Undoing Multiplication and Division to Solve Equations | Grade 6 Math | LevelKite

You already know how to solve an equation such as $x + 7 = 12$ . The letter $x$ is a variable: it stands for a number you do not know yet. You subtract $7$ from both sides, because subtraction undoes addition. That leaves $x = 5$ . This value is the solution, the number that makes the equation true.

Now you will solve equations that multiply or divide the unknown number, such as $6x = 54$ and $x \div 8 = 7$ . In algebra, $6x$ means $6 \times x$ . The plan stays the same. Find the operation that undoes what the equation does to $x$ , then use it on both sides.

Two operations are inverse operations when each one undoes the other. Multiplying by a number and dividing by that same number are inverse operations. The number must be nonzero, which means it is not $0$ .

Equal groups show why. Let $n$ stand for any number. Multiplying $n$ by $5$ makes $5$ equal groups of $n$ . Dividing that total by $5$ splits it into $5$ equal groups again, and each group is $n$ . You are back where you started.

The other order works too. Dividing $n$ by $4$ gives the size of one of $4$ equal parts of $n$ . Multiplying that part by $4$ joins the $4$ parts, and together they make $n$ again.

The table shows both orders with whole numbers, fractions and decimals.

An operation followed by its inverse returns the start number

Start number	Operation	Result	Inverse operation	Back to the start
$3$	multiply by $5$	$3 \times 5 = 15$	divide by $5$	$15 \div 5 = 3$
$0.6$	multiply by $5$	$0.6 \times 5 = 3$	divide by $5$	$3 \div 5 = 0.6$
$20$	divide by $4$	$20 \div 4 = 5$	multiply by $4$	$5 \times 4 = 20$
$3$	divide by $4$	$3 \div 4 = \frac{3}{4}$	multiply by $4$	$\frac{3}{4} \times 4 = 3$
$12$	multiply by $0.5$	$12 \times 0.5 = 6$	divide by $0.5$	$6 \div 0.5 = 12$

Five rows cannot prove that this happens for every number. Examples that match can suggest a rule, but they cannot prove it. The meaning of division proves it.

In $15 \div 5 = 3$ , the dividend is $15$ , the divisor is $5$ and the quotient is $3$ . Division asks a question about multiplication: what number times the divisor gives the dividend? So when the divisor is nonzero, the exact quotient times the divisor equals the dividend. This holds for whole numbers, fractions and decimals alike. That is why multiplying by a nonzero number and dividing by the same number always undo each other, in either order.

The divisor can never be $0$ . Multiplying by $0$ turns every number into $0$ , so you cannot tell which number you started with. No operation can undo that, so you can never divide by $0$ .

In symbols, let $n$ be any number and let $a$ be any nonzero number. The two statements below are true for every $n$ and every nonzero $a$ . In each statement, the two sides are equivalent expressions. That means they have the same value for every allowed value of the variables.

(n \times a) \div a = n \qquad\qquad (n \div a) \times a = n \qquad (a \neq 0)

Now connect this to equations. An equation says that its two sides are equal amounts. Suppose you multiply both sides by the same nonzero number, or divide both sides by the same nonzero number. You have done the same thing to equal amounts, so the results are still equal. The new equation has the same solution as the old one.

This gives you a way to solve. Choose the inverse operation that undoes what the equation does to the variable. Apply it to both sides. The variable is then alone on one side, and the other side shows its value.

Key concept

Undo the operation on both sides

To solve a one-step multiplication or division equation, look at what the equation does to the variable. Then apply the inverse operation to both sides.

If the equation multiplies the variable by a nonzero number, divide both sides by that number.
If the equation divides the variable by a nonzero number, multiply both sides by that number.

The same operation on both sides keeps them equal, and the inverse operation leaves the variable alone. Then check: an exact solution makes the two sides of the original equation equal.

In the formula, $x$ is the unknown number, and $a$ and $b$ are known numbers with $a \neq 0$ .

\text{For } a \neq 0:\quad ax = b \text{ gives } x = b \div a, \qquad x \div a = b \text{ gives } x = b \times a

Worked example

A multiplication equation with whole numbers

Solve $6x = 54$ . Here $x$ stands for an unknown number, and $6x$ means $6 \times x$ .

Find what the equation does to x. It multiplies x by 6, so 6 equal groups of x make 54.
$6 \times x = 54$
Division by 6 undoes multiplication by 6. You may divide by 6 because 6 is not zero. Divide both sides by 6 so that the two sides stay equal.
$6x \div 6 = 54 \div 6$
Simplify each side. On the left, dividing 6 equal groups of x by 6 leaves one group, which is x. On the right, 54 divided by 6 is 9, because 6 times 9 is 54.
$x = 9$
Check by putting 9 in place of x in the original equation. The quotient 9 is exact, so 6 times 9 must be exactly 54.
$6 \times 9 = 54$

Answer

The solution is $x = 9$ , because $6 \times 9 = 54$ .

Worked example

A division equation with whole numbers

Solve $\frac{x}{8} = 7$ . Here $x$ stands for an unknown number, and $\frac{x}{8}$ means $x \div 8$ .

