Unit 6 Expressions, Equations, and Inequalities — Unit Plan

Tape diagrams are useful for representing how quantities are related and can help us answer questions about a situation. 

Example: A school receives 46 copies of a popular book. The library takes 26 copies and the remainder are split evenly among 4 teachers. How many books does each teacher receive?

This situation involves a total formed by 4 equal parts and one other part. We can represent the situation with a diagram labeled 46 for the total number of books. That total length is divided into parts—one long part labeled 26 for the books given to the library and 4 equal-sized parts for the books split among 4 teachers. We label each of those parts with a variable, $x$, because we don’t know how many books each teacher got. Using the same variable, $x$, in each part means that the same number is represented four times.

Some situations have parts that are all equal, but each part has been increased from an original amount:

Example: A company manufactures a special type of sensor, and packs them in boxes of 4 for shipment. Then a new design increases the weight of each sensor by 9 grams. The new package of 4 sensors weighs 76 grams. How much did each sensor weigh originally?

We can represent this situation with a rectangle representing a total of 76 split into 4 equal parts. Each part shows that the new weight, $x+9$, is 9 more than the original weight, $x$.

We have seen how tape diagrams represent relationships between quantities. Because of the meaning and properties of addition and multiplication, more than one equation can often be used to represent a single tape diagram.

Let’s take a look at two tape diagrams.

We can represent this diagram with several different equations. Here are some of them:

• $26 + 4x=46$, because the parts add up to the whole.
• $4x+26=46$, because addition is commutative.
• $46=4x+26$, because if two quantities are equal, it doesn’t matter how we arrange them around the equal sign.
• $4x=46-26$, because one part (the part made up of 4 $x$’s) is the difference between the whole and the other part.

Here are some equations that represent this diagram:

• $4(x+9)=76$, because multiplication means having multiple groups of the same size.
• $(x+9)\boldcdot 4=76$, because multiplication is commutative.
• $76\div4=x+9$, because division tells us the size of each equal part.

Many situations can be represented by equations. Writing an equation to represent a situation can help us express how quantities in the situation are related to each other, and can help us reason about unknown quantities whose value we want to know. Here are two situations:

1. A camp counselor has a large bag that contains 34 cups of coconut. She uses 10 cups to make some trail mix. Then she uses the rest of the bag to make 144 identical granola bars. Campers want to know how much coconut is in each bar.

2. Kiran is trying to save $144 to buy a new guitar. He has $34 and is going to save $10 a week from money he earns mowing lawns. He wants to know how many weeks it will take him to have enough money to buy the guitar.

We see the same three numbers in the situations: 10, 34, and 144. How could we represent each situation with an equation?

In the camp situation, there is one part of 10 and then 144 equal parts of unknown size that all add together to 34. This can be represented by the equation $10+144x=34$. Since 24 is needed to get from 10 to 34, the value of $x$ is $(34−10) \div 144$ or $\frac16$. There is $\frac16$ cup of coconut in each serving.

In Kiran’s situation, there is one part of 34 and then an unknown number of equal parts of size 10 that all add together to 144. This can be represented by the equation $34+10x=144$. Since it takes 11 groups of 10 to get from 34 to 144, the value of $x$ in this situation is $(144-34)\div{10}$, or 11. It will take Kiran 11 weeks to raise the money for the guitar.

In this lesson, we encountered two main types of situations that can be represented with an equation. Here is an example of each type:

1. After adding 8 students to each of 6 same-sized teams, there were 72 students altogether.

2. After adding an 8-pound box of tennis rackets to a crate with 6 identical boxes of table tennis paddles, the crate weighed 72 pounds.

The first situation has all equal parts, since additions are made to each team. An equation that represents this situation is $6(x+8)=72$, where $x$ represents the original number of students on each team. Eight students were added to each group, there are 6 groups, and there are a total of 72 students.

In the second situation, there are 6 equal parts added to one other part. An equation that represents this situation is $6x+8=72$, where $x$ represents the weight of each box of table tennis paddles. There are 6 boxes of table tennis paddles, an additional box that weighs 8 pounds, and the crate weighs 72 pounds altogether.

In the first situation, there were 6 equal groups, and 8 students added to each group. $6(x+8)=72$.

In the second situation, there were 6 equal groups, but 8 more pounds in addition to that. $6x+8=72$.

In this lesson, we worked with two ways to show that two amounts are equal: a balanced hanger and an equation. We can use think about the weights on a balanced hanger to understand steps we can use to find an unknown amount in a matching equation.

This hanger diagram shows a total weight of 7 units on one side that is balanced with 3 equal, unknown weights and a 1-unit weight on the other. An equation that represents the relationship is $7=3x+1$.

