Equation Lab #2 – Solving Equations by Multiplication and Division

Yesterday, I went over how to solve equations by addition and subtraction. You can solve an additive equation by subtracting the same constant being added, and solve a subtractive equation by adding the same constant being subtracted. There’s still two other basic operators to cover, which is exactly what today’s lesson is going to be about.

Before I go over how to solve multiplicative and divisive equations, let’s go over coefficients. A coefficient is the numerical multiplier to a variable. Take for instance, $3x$. While $x$ remains to be unknown, $3x$ is $x$ times $3$. In other words, $3x=x+x+x$. Just letting you know what this also means, if $x=1$, then $3x=3$. If $x=2$, then $3x=6$. If $x=3$, then $3x=9$. If $x=4$, then $3x=12$. There are also fractions. In these terms, the denominator is the divisor. So when you see $\frac{x}{3}$, that is $x$ divided by $3$.

Solving Equations by Multiplication and Division:

Solving equations by multiplication and division have the same rules as solving equations by addition and subtraction. When you apply one operation, you apply it to both sides of the equation. In other words, when $a=b$, then here are the properties:

• Multiplicative Law of Equality:
  • $a\cdot c=b\cdot c$
• Divisive Law of Equality:
  • $a/c=b/c$

Like yesterday, the three questions to ask when solving an equation are:

• What is the variable?
• What number is being applied to the variable?
• How is the number being applied to the variable?

In regards to the third question, if you see a coefficient, the operation is multiplication, which can be reversed by division. And if you see a denominator, the operation is division, which can be reversed by multiplication. Remember, same number, opposite operation.

Let’s start with the equation $4t=36$. What is the variable? The answer is $t$. We want to solve for $t$. What number is being applied to $t$? The answer is $4$. How is $4$ applied to $t$? The answer is multiplication. If you multiply $t$ by $4$, you get $36$. So, in order to find the value of $t$, we must divide both sides by $4$. That is, we must divide $4t$ by $4$, and $36$ by $4$.

Starting with the first step, $4t/4=t$ using the inverse and identity properties of multiplication. By dividing $4$ by $4$, you get $1$, which is the multiplicative equivalent of adding $0$. You are left with $t$. But $36/4=9$. As a result, $t=9$.

1. $4t=36$
2. $4t/4=36/4$
3. $t=9$

Now let’s check our work. Now that we found that $t=9$, we can substitute $t$ in $4t=36$ with $9$. The new equation is $4\cdot 9=36$, which becomes $36=36$. The answer is correct, so $t=9$ when $4t=36$.

Let’s go to the next equation, $\frac{r}{3}=8$. What is the variable? The answer is $r$. We want to solve for $r$. What number is being applied to $r$? The answer is $3$. How is $3$ applied to $r$? The answer is division. If you divide $r$ by $3$, you will get $8$. So, in order to find the value for $r$, we must multiply both sides by $3$.

Starting with the first step, $\frac{r}{3}\cdot 3=r$. A fun fact is that $\frac{r}{3}$ is the same as $\frac{1}{3} \normalsize r$, and when you multiply a fraction by a whole number, you multiply the numerator. The fraction becomes $\frac{3}{3}$, which reduces to $1$. This should leave you with $r$. At the same time, $8\cdot 3=24$. As a result, $r=24$.

1. $\frac{r}{3}=8$
2. $\frac{r}{3}\cdot 3=8\cdot 3$
3. $r=24$

Now let’s check our work. Now that we found that $r=24$, we can substitute $r$ with $24$ in $\frac{r}{3}=8$. The new equation is $\frac{24}{3}=8$. This results in $8=8$. The answer is correct. So, $r=24$ when $\frac{r}{3}=8$.

Now that I went over how to solve equations by multiplication and division, it’s time to go over a word problem.

Word Problem #2:

Martin has collected coins in the past. When he collects coins, he puts them on a stack until it’s too tall for him. He now has 216 coins in total. On his desk, there are 12 stacks of coins. How many coins are in each stack?

First, let’s define the variable. We’ll let $c$ be the number of coins in a stack since the word “coin” starts with a ‘c’. We are given the number of coins in total, which is 216, and the number of stacks of coins, which is 12. This sounds like a multiplication problem. So the equation is $12c=216$, where $c$ is the number of coins per stack, $12$ is the number of stacks, and $216$ is the number of coins in total.

Now we can solve the equation. Since this is a multiplicative equation, we must solve by division. And since $12$ is the number being applied to $c$, we must divide both sides by $12$. $12c/12=c$, but $216/12=18$. As a result, $c=18$.

1. Define $c$ as number of coins per coin stack.
2. There are $216$ total coins and $12$ stacks of
   coins.
3. $12c=216$
4. $12c/12=216/12$
5. $c=18$

There are 18 coins per stack.

Now is our answer correct? We can multiply $18$ by $12$ to see if we can get $216$. Since $18\cdot 12=216$, the answer is correct.

Quiz #2:

Now that the lesson is over, let’s see if you can complete this quiz. Whether you put down the missing value or the variable with the missing value is fine for as long as you use the correct variable and correct answer. For instance, when the question is $b+3=6$, you can use $3$ or $b=3$ as your answer, but not $a=3$.