3.1: Solving Two Step Equations
Introduction
The Marching Band
The marching band at Floyd Middle School is excellent and well respected throughout the community. Each year the band continues to grow, and the students who participate make a huge commitment to practices, football games and parades.
When the students gathered for their first rehearsal, Mrs. Kline the band director gathered them altogether for a few announcements.
“We will be marching in the big parade this year once again,” she said smiling from ear to ear.
“I am so glad,” Keri said leaning over to Anica. “I was hoping that we would.”
“We will also be adding four new students this year. We started with 140 students and we will add four. This means that we need to redo our formation for the big finale. We need to reorganize the band into eight even rows. Let’s take a look at what this will look like. Please take out a piece of paper and a pencil,” Mrs. Kline said turning to the blackboard.
“I can figure out the number of students in each row with an equation,” Anica said smiling.
“Yes, and don’t forget to count Jake as the Drum Major in the lead,” Keri added.
Do you know what this equation needs to look like? We have been given the sum of the students, a Drum Major and we know that we need eight even rows. As Anica said, we will need an equation to figure out the number of students in each row. This lesson will teach you all that you need to know about two-step equations so that you will know how to solve this in the end.
What You Will Learn
In this lesson you will learn how to complete the following skills.
- Solve equations involving inverse properties of addition and multiplication.
- Solve equations involving inverse properties of addition and division.
- Solve equations involving inverse properties of subtraction and multiplication.
- Solve equations involving inverse properties of subtraction and division.
- Model and solve real-world problems using two step equations.
Teaching Time
I. Solve Equations Involving Inverse Properties of Addition and Multiplication
Let’s start this lesson by thinking about what we already know about equations.
We know that an equation is a statement with an equal sign where the quantity on one side of the equals is the same as the quantity on the other side of the equals.
Example
\begin{align*}4 + 11 = 15\end{align*}
This is an example of an equation. The quantity on one side of the equals is the same as the quantity on the other side.
If we don’t know one of the pieces of an equation, we can use a variable. A variable is a letter used to represent an unknown quantity. An equation that includes at least one variable is called an algebraic equation.
Example
\begin{align*}x+11=15\end{align*}
Here we have an equation with a variable where \begin{align*}x\end{align*} is the unknown quantity. To solve this, the natural thing to do is to perform an inverse operation or opposite operation. We subtract eleven from 15 which leaves us with 4. That is the value of the variable.
Most of the time, you don’t even think about performing an inverse operation, your mind naturally solves the problem in this way.
When you have an equation with one variable, it is called a one step equation. It only takes one operation or one inverse operation to solve it. You have had a lot of practice solving one step equations.
In this lesson, you are going to learn how to solve two step equations. Let’s begin.
To solve a two-step equation, we will need to use more than one inverse operation. Let's take a look at how to solve a two-step equation now. When we perform inverse operations to find the value of a variable, we work to get the variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations. You can use isolating the variable whether you are solving one step or two step equations.
Let’s look at an example of a two step equation.
Example
Solve for \begin{align*}a\end{align*}: \begin{align*}3a + 12 = 45\end{align*}.
We can call each piece of the equation a term. There is a term with a variable and there is a term without a variable. Notice that there are two terms on the left side of the equation, \begin{align*}3a\end{align*} and 12.
Our first step is to use inverse operations to get the term that includes a variable, \begin{align*}3a\end{align*}, by itself on one side of the equal (=) sign. Because the three is connected to the variable, we perform the other inverse operation first. We work with the number connected with the variable last.
In the equation, 12 is added to \begin{align*}3a\end{align*}. So, we can use the inverse of addition—subtraction. We can subtract 12 from both sides of the equation.
Let's see what happens when we subtract 12 from both sides of the equation.
\begin{align*}3a + 12 & = 45\\ 3a + 12 - 12 & = 45 - 12\\ 3a + 0 & = 33\\ 3a & = 33\end{align*}
Now, the term that includes a variable, \begin{align*}3a\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}a\end{align*} by itself. Since \begin{align*}3a\end{align*} means \begin{align*}3 \times a\end{align*}, we can use the inverse of multiplication—division. We can divide both sides of the equation by 3. Let's see what happens when we divide both sides of the equation by 3.
\begin{align*}3a & = 33\\ \frac{3a}{3} & = \frac{33}{3}\\ 1a & = 11\\ a & = 11\end{align*}
The value of \begin{align*}a\end{align*} is 11.
