Lesson 43: Equivalent Division Problems

New Concept

You can create an equivalent division problem by multiplying or dividing the dividend and divisor by the same number. This simplifies the problem, just like reducing a fraction.

What’s next

Next, we'll review worked examples showing how to simplify division with whole numbers and fractions, and also solve for unknowns in equations.

Property

Two division problems that have the same quotient are called equivalent division problems. You can create one by multiplying or dividing both the dividend and the divisor by the same number.

Examples

Example: Simplify by dividing both numbers by 4: $1200 \div 16 \rightarrow (1200 \div 4) \div (16 \div 4) = 300 \div 4 = 75$.
Example: Simplify by doubling both numbers to remove fractions: $7\frac{1}{2} \div 2\frac{1}{2} \rightarrow (7\frac{1}{2} \cdot 2) \div (2\frac{1}{2} \cdot 2) = 15 \div 5 = 3$.
Example: Simplify by multiplying by 10 to remove decimals: $6.4 \div 0.8 \rightarrow (6.4 \cdot 10) \div (0.8 \cdot 10) = 64 \div 8 = 8$.

Explanation

Tired of dividing by tricky numbers? You can simplify a division problem by multiplying or dividing both the dividend and the divisor by the same number. It's just like reducing a fraction, but for division! This trick makes mental math a breeze by turning complex problems into much simpler ones.

Property

If you are unsure how to find the solution to a problem with fractions or decimals, try making up a similar, easier problem with whole numbers to help you determine how to find the answer.

Examples

Example: To solve $d - 5 = 3.2$, think of $d - 5 = 3$. You'd add, so $d = 3.2 + 5 = 8.2$.
Example: To solve $f + \frac{1}{5} = \frac{4}{5}$, think of $f + 1 = 4$. You'd subtract, so $f = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}$.
Example: To solve $0.5x = 10$, think of $2x = 10$. You'd divide, so $x = 10 \div 0.5 = 20$.

Explanation

Solving for an unknown variable with fractions or decimals can feel tricky. The secret is to think of a simpler, whole-number version of the problem first. Figure out the steps for the easy problem (like $x+2=5$), then apply the exact same logic to solve the original, more complex one.

Property

When two numbers are multiplied and the product is 1, the two factors must be reciprocals. To solve an equation like $\frac{a}{b}n = 1$, find the reciprocal of the known factor. The reciprocal of $\frac{a}{b}$ is $\frac{b}{a}$.

Examples

Example: In the equation $\frac{3}{8}n = 1$, the unknown $n$ must be the reciprocal of $\frac{3}{8}$, so $n = \frac{8}{3}$.
Example: To solve $\frac{7}{4}x = 1$, you simply flip the fraction $\frac{7}{4}$ to find the answer. Thus, $x = \frac{4}{7}$.
Example: For $5y=1$, think of 5 as $\frac{5}{1}$. The reciprocal is $\frac{1}{5}$, so $y = \frac{1}{5}$.

Explanation

When a fraction multiplied by an unknown number equals 1, you've found a shortcut! The unknown number must be the reciprocal, or the 'flipped' version, of the known fraction. This happens because multiplying a number by its reciprocal always cancels everything out perfectly, resulting in the number 1.