We can remove a weight of 1 unit from each side and the hanger will stay balanced. This is the same as subtracting 1 from each side of the equation. 

Balanced hanger, left side, 6 blue squares and one red squared being removed. Right side, 3 green squares, and one red square being removed. To the side, an equation 7 minus 1 = 3 x  + 1 minus 1, with each minus 1 written in red.

An equation for the new balanced hanger is $6=3x$.

We can make 3 equal groups on each side and the hanger will stay balanced. This is the same as dividing each side of the equation by 3 (or multiplying each side by $\frac13$ ). In other words, the hanger will balance with $\frac13$ of the weight on each side.

Balanced hanger, left side 6 blue squares, right side, three green circles.  A dotted line is drawn around each of 3 groups, each groups consists of two blue square s from the left side and one green circle from the right side. To the side, an equation says 1/3 6 = 1/3 3 x.

The two sides of the hanger balance with two 1-unit weights on one side and 1 weight of unknown size on the other side. So, the unknown weight is 2 units.

Here is a concise way to write the steps above:

$\begin{aligned} 7&=3x+1 & \\ 6&=3x & \text{after subtracting 1 from each side} \\ 2 &= x & \text{after multiplying each side by } \tfrac13 \\ \end{aligned}$

The balanced hanger diagram shows the amounts on the left equal the amounts on the right. The left side has 3 pieces that each have unknown weight $x$ and 3 pieces that each weigh 2 units. So, the left side shows 3 $x$’s plus 6 units. The right side shows 18 units.  We could represent this diagram with an equation and solve the equation the same way we did before.

$\begin{aligned} 3x+6&=18 \\ 3x&=12 \\ x&=4 \\ \end{aligned}$

Since there are 3 groups of $x+2$ on the left, we could represent this hanger with a different equation: $3(x+2)=18$.

The two sides of the hanger balance with these weights: 3 groups of $x+2$ on one side, and 18, or 3 groups of 6, on the other side.

Balanced hanger, left side, circle labeled x, square labeled 2, circle labeled x, square labeled 2, circle labeled x, square labeled 2, right side, rectangle not labeled.  A dotted line is drawn around three groups, each group contains one circle and one square from the left side and a third of the rectangle on the right side.  To the side, an equation says 1/3 3 ( x + 2 ) = 1/3 18.

The two sides of the hanger will balance with $\frac13$ of the weight on each side:

We can remove 2 units of weight from each side, and the hanger will stay balanced. This is the same as subtracting 2 from each side of the equation.

Balanced hanger. Left side, circle labeled x and square labeled 2, the square appears to be loose from the hanger. Right side, rectangle labeled 4 and square labeled 2, the square appears to be loose from the hanger.  To the side, an equation says x + 2 - 2= 6 - 2.

An equation for the new balanced hanger is $x=4$. This gives the solution to the original equation.

Here is a concise way to write the steps above:

$\begin{aligned} 3(x+2) &= 18 \\ x + 2 &= 6 & \text{after multiplying each side by } \tfrac13 \\ x &= 4 & \text{after subtracting 2 from each side} \\ \end{aligned}$

To find a solution to some equations, we can just think about what value in place of the variable would make the equation true. Sometimes we also draw diagrams to reason about the solution. Using balanced hanger diagrams helped us understand that doing the same thing to each side of an equation keeps the equation true. So, another way to solve an equation is to perform the same operation on each side in order to get the variable alone on one side.

Doing the same thing to each side of an equation also works when an equation involves negative numbers. Here are some examples of equations that have negative numbers and steps we could take to solve them.

Example:

$\begin{aligned} 2(x-5) &= \text-6 \\ \tfrac12 \boldcdot 2(x-5) &= \tfrac12 \boldcdot (\text-6) & \text{Multiply each side by }\tfrac12 \\ x-5 &= \text-3 \\ x-5+5 &= \text- 3 + 5 & \text{ Add 5 to each side} \\ x &= 2 \\ \end{aligned}$

Example:

$\begin{aligned} \text-2x + \text-5 &= 6 \\ \text-2x + \text-5 - \text-5 &= 6 - \text-5 & \text{Subtract -5 from each side} \\ \text-2x &= 11 \\ \text-2x \div \text-2 &= 11 \div \text-2 & \text{Divide each side by -2} \\ x&=\text- \tfrac{11}{2}\\ \end{aligned}$

Doing the same thing to each side maintains equality even if it is not helpful for finding the solution. For example, we could take the equation $\text-3x +7=\text-8$ and add -2 to each side: 

$\begin{aligned} \text-3x+7 &= \text-8 \\ \text-3x + 7 + \text-2 &= \text-8 + \text-2 & \text{Add -2 to each side}\\ \text-3x+5 &= \text-10 \\ \end{aligned}$

If $\text-3x+7=\text-8$ is true then $\text-3x+5=\text-10$ is also true, but we are no closer to a solution than we were before adding -2. We can use moves that maintain equality to make new equations that all have the same solution. Helpful combinations of moves will eventually lead to an equation like $x=5$, which gives the solution to the original equation (and every equation we wrote in the process of solving).