Find what the equation does to x. It divides x by 8, a nonzero number, and the quotient is 7.
$x \div 8 = 7$
Multiplication by 8 undoes division by 8. Multiply both sides by 8 so that the two sides stay equal.
$\frac{x}{8} \times 8 = 7 \times 8$
Simplify each side. On the left, the exact quotient times the divisor 8 gives back the dividend, which is x. On the right, 7 times 8 is 56.
$x = 56$
Check by putting 56 in place of x in the original equation. The division is exact, with nothing left over.
$\frac{56}{8} = 56 \div 8 = 7$

Answer

The solution is $x = 56$ , because $56 \div 8 = 7$ .

The same two moves work when the numbers are fractions or decimals. The reasoning does not change. Multiplying and dividing by the same nonzero number undo each other, whether that number is a whole number, a fraction or a decimal. Only the arithmetic takes more care.

One reminder will help. To divide by a nonzero fraction, multiply by its reciprocal. The reciprocal swaps the top number (the numerator) and the bottom number (the denominator), so the reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$ .

Worked example

A fraction multiplies the variable

Solve $\frac{3}{4}x = 6$ , where $x$ stands for an unknown number.

The equation multiplies x by 3/4. Division by 3/4 undoes that, and 3/4 is not zero. Divide both sides by 3/4 so that the sides stay equal.
$\frac{3}{4}x \div \frac{3}{4} = 6 \div \frac{3}{4}$
The left side is now x alone. On the right, divide by 3/4 by multiplying by its reciprocal, 4/3.
$x = 6 \div \frac{3}{4} = 6 \times \frac{4}{3}$
Multiply 6 by 4 to get 24 thirds. Then divide 24 by 3 to get 8.
$x = \frac{6 \times 4}{3} = \frac{24}{3} = 8$
Check in the original equation. Three fourths of 8 is 24 fourths, and 24 divided by 4 is 6.
$\frac{3}{4} \times 8 = \frac{24}{4} = 6$

Answer

The solution is $x = 8$ , because $\frac{3}{4} \times 8 = 6$ .

Worked example

A decimal multiplies the variable

Solve $0.8x = 3.6$ , where $x$ stands for an unknown number.

The equation multiplies x by 0.8, and 0.8 is not zero. Divide both sides by 0.8 to undo the multiplication and keep the sides equal.
$0.8x \div 0.8 = 3.6 \div 0.8$
The left side is x alone. On the right, multiply the dividend 3.6 and the divisor 0.8 by the same nonzero number, 10. The divisor becomes a whole number and the quotient does not change.
$x = 3.6 \div 0.8 = 36 \div 8$
Divide 36 by 8. Four 8s make 32, so 4 is left. Share that 4 as well: 4 divided by 8 is 0.5. The exact quotient is 4.5.
$x = 36 \div 8 = 4.5$
Check in the original equation. Eight tenths times 4.5 is 36 tenths, which is 3.6.
$0.8 \times 4.5 = 3.6$

Answer

The solution is $x = 4.5$ , because $0.8 \times 4.5 = 3.6$ .

Four more equations solved the same way. In every row, x stands for an unknown number.

Equation	What the equation does to x	Inverse step on both sides	Solution	Check
$8x = 5.6$	multiplies by $8$	divide by $8$ : $5.6 \div 8 = 0.7$	$x = 0.7$	$8 \times 0.7 = 5.6$
$5x = \frac{1}{2}$	multiplies by $5$	divide by $5$ : $\frac{1}{2} \div 5 = \frac{1}{10}$	$x = \frac{1}{10}$	$5 \times \frac{1}{10} = \frac{5}{10} = \frac{1}{2}$
$\frac{x}{2.5} = 3.2$	divides by $2.5$	multiply by $2.5$ : $3.2 \times 2.5 = 8$	$x = 8$	$8 \div 2.5 = 3.2$
$x \div \frac{3}{5} = 10$	divides by $\frac{3}{5}$	multiply by $\frac{3}{5}$ : $10 \times \frac{3}{5} = 6$	$x = 6$	$6 \div \frac{3}{5} = 6 \times \frac{5}{3} = 10$

A solution is an exact number, and it does not have to be whole. You have already seen solutions such as $4.5$ , $0.7$ and $\frac{1}{10}$ .

A remainder answers a different kind of question. Suppose you share $30$ marbles equally among $4$ friends, and you cannot cut a marble. Each friend gets $7$ marbles, and $2$ marbles are left over. That check is divisor times whole quotient plus remainder: $4 \times 7 + 2 = 30$ .

The equation $4x = 30$ asks something else. Here $x$ is an unknown number that can be a fraction or a decimal, so nothing is left over. Divide both sides by $4$ . The solution is the exact number $x = 30 \div 4 = 7.5$ , and the check is exact: $4 \times 7.5 = 30$ . If you rounded $7.5$ to $8$ , the check would only come close: $4 \times 8 = 32$ , not $30$ . Keep exact values.

Takeaway: Multiplying by a nonzero number and dividing by that same number undo each other. To solve a one-step multiplication or division equation, apply the inverse operation to both sides so that the variable is alone. Then check your exact solution in the original equation. This one plan works for whole numbers, fractions and decimals.

Next, you will write one-step multiplication and division equations from words and use them to solve word problems.