Solving One-Step Equations with Multiplication — Examples

Multiplication is another prong on the equation-solving fork, along with division, subtraction, and addition. (Yes, that analogy seems a bit odd, but at least it'll be hard to forget).

Multiplication and division are inverse operations. This means that they are opposite processes. Another way to think about it is that they undo each other. So, whenever you have an equation with a fraction (which is basically built-in division), using multiplication can help get rid of that and make your equation easier to solve.

Let's look at a few examples of using multiplication to solve equations.

This explanation includes:

• One-step equations
• Two-step equations
• Two step equations with multiple operations (addition and subtraction as well as multiplication)

Solving one-step equations using multiplication

Take a look at this one-step equation:

$$\frac{x}{6} = 3$$

There's a fraction there which, like we mentioned above, is basically like division built in to the equation. The way to eliminate this fraction is to apply the inverse operation: multiply each side of the equation by $6$.

$$\begin{align} \frac{x}{6} &= 3\\ \frac{x}{6} \cdot 6 &= 3 \cdot 6\\ x &= 18 \end{align}$$

Let's look at how we would model this.

Solving two-step equations using multiplication

Let's make our equation a little more challenging by adding a second step:

$$\frac{x}{6} + 4 = 3$$

Before we can do our multiplication, we need to get rid of the $4$ hanging out next to the fraction.

$$\begin{align} \frac{x}{6} + 4 &= 3\\ \frac{x}{6} + 4 - 4 &= 3 - 4\\ \frac{x}{6} &= -1 \end{align}$$

Now we can multiply to solve the equation!

$$\begin{align} \frac{x}{6} &= -1\\ \frac{x}{6} \cdot 6 &= -1 \cdot 6\\ x &= -6 \end{align}$$

Solving two-step equations using multiplication and other operations

For the last example, let's add in another operation to our equation:

$$2\left(\frac{x}{6} + 1\right) + 4 = 3$$

We've got parentheses looking back at us now. We need to remember PEMDAS here and take care of those parentheses first by using the distributive property.

$$\begin{align} 2\left(\frac{x}{6} + 1\right) + 4 &= 3\\ \frac{2x}{6} + 2 + 4& = 3 \end{align}$$

(Note that we multiplied the numerator by $2$ but not the denominator. It helps to think about the $2$ outside the parentheses as "$\frac{2}{1}$").

Now we are set to isolate the $x$ term by itself on one side of the equation.

$$\begin{align} \frac{2x}{6} + 2 + 4 &= 3\\ \frac{2x}{6} + 6 &= 3\\ \frac{2x}{6} + 6 -6 &= 3-6\\ \frac{2x}{6} &= -3 \end{align}$$

We can (almost) finish solving the equation by multiplying both sides by $6$: $$\begin{align} \frac{2x}{6} &= -3\\ \frac{2x}{6} \cdot 6 &= -3 \cdot 6\\ 2x &= -18 \end{align}$$

The very last thing to do, in order to get $x$ totally alone, is to divide both sides of the equation by $2$.

$$\begin{align} 2x &= -18\\ \frac{2x}{2} &= \frac{-18}{2}\\ x &= -9 \end{align}$$

Using Multiplication to Isolate a Variable

Solving an algebraic equation means solving for the correct value for the variable in the equation. For any of these equations, the solution set will be a single value. That is, these equations only have one solution.

There are two kinds of equations that involve division that will require multiplication to solve for the variable:

1. Your variable is being divided by something. For example: $\frac{x}{3} = 5$
2. Your variable is dividing something. For example: $\frac{36}{x} = 4$

In the next sections, I'll show you how to figure out the answer to each one.

Finding the answer when your variable is being divided

Sometimes, the variable you're trying to find will be on the top (numerator) of a fraction—in other words, it is being divided by another number. To solve these equations, all you have to do is multiply both sides by the number in the bottom of the fraction (denominator). Let's try it with the example equation from above: $$\frac{x}{3} = 5$$

Our first step is to multiply both sides by $3$. This will cancel out the division on the left side of the equation: $$\begin{align} \frac{x}{3} \times 3&= 5\times 3\\ \frac{x\times 3}{3} &= 5\times 3\\ \frac{x}{1} &= 5\times 3\\ x &= 5\times 3 \end{align}$$

$$5 \times 3 = 15\\ \text{so}\\ x = 15$$

Here's a graphic with another example!

Next, we'll look at the other type of equation.

Finding the answer when your variable is doing the dividing

Other times, the variable you're trying to find will be on the bottom (denominator) of an equation—it will be the thing dividing another number. To solve these equations, you first have to get the variable out of the denominator, and then get it alone on one side of the equation. Let's take a look at our example equation to get a better idea of how this works: $$\frac{36}{x} = 4$$

Our first step is to multiply both sides by $x$. This will cancel out the division on the left side of the equation: $$\begin{align} \frac{36}{x} \times x&= 4\times x\\ \frac{36\times x}{x}&= 4\times x\\ 36&= 4x \end{align}$$

Awesome, $x$ is out of the denominator. But we don't actually have an answer yet, since $x$ isn't alone on one side of the equation! Now, though, the problem looks like a multiplication equation that requires division to solve.

$$\begin{align} \frac{36}{4} &= \frac{4x}{4}\\ 9 &= \frac{x}{1}\\ 9 &= x \end{align}$$

Done!

Here's another graphic with an example of this type of problem.

(Video) Solving Basic Equations pt. 2

by Math Antics

This video by Math Antics shows you how to solve equations using division and Multiplication.

$\large{\text{Summary}}$

The name of the game when solving basic algebraic equations is to get all unknown variables on one side of the equal sign, and all the known numbers on the other side of the equal sign. How do we rearrange the equations though? We can use operations such as addition, subtraction, multiplication and division to help us on the road to solving for a variable!

When you are solving an algebraic equation, you should think of it like a Balanced Scale: where each side of the equation holds the same amount of weight. This means that whatever is on one side of the equal sign must have the SAME EXACT value as the other side of the equal sign, otherwise the equation would not be true. We need to make sure that the sides are balanced because as we rearrange equations, we may accidentally make a mistake that would change one side more than the other, causing the equation to be “out of balance”, and therefore false.

A key to keeping the equations “balanced” is to make the exact same changes at the exact same time to both sides of the equation. In summary, if we want to add/subtract something from one side of the equation, we’ll have to perform the same exact operation to the other side of the equation. The same thing can be said for if we want to multiply/divide something from one side of the equation.

Multiplying

He goes over the problem $\frac{x}{10} = 15$. We can isolate $x$ by multiplying both sides by $10$. $$\begin{align} \frac{x}{10}&=15\\ \frac{x}{\cancel{10}} {\color{#29A603}{\cdot \cancel{10}}} &= 15 \,{\color{#29A603}{\cdot 10}}\\ x&=150\end{align}$$