Equations can be solved in many ways. In this lesson, we focused on equations with a specific structure, and two specific ways to solve them. 

Suppose we are trying to solve the equation $\frac45(x+27)=16$. Two useful approaches are:

• Divide each side by $\frac45$.
• Apply the distributive property.

In order to decide which approach is better, we can look at the numbers and think about which would be easier to compute. We notice that $\frac45 \boldcdot 27$ will be hard, because 27 isn't divisible by 5.  So, distributing the $\frac45$ is not the best method. But $16 \div \frac45$ gives us $16 \boldcdot \frac54$, and 16 is divisible by 4. So, dividing each side by $\frac45$ is a good choice.

$\begin{aligned} \tfrac45 (x+27) &= 16 \\ \tfrac54 \boldcdot \tfrac45 (x+27) &= 16 \boldcdot \tfrac54 \\ x+27 &= 20 \\ x &= \text- 7 \\ \end{aligned}$

Sometimes the calculations are simpler if we first use the distributive property. Let's look at the equation $100(x+0.06)=21$. If we first divide each side by 100, we get $\frac{21}{100}$  or 0.21 on the right side of the equation. But if we use the distributive property first, we get an equation that only contains whole numbers. 

$\begin{aligned} 100(x+0.06) &= 21 \\ 100x+6 &= 21 \\ 100x &= 15 \\ x &= \tfrac{15}{100} \\ \end{aligned}$

Many problems can be solved by writing and solving an equation. Here is an example:

Clare ran 4 miles on Monday. Then for the next 6 days, she ran the same distance each day. Clare ran a total of 22 miles during the week. How many miles did she run on each of the 6 days?

One way to solve the problem is to represent the situation with an equation, $4+6x = 22$, where $x$ represents the distance, in miles, Clare ran on each of the 6 days. Solving the equation gives the solution to this problem.

$\begin{aligned} 4+6x &= 22 \\ 6x &= 18 \\ x &= 3 \\ \end{aligned}$

Clare ran 3 miles each day.

We can solve problems where there is a percent increase or decrease by using what we know about equations. For example, a camping store increases the price of a tent by 25%. A customer then uses a $10 coupon for the tent and pays $152.50. We can draw a diagram that shows first the 25% increase and then the $10 coupon.

Three tape diagrams of unequal length. Top diagram, original price, one part labeled p. Middle diagram, labeled 25% increase, 4 equal parts which total to the same length as p above, with another equal part on the end labeled point 25 p. Third diagram, same total length as diagram above, labeled 10 dollar coupon, first part labeled 152 point 50, second part, dotted outline, labeled 10.

The price after the 25% increase is $p+0.25p$ or $1.25p$. An equation that represents the situation including the $10 off for the coupon is LATEX2. To find the original price before the increase and discount, we can add 10 to each side and divide each side by 1.25, resulting in LATEX3. The original price of the tent was $130.

Inequalities can be used to describe a range of numbers. For example, in many places, people are eligible to get a driver’s license when they are at least 16 years old. If $h$ is the age of a person, then we can check if they are eligible to get a driver’s license by checking if their age makes the inequality $h>16$ (they are older than 16) or the equation $h=16$ (they are 16) true. The symbol $\geq$, pronounced “greater than or equal to,” combines these two cases and we can just check if $h \geq 16$ (their age is greater than or equal to 16).

The inequality $h \geq 16$ can be represented on a number line. The closed, or filled in, circle at 16 shows that 16 is a solution. The shading and arrow pointing right from 16 shows that all numbers greater than 16 are also solutions.

We can write and solve inequalities to solve problems.

Example:  Elena has $5 and sells pens for $1.50 each. Her goal is to save $20. We could solve the equation LATEX0 to find the number of pens, LATEX1, that Elena needs to sell in order to save exactly $20. Adding $-5$ to both sides of the equation gives us $1.5x=15$, and then dividing both sides by 1.5 gives the solution $x=10$ pens.

What if Elena wants to save more than $20? The inequality LATEX5 tells us that the amount of money Elena saves needs to be greater than $20. The solution to the previous equation will help us understand what the solutions to the inequality will be. We know that if Elena sells 10 pens, she will save exactly $20. Since each pen gives her more money, she needs to sell more than 10 pens to save more than $20. So, we can represent all the solutions to the inequality with another inequality: $x>10$.  A solution to an inequality is a number that can be used in place of the variable to make the inequality true.