That is a great question. If you take your answer and substitute it back into the original equation, then solve the equation. If both sides of the equation balance, then your work is accurate. Let’s try it out.
\begin{align*}3(11) + 12 & = 45\\ 33 + 12 & = 45\\ 45 & = 45\end{align*}
Here our work is accurate because the quantity on one side of the equals is the same as the quantity on the other side of the equals. When in doubt, check it out!!
Notice that this problem used addition and multiplication.
Let's review the properties of equality that can help us understand the strategy used in the example.
Notice that in the example, our first step was to subtract 12 from both sides of the equation. We needed to subtract that number from both sides of the equation because of the subtraction property of equality, which states:
if \begin{align*}a = b\end{align*}, then \begin{align*}a - c = b - c\end{align*}.
So, if you subtract a number, \begin{align*}c\end{align*}, from one side of an equation, you must subtract that same number, \begin{align*}c\end{align*}, from the other side, too, to keep the values on both sides equal.
In the example, our second step was to divide both sides of the equation by 3. We needed to divide both sides of the equation by that number because of the division property of equality, which states:
if \begin{align*}a = b\end{align*} and \begin{align*}c \neq 0\end{align*}, then \begin{align*}\frac{a}{c} = \frac{b}{c}\end{align*}.
So, if you divide one side of an equation by a nonzero number, \begin{align*}c\end{align*}, you must divide the other side of the equation by that same number, \begin{align*}c\end{align*}, to keep the values on both sides equal.
Let’s review our steps to solving this two step equation.
Take a few minutes to write these steps in your notebook.
II. Solve Equations Involving Inverse Properties of Addition and Division
We can apply the same steps as we did when working with addition and multiplication when we work with other types of two step equations. We will focus on isolating the variable and on using inverse operations to do that. Let’s look at an example.
Example
Solve for \begin{align*}c\end{align*}: \begin{align*}5 + \frac{c}{4} = 15\end{align*}.
Notice that there are two terms on the left side of the equation, 5 and \begin{align*}\frac{c}{4}\end{align*}.
Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{c}{4}\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 5 is added to \begin{align*}\frac{c}{4}\end{align*}. So, we can use the inverse of addition—subtraction. We can subtract 5 from both sides of the equation.
\begin{align*}5 + \frac{c}{4} & = 15\\ 5 - 5 + \frac{c}{4} & = 15 - 5\\ 0 + \frac{c}{4} & = 10\\ \frac{c}{4} & = 10\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{c}{4}\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}c\end{align*} by itself. Since \begin{align*}\frac{c}{4}\end{align*} means \begin{align*}c \div 4\end{align*}, we can use the inverse of division—multiplication. We can multiply both sides of the equation by 4.
\begin{align*}\frac{c}{4} & = 10\\ \frac{c}{4} \times 4 & = 10 \times 4\\ \frac{c}{4} \times \frac{4}{1} & = 40\end{align*}
The number 4, or \begin{align*}\frac{4}{1}\end{align*}, is the multiplicative inverse, or reciprocal, of \begin{align*}\frac{1}{4}\end{align*}. You can find the multiplicative inverse of a number by flipping its numerator and its denominator. So, the multiplicative inverse of \begin{align*}\frac{4}{1}\end{align*} is \begin{align*}\frac{1}{4}\end{align*}. When a number is multiplied by its multiplicative inverse, the product is 1.
\begin{align*}\frac{c}{4} \times \frac{4}{1} & = 40\\ c \times \left ( \frac{1}{4} \times \frac{4}{1} \right ) & = 40\\ c \times 1 & = 40\\ c & = 40\end{align*}
The work above shows how multiplying each side of the equation by 4 isolates the variable.
