Equation Lab #3 – Solving Two-Step Equations

In the last two days, I went over how to solve one-step equations, with the first lesson on addition and subtraction, and the second lesson on multiplication and division. Today, you will get to learn how to solve equations that involve both addition/subtraction and multiplication/division. This lesson is on two-step equations.

Solving Two-Step Equations:

Let’s do a recall of the equation-solving properties. As we know, if you apply an operation to an equation, you apply the same operation to both sides. Recalling our four properties when $a=b$:

• Additive Law of Equality:
  • $a+c=b+c$
• Subtractive Law of Equality:
  • $a-c=b-c$
• Multiplicative Law of Equality:
  • $a\cdot c=b\cdot c$
• Divisive Law of Equality:
  • $a/c=b/c$

Since we are doing two steps now, here are some questions to ask:

• What is the variable?
• What two numbers are being applied to the variable?
• How are these two numbers applied to the variable?

Remember, we use the same numbers, but opposite operations.

Let’s start with the equation $\frac{p}{7}+6=9$. What is the variable? The answer is $p$. We want to solve for $p$. What two numbers are applied to $p$? The answer is $7$ and $6$. How are $7$ and $6$ applied to $p$? $7$ is applied to $p$ by division, and $6$ is applied to $\frac{p}{7}$ by addition. If you divide $p$ by $7$ and then add $6$, you will get $9$. So, in order to solve for $p$, you’ll have to subtract $6$ and multiply by $7$. Both operations must be done to both sides of the equation. If you remember the order of operations, multiplication and division always comes before addition and subtraction. But when we’re solving equations, we must take the opposite direction. That is, we add or subtract before we multiply or divide.

Starting with the first step, $\frac{p}{7}+6-6 =\frac{p}{7}$. A number minus itself always equals $0$, which will never change the result when being added to or subtracted from a number. Meanwhile, $9-6=3$. The equation is now $\frac{p}{7} \normalsize =3$. Now we can multiply. $\frac{p}{7}\cdot 7 = p$. A number divided by itself always equals $1$, which will never change the result when multiplied to or divided from a number. Meanwhile, $3\cdot 7=21$. As a result, $p=21$.

1. $\frac{p}{7}+6=9$
2. $\frac{p}{7}+6-6=9-6$
3. $\frac{p}{7}= 3$
4. $\frac{p}{7}\cdot 7=3\cdot 7$
5. $p=21$

Like always, we check our work after solving the equation. We found that $p=21$, let’s prove that this is the answer to $\frac{p}{7}+6=9$. Substituting $p$ with $21$, we get $\frac{21}{7}+6=9$. $\frac{21}{7}=3$, and $3+6=9$. The result becomes $9=9$. Therefore, $p=21$ when $\frac{p}{7}+6=9$.

Let’s do the opposite operations. The next equation is $5b-8=22$. What is the variable? The answer is $b$. We want to solve for $b$. What two numbers are applied to $b$? The answer is $5$ and $8$. How are $5$ and $8$ applied to $b$? $5$ is applied to $b$ by multiplication, and $8$ is applied to $5b$ by subtraction. If you multiply $b$ by $5$ and then subtract $8$, you will get $22$. So, in order to solve for $b$, you must add $8$ to both sides and divide both sides by $5$. We start with addition before getting to division.

Starting with the first step, $5b-8+8=5b$. Even with the variable term having a coefficient, the variable term is on its own now that $8$ is eliminated. Meanwhile, $22+8=30$. The equation is now $5b=30$. Now we can divide. $5b/5=b$, giving us a bald variable of $b$. Meanwhile, $30/5=6$. As a result, $b=6$.

1. $5b-8=22$
2. $5b-8+8=22+8$
3. $5b=30$
4. $5b/5=30/5$
5. $b=6$

Now let’s check our work. Given that $b=6$, we can substitute $b$ with $6$ in $5b-8=22$. We should have $5\cdot 6-8=22$. $5\cdot 6=30$, and $30-8=22$. The result becomes $22$. Therefore, $b=6$ when $5b-8=22$.

With enough practice on the two-step equations, let’s try putting this into practice with word problems.

Word Problem #3:

Sally is saving up to buy a plasma lamp. A plasma lamp at the local store costs $47. She currently owns $12. For that reason, she started doing chores for her father, who will pay her $5 for every chore she gets done. How many chores does she need to do to have enough money for a plasma lamp?

First, let’s define the variable. We’ll let $c$ be the number of chores Sally needs to do to get enough money for the plasma ball. We are given how much money she’ll be paid for every chore she gets done, which is $5. Since this sounds like a multiplication problem, we can write the variable term as $5c$, which represents how much money she has from doing chores. We are also given the initial amount of money she has, which is $12. The total amount of money is $5c+12$, where $c$ is the number of chores Sally needed to do, $5$ is how many dollars she gets paid per chore, and $12$ is the initial amount of money she had. The goal is to get the plasma lamp, which costs $47. The equation should be $5c+12=47$.

Now we can solve the equation. Since the numbers are applied by addition and multiplication, we must use division and subtraction to solve the equation, and it begins with subtracting $12$ from both sides of the equation. $5c+12-12=5c$, and $47-12=35$. The equation is now $5c=35$. The next step is to divide both sides of the equation by $5$. $5c/5=c$, and $35/5=7$. The result is $c=7$.

1. Define $c$ as the number of chores to do.
2. The price is 47 dollars. Sally can only
   earn 5 dollars per chore, as she initially
   has 12 dollars.
3. $5c+12=47$
4. $5c+12-12=47-12$
5. $5c=35$
6. $5c/5-35/5$
7. $c=7$

Sally needs to do 7 chores to earn enough money to buy the plasma lamp.

Now is our answer correct? We can multiply $7$ by $5$ and add $12$ to see if we can get $47$. $7\cdot 5=35$, and $35+12=47$. Our answer is correct.

Quiz #3:

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 $a=3$.