Here is an inequality: $3(10-2x) < 18$. The solution set for this inequality is all the values that can be used in place of $x$ to make the inequality true. Each solution is one value that makes the inequality true.

In order to solve this inequality, we can first solve the related equation $3(10-2x) = 18$ to get the solution $x = 2$. That means 2 is the boundary between values of $x$ that make the inequality true and values that make the inequality false.

To solve the inequality, we can check numbers greater than 2 and less than 2 and see which ones make the inequality true.

Let’s check a number that is greater than 2: $x= 5$. Replacing $x$ with 5 in the inequality, we get $3(10-2 \boldcdot 5) < 18$ or just $0 < 18$. This is true, so $x=5$ is a solution. This means that all values greater than 2 make the inequality true. We can represent the solutions as $x > 2$ and also represent the solutions on a number line:

Notice that 2 itself is not a solution because it's the value of $x$ that makes $3(10-2x)$ ​equal to 18, and so it does not make $3(10-2x) < 18$ true.

For confirmation that we found the correct solution, we can also test a value that is less than 2. If we test $x=0$, we get $3(10-2 \boldcdot 0) < 18$ or just $30 < 18$. This is false, so $x = 0$ and all values of $x$ that are less than 2 are not solutions.

Many real-world problems can be represented and solved by using inequalities. Writing inequalities is very similar to writing equations to represent a situation. The expressions that make up the inequalities can be thought of in much the same way as the expressions that make up equations. For inequalities, we also have to think about how expressions compare to each other—which one has a greater value, which one has a lesser value, and can they also be equal?

For example, a school fundraiser has a minimum target of $500. Faculty have donated $100 and there are 12 student clubs that are participating with different activities. How much money would each club need to raise if the 12 clubs shared the responsibility of meeting the goal equally? If $n$ is the amount of money that each club raises, then the solution to $100+12n=500$ is the amount each club has to raise to meet the goal. It is more realistic, though, to use the inequality $100+12n\geq500$, since the more money raised, the more successful the fundraiser. There are many solutions because there are many different amounts of money the clubs could raise that would get them above their minimum goal of $500.

We can write inequalities to represent situations and solve problems. First, it’s important to decide what quantity we are representing with a variable. Next, we can connect the quantities in the situation to write an expression. Then we choose an inequality symbol and complete the inequality.

When solving the inequality to answer a question about the situation, it’s important to keep the meaning of each quantity in mind. This helps us decide if the solution to the inequality makes sense for the situation.

Example: Han has 50 centimeters of wire and wants to make a square picture frame with a loop to hang it. He uses 3 centimeters for the loop. If Han wants to use all the wire, this situation can be represented by the equation $3+4s=50$, where $s$ is the length of each side in centimeters.

If Han doesn’t need to use all the wire, we can represent the situation with the inequality $3+4s\leq50$. The solution to this inequality is $s \leq 11.75$. However, not all solutions to this inequality make sense for the situation. For example, we cannot have negative lengths or a side length of 0 centimeters.

In other situations, the variable may represent a quantity that increases by whole numbers, such as numbers of magazines, loads of laundry, or students. In those cases, only whole-number solutions make sense.

Properties of operations can be used in different ways to rewrite expressions and create equivalent expressions. For example, the distributive property can be used to expand an expression such as $3(x+5)$ to get $3x+15$.

The distributive property can also be used in the other direction to factor an expression such as $12x−8$. In this case, we know the product and need to find the factors.

The terms of the product go inside:

Think of a factor each term has in common: $12x$ and $-8$ each have a factor of 4. The common factor can be placed on one side of the large rectangle:

Now think: "4 times what is 12x?" and "4 times what is -8?" Write the other factors on the other side of the rectangle:

So, $12x-8$ is equivalent to $4(3x-2)$.

There are many ways to write equivalent expressions, and they may look very different from each other. One way to determine if two expressions are equivalent or not is to substitute the same number for the variable in both expressions.

For example, when $x$ is 1, the expression $2(\text-3+x)+8$ equals 4 and the expression $2x+5$ equals 7. This means $2(\text-3+x)+8$ and $2x+5$ are not equivalent.

If two expressions are equal when many different values are substituted for the variable, then the expressions may be equivalent—it is impossible to compare the two expressions for all values. To know for sure, we use properties of operations. For example, $2(\text-3+x)+8$ is equivalent to $2x+2$ because:

$\begin{aligned} 2(\text-3+x)+8\\ \text-6+2x+8 & \quad\text{by the distributive property}\\ 2x+\text-6+8 & \quad\text{by the commutative property}\\ 2x+(\text-6+8) & \quad\text{by the associative property} \\ 2x+2 \\ \end{aligned}$