Because 4 is the multiplicative inverse, or reciprocal, of \begin{align*}\frac{1}{4}\end{align*}, we could also have solved this problem by canceling out the 4's like this:
\begin{align*}\frac{c}{\bcancel{4}} \times \frac{\bcancel{4}}{1} & = 40\\ \frac{c}{1} & = 40\\ c & = 40\end{align*}
The answer is that \begin{align*}c\end{align*} is equal to 40.
Yes. You can substitute the answer into the original equation and then solve it. If both sides are equal, then your work is accurate.
\begin{align*}5 + \frac{40}{4} & = 15\\ 5 + 10 & = 15\\ 15 & = 15\end{align*}
Both sides are equal, so our equation is balanced.
In the first step of example, we subtracted 5 from both sides of the equation because of the subtraction property of equality, which was reviewed in the first section of this lesson.
In the second step of Example 2, we multiplied both sides of the equation by 4. We must multiply both sides of the equation by that number because of the multiplication property of equality, which states:
if \begin{align*}a = b\end{align*} and \begin{align*}c \neq 0\end{align*}, then \begin{align*}a \times c = b \times c\end{align*}.
So, if you multiply one side of an equation by a nonzero number, \begin{align*}c\end{align*}, you must multiply the other side of the equation by that same number, \begin{align*}c\end{align*}, to keep the values on both sides equal.
III. Solve Equations Involving Inverse Properties of Subtraction and Multiplication
Now, let's take a look at another two-step equation. This one involves subtraction.
Example
Solve for \begin{align*}x\end{align*}: \begin{align*}2x - 9 = 17\end{align*}.
Notice that there are two terms on the left side of the equation, \begin{align*}2x\end{align*} and 9. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}2x\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 9 is subtracted from \begin{align*}2x\end{align*}. So, we can use the inverse of subtraction—addition. We can subtract 9 from both sides of the equation.
\begin{align*}2x - 9 & = 17\\ 2x(-9 + 9) & = 17 + 9\\ 2x & = 26\end{align*}
Notice how we rewrote the problem above. Since we are adding a positive number, 9, to a number that is being subtracted from \begin{align*}2x\end{align*}, we can represent this as adding 9 to -9 as we did above: (-9 + 9).
The number 9 is the additive inverse, or opposite, of -9.
We can now use inverse operations to get the \begin{align*}x\end{align*} by itself. Since \begin{align*}2x\end{align*} means \begin{align*}2 \cdot x\end{align*}, we can use the inverse of multiplication—division. We can divide both sides of the equation by 2.
\begin{align*}2x & = 26\\ \frac{2x}{2} & = \frac{26}{2}\\ x & = 13\end{align*}
The value of \begin{align*}x\end{align*} is 13.
Now we can check our work too.
\begin{align*}2 (13) - 9 & = 17\\ 26 - 9 & = 17\\ 17 & = 17\end{align*}
Our work is accurate.
IV. Solve Equations Involving Inverse Properties of Subtraction and Division
By now you are probably beginning to see some patterns for solving two step equations. You will notice that the steps really don’t change, the inverse operations do depending on the operations in the equations, but the steps to solving the equations remain the same.
Now let’s look at solving an equation that has subtraction and division in it.
Example
Solve for \begin{align*}z\end{align*}: \begin{align*}\frac{z}{6} - 7 = 3\end{align*}.
Notice that there are two terms on the left side of the equation, \begin{align*}\frac{z}{6}\end{align*} and 7. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{z}{6}\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 7 is subtracted from \begin{align*}\frac{z}{6}\end{align*}. So, we can use the inverse of subtraction—addition. We can add 7 to both sides of the equation, like this:
\begin{align*}\frac{z}{6} - 7 & = 3\\ \frac{z}{6} - 7 + 7 & = 3 + 7\\ \frac{z}{6} + (-7 + 7) & = 10\\ \frac{z}{6} + 0 & = 10\\ \frac{z}{6} &= 10\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{z}{6}\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}z\end{align*} by itself. Since \begin{align*}\frac{z}{6}\end{align*} means \begin{align*}z \div 6\end{align*}, we can use the inverse of division—multiplication. We can multiply both sides of the equation by 6, like this:
\begin{align*}\frac{z}{6} &= 10\\ \frac{z}{6} \times 6 & = 10 \times 6\\ \frac{z}{\bcancel{6}} \times \frac{\bcancel{6}}{1} & = 60\\ \frac{z}{1} & = 60\\ z & = 60\end{align*}
The value of \begin{align*}z\end{align*} is 60.
