution

Write a situation that could represent the equation \dfrac{x}{4}=52.

Worked Solution

Idea summary

Multiplication property of equality | \text{If } a=b \text{ then } a\cdot c=b \cdot c |
---|---|

Division property of equality | \text{If } a=b, \text{ and } c \neq 0, \text{ then } \dfrac{a}{c}=\dfrac{b}{c} |

Inverse property of multiplication | a \cdot \dfrac{1}{a}=1 \text{ and } \dfrac{1}{a} \cdot a=1 |

Identity property of multiplication | a \cdot 1=a \text{ and } 1 \cdot a = a |

Substitution property | \text{If } {a=b} \text{, then } b \text{ can be substituted for } a \text{ in any expression,}\\\text{equation, or inequality.} |