Now let’s check our work.
\begin{align*}\frac{60}{6} - 7 & = 3\\ 10 - 7 & = 3\\ 3 & = 3\end{align*}
Our work is accurate.
V. Model and Solve Real – World Problems Using Two Step Equations
We can also use equations to solve real-world problems. Not every problem or dilemma can be solved with only one operation. This is where solving two step equations can be very useful.
The key words in the chart below can help you translate words into algebraic language.
Key Words for Addition | Key Words for Multiplication | Key Words for Subtraction | Key Words for Division |
---|---|---|---|
sum of 1 and \begin{align*}a\end{align*} \begin{align*}1 + a\end{align*} | product of 8 and \begin{align*}b\end{align*} \begin{align*}8b\end{align*} | 4 less \begin{align*}v\end{align*} \begin{align*}4 - v\end{align*} | 7 divided into \begin{align*}r\end{align*} equal groups \begin{align*}\frac{7}{r}\end{align*} |
10 plus \begin{align*}z\end{align*} \begin{align*}10 + z\end{align*} | 3 times \begin{align*}x\end{align*} \begin{align*}3x\end{align*} | 5 less than \begin{align*}t\end{align*} \begin{align*}t - 5\end{align*} | \begin{align*}p\end{align*} shared equally by 4 \begin{align*}\frac{p}{4}\end{align*} |
2 more than \begin{align*}y\end{align*} \begin{align*}2 + y\end{align*} | twice \begin{align*}w\end{align*} \begin{align*}2w\end{align*} | 6 fewer than \begin{align*}s\end{align*} \begin{align*}s - 6\end{align*} | half of \begin{align*}n\end{align*} \begin{align*}\frac{n}{2}\end{align*} |
9 and \begin{align*}b\end{align*} \begin{align*}9 + b\end{align*} | one-third of \begin{align*}m\end{align*} \begin{align*}\frac{m}{3}\end{align*} |
Other Words for Addition or Multiplication | Other Words for Subtraction | Other Words for Division |
---|---|---|
how many total | how many left | how many in each |
how many all together | how many more | |
how many in all | how many fewer | |
how much change |
The bolded key words in the phrases above provide clues that can help you translate word problems into algebraic equations. Remember, key words can be a helpful guide, but you should always think about which operation makes sense for a particular situation. Key words alone will not always be enough to help you write an equation to represent a problem.
Now let’s look at working with these key words while solving a problem.
Example
Brandon and Felicia sold rolls of wrapping paper for a school fundraiser. Brandon sold 3 less than half the number of rolls that Felicia sold. Brandon sold a total of 9 rolls of wrapping paper.
a. Write an algebraic equation to represent \begin{align*}f\end{align*}, the number of rolls of wrapping paper that Felicia sold.
b. Find the number of rolls of wrapping paper that Felicia sold.
Consider part a first.
Use a number, an operation sign, a variable, or an equal sign to represent each part of that problem. Since Brandon sold 9 rolls of wrapping paper, represent the number of rolls Brandon sold as 9. Use the key words from the chart to help you translate the rest of the problem into an equation. For example, you can translate “half the number of rolls...Felicia sold” as \begin{align*}\frac{f}{2}\end{align*}.
\begin{align*}& \underline{Brandon} \ \underline{sold} \ \underline{3} \ \underline{less \ than} \ \underline{half \ the \ number \ldots \ Felicia \ sold.}\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \searrow \qquad \downarrow \qquad \qquad \quad \swarrow\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \searrow \qquad \downarrow \qquad \qquad \quad \swarrow\\ & \quad \ \downarrow \qquad \ \downarrow \quad \ \swarrow \qquad \downarrow \qquad \qquad \quad \searrow\\ & \quad \ \ 9 \qquad = \ \ \frac{f}{2} \qquad - \qquad \qquad \quad 3\end{align*}
So, this equation, \begin{align*}9 = \frac{f}{2} - 3\end{align*}, represents \begin{align*}f\end{align*}, the number of rolls of wrapping paper that Felicia sold.
Next, consider part b.
Solve the equation for \begin{align*}f\end{align*} to find the number of rolls that Felicia sold.
Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{f}{2}\end{align*}, by itself on one side of the equal (=) sign. In the equation, 3 is subtracted from \begin{align*}\frac{f}{2}\end{align*}. So, we can use the inverse of subtraction and add 3 to both sides of the equation, like this:
\begin{align*}9 &= \frac{f}{2}-3\\ 9 + 3 & = \frac{f}{2} - 3 + 3\\ 12 & = \frac{f}{2} + (-3 + 3)\\ 12 & = \frac{f}{2} + 0\\ 12 & = \frac{f}{2}\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{f}{2}\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}f\end{align*} by itself. Since \begin{align*}\frac{f}{2}\end{align*} means \begin{align*}f \div 2\end{align*}, we can use the inverse of division and multiply both sides of the equation by 2, like this:
\begin{align*}12 &= \frac{f}{2}\\ 12 \times 2 & = \frac{f}{2} \times 2\\ 24 & = \frac{f}{\bcancel{2}} \times \frac{\bcancel{2}}{1}\\ 24 & = \frac{f}{1}\\ 24 & = f\end{align*}
The value of \begin{align*}f\end{align*} is 24, so Felicia sold 24 rolls of wrapping paper for the fundraiser.
Example
A gardener charges $20 for each gardening job plus $15 for each hour worked. He charged $80 for a gardening job he did yesterday.
a. Write an algebraic equation to represent \begin{align*}h\end{align*}, the number of hours that the gardener worked on that $80 job.
b. Find the number of hours that the gardener worked on that $80 job.
Consider part a first.
Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The gardener earned $15 for each hour worked on that job, so you could multiply $15 by \begin{align*}h\end{align*}, the number of hours worked, to find how much money the gardener charged for his worktime.
\begin{align*}& \underline{\$ 20} \ for \ each \ gardening \ job \ \underline{plus} \ \underline{\$15 \ for \ each \ hour} \ worked \ldots \underline{charged} \ \underline{\$80} \ for \ one \ldots job.\\ & \downarrow \qquad \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \downarrow\\ & 20 \qquad \qquad \qquad \qquad \qquad \quad \ + \qquad \qquad \quad 15h \qquad \qquad \qquad \qquad = \quad \ 80\end{align*}
So, this equation, \begin{align*}20 + 15h = 80\end{align*}, represents \begin{align*}h\end{align*}, the number of hours the gardener worked on the $80 job.
Next, consider part b.
Solve the equation to find the number of hours the gardener worked on that job.
Since 20 is added to the term that includes a variable, \begin{align*}15h\end{align*}, we can use the inverse of addition—subtraction. We can subtract 20 from both sides of the equation, like this:
\begin{align*}20 + 15h & = 80\\ 20 - 20 + 15h & = 80 - 20\\ 0 + 15h & = 60\\ 15h & = 60\end{align*}
Since 15 is multiplied by the variable, \begin{align*}h\end{align*}, we can use the inverse of multiplication—division. We can divide both sides by 15 to solve for \begin{align*}h\end{align*}, like this:
\begin{align*}15h & = 60\\ \frac{15h}{15} & = \frac{60}{15}\\ 1h & = 4\\ h & = 4\end{align*}
The gardener worked four hours on the job he did yesterday.
Now let’s go back to the problem from the introduction and work on writing and solving the equation.
Real-Life Example Completed
The Marching Band
Here is the original problem once again. First reread it. Then make some notes about the given information. Finally write an equation and then solve it for the number of students in each row. There will be two parts to your answer.
The marching band at Floyd Middle School is excellent and well respected throughout the community. Each year the band continues to grow, and the students who participate make a huge commitment to practices, football games and parades.
When the students gathered for their first rehearsal, Mrs. Kline the band director gathered them altogether for a few announcements.
“We will be marching in the big parade this year once again,” she said smiling from ear to ear.
“I am so glad,” Keri said leaning over to Anica. “I was hoping that we would.”
“We will also be adding four new students this year. We started with 140 students and we will add four. This means that we need to redo our formation for the big finale. We need to reorganize the band into eight even rows. Let’s take a look at what this will look like. Please take out a piece of paper and a pencil,” Mrs. Kline said turning to the blackboard.
“I can figure out the number of students in each row with an equation,” Anica said smiling.
“Yes, and don’t forget to count Jake as the Drum Major in the lead,” Keri added.
Now it’s time for you to apply what you have learned. Remember, there are two parts to your answer.
Solution to Real – Life Example
First, let’s look at what our given information is.
There are 144 students in the band.
There is also one Drum Major.
We need to organize the students into eight even rows.
Here is our equation.
\begin{align*}8x=144\end{align*}
Now you may be wondering why we didn’t include the Drum Major. Well, as Keri points out, the Drum Major is in the lead. In this case, Jake is not included in the equation since he is not in the rows.
We have a one-step equation here. We can solve the equation now.
\begin{align*}x=18 \ students \end{align*}
There will be 18 students in each row.
Vocabulary
Here are the vocabulary words that are found in this lesson.
- Equation
- a mathematical statement with an equal sign where the quantity on one side of the equation is equal to the quantity on the other side.
- Variable
- a letter used to represent an unknown quantity.
- Algebraic Equation
- An equation with at least one variable in it.
- One Step Equation
- An algebraic equation with one operation in it.
- Two Step Equation
- An algebraic equation with two operations in it.
- Term
- the different numbers/variables in an equation.
Time to Practice
Directions: Solve the following two step equations that have addition and multiplication in them.
- \begin{align*}3x + 4 = 22\end{align*}
- \begin{align*}4y + 3 = 15\end{align*}
- \begin{align*}6x + 5 = 35\end{align*}
- \begin{align*}7x + 2 = 16\end{align*}
- \begin{align*}9y + 8 = 80\end{align*}
- \begin{align*}12x + 15 = 51\end{align*}
- \begin{align*}14y + 2 = 30\end{align*}
- \begin{align*}7y + 5 = 40\end{align*}
Directions: Solve the following two step equations that have addition and division in them.
- \begin{align*}\frac{x}{3} + 4 = 8\end{align*}
- \begin{align*}\frac{x}{5} + 8 = 10\end{align*}
- \begin{align*}\frac{a}{6} + 7 = 13\end{align*}
- \begin{align*}\frac{a}{9} + 4 = 30\end{align*}
- \begin{align*}\frac{b}{8} + 6 = 15\end{align*}
- \begin{align*}\frac{c}{12} + 9 = 18\end{align*}
- \begin{align*}\frac{x}{7} + 7 = 21\end{align*}
- \begin{align*}\frac{x}{11} + 5 = 12\end{align*}
- \begin{align*}\frac{x}{12} + 9 = 16\end{align*}
- \begin{align*}\frac{a}{14} + 6 = 8\end{align*}
Directions: Solve each two step equation that has multiplication and subtraction in it.
- \begin{align*}4x - 3 = 13\end{align*}
- \begin{align*}5y - 8 = 22\end{align*}
- \begin{align*}7x - 11 = 31\end{align*}
- \begin{align*}8y - 15 = 25\end{align*}
- \begin{align*}9x - 12 = 42\end{align*}
- \begin{align*}12y - 9 = 99\end{align*}
Directions: Solve each two step equation that has division and subtraction in it.
- \begin{align*}\frac{x}{5} - 4 = 8\end{align*}
- \begin{align*}\frac{y}{6} - 3 = 8\end{align*}
- \begin{align*}\frac{x}{7} - 7 = 10\end{align*}
- \begin{align*}\frac{x}{8} - 4 = 12\end{align*}
- \begin{align*}\frac{y}{7} - 5 = 11\end{align*}
- \begin{align*}\frac{x}{4} - 10 = 12\end{align*}