Mathematics: Levels E, M, D, and A Study Guide for the TABE Test

Expressions and Equations: Part 1

Percentage of Test Level Specifically Assessing Expressions and Equations (— = Assumed)

L	E	M	D	A
not tested	not tested	15%	18%	—

Expressions and equations are at the very heart of algebra, especially if they involve variables. This section is about how to write and rewrite equations and expressions in an organized way to simplify or solve them.

Vocabulary

Here we will take a look at terms you should know that are used again and again in algebra.

Expression

Math expressions are combinations of numbers, variables, and symbols for operations such as $+, -, \times, \div,$ and others. An expression can be as simple as $5+3$ or $2x-10$, or even just a single number like $9$. A couple of less simple expressions are $x^2-4x+4$ and $6x+2y -13+ 5(3x-y)$. Pretty much any combination of numbers, variables, and signs of operation that you can imagine is a mathematical expression. An expression has no equality sign.

Equation

If you write two expressions with an equal sign between them, you have an equation. Equations can be really simple, such as $x=12$, or not so simple, such as $3x^2 +5x -11= 2x^2 -3x$.

Inequality

You need to be familiar with the two inequality signs.

The symbol $<$ means less than. Example: $x<5$ means $x$ has a value less than $5$. So, could $x=-2$? Yes, because $-2$ is less than $5$.

The symbol $>$ means greater than. Example: $x>10$ means $x$ has a value greater than $10$. So, could $x=-6$? No, because $-6$ is less than $10$, not greater.

Solving an inequality is a lot like solving an equation. The big difference is that, instead of finding the value of $x$, you will find a range of values for $x$. It may help to take a look at inequalities and number lines.

The inequality $x<-2$ means that $x$ can be any value less than $-2$. The number line below shows the same thing. The arrow going to the left means that $x$ can be any value less than $-2$ and the hollow dot at $-2$ means that $-2$ itself is not included in the values for $x$.

70 Showing an Inequality on a Number Line A.png

The inequality $x>-2$ means that $x$ can be any value greater than $-2$. Again, the number line shows the same thing. The arrow going to the right means that $x$ can be any value greater than $-2$, but the hollow dot at $-2$ again means that $-2$ is not included.

71 Showing an Inequality on a Number Line B.png

Here’s a sample problem that can be solved using an inequality.

The length of a rectangle is twice its width. If its perimeter must be less than $60$, what values can the width have?

First, write an expression for the perimeter. Let the width be $w$ and the length be $2w$. The perimeter expression for the four sides would be:

\[w + w + 2w + 2w\]

We can write that as $6w$.

The perimeter must be less than $60$, so we can write:

\[6w<60\]

Dividing both sides by $6$, we get:

\[w<10\]

So, the width can only have values less than $10$. It could be $3$. It could be $8.23$. One thing it couldn’t be is negative, because geometric figures never have negative dimensions. The best answer for this problem is actually a more complicated expression: $x>0$ and $x<10$, meaning $x$ has to be greater than $0$ and less than $10$.

Constant

As mentioned earlier in this guide, a constant is simply a number like $22$ or $-8$ or $\sqrt{3}$. The only reason they have the special name constant is to distinguish them from variables.

Variable

Variables have come up before in this guide, where we saw that a variable is a letter that stands for a number. Popular variables are $x$ and $y$. Because $x$ and $y$, or any other letter, can stand for any number, they are not constant and are called variables. The number $5$ always means five, never $6$ or $108$, but $x$ could mean any number. Algebra is the way we find out the value of $x$ in a particular problem.

In an equation with two variables, sometimes it’s useful to think of one variable as the independent variable and the other one as the dependent variable. This example should help show how that works.

Suppose a man is sawing a tree trunk into sections. Out of curiosity, he weighs a few of them. He finds that a $1$-foot long piece weighs $30$ pounds, a $3$-foot piece weighs $90$ pounds, and a $6$-foot piece weighs $180$ pounds. Letting $w$ be weight and $l$ be length, we can make a table:

\[\begin {array}{|c|c|} \hline l & w \\ \hline 1 & 30\\ \hline 3 & 90\\ \hline 6 & 180\\ \hline \end{array}\]

Can we write a simple equation with two variables that shows the relationship between weight and length? Yes. We can see that the weight is always $30$ times the length, so we can write this: $w = 30\ l$. It tells us how changing the length with a saw will make the weight change. The weight depends on the length in a specific way. For that reason, $w$, the weight, is called the dependent variable. That makes $l$, the length, the independent variable.

A good way to think of this is to think of cause and effect. Changing the value of $l$, the cause, produces a change in $w$, the effect.

The value that is deliberately changed is the independent variable, and the value produced is the dependent variable.

Evaluating an Expression

To evaluate an expression is to reduce the expression to a single numeric value. This is done by substituting a number for the variable(s) in the expression and then simplifying the expression. An example should help make this clear.

If $x=3$, evaluate the expression $3(x^2 -11)$.

Substitute $3$ for $x$:

\[3(3^2 -11)\] \[3(9-11)\] \[3(-2)\] \[-6\]

You will always be given the value of the variable and the expression.

Solving an Equation

Basic equation solving means using the rules below to end up with the equation $x= \text{some number}$. In other words, you want to get $x$ by itself on one side and a single number on the other side.

You can:

Add the same number to both sides of the equation.
Subtract the same number from both sides of the equation.
Multiply the same number times both sides of the equation.
Divide both sides of the equation by the same number.

Let’s do a few examples.

Solve $x+14 = 21$

Since there is a $14$ on the side with the $x$, we need to get rid of it. All we need to do is subtract $14$ from the left side, but we also have to subtract it from the right side. We can show it like this:

\[x+14-14=21-14\]

Then:

\[x+0 = 7\]

Which we can write without the zero:

\[x=7\]

This is the solution to the equation.

Here’s another example:

Solve $9x=45$

There is a $9$ on the left side with the $x$, so we have to get rid of it. Subtracting does not work in this case because the $9$ is multiplied by the $x$. To make the $9$ go away in this case, we need to divide both sides by $9$.

\[9x \div 9 = 45 \div 9\] \[x = 5\]

Summary: To make a number go away, you always need to do the opposite of what is happening in the equation:

To get rid of an added number, you subtract it from both sides.
To get rid of a subtracted number, you add it to both sides.
To get rid of a multiplied number, you divide both sides by it.
To get rid of a divided number, you multiply both sides by it.

Very often, you will need to use two or more of the rules to solve an equation.

Solve: $2x+4 = 22$

First, subtract $4$ from both sides:

\[2x+4-4=22-4\] \[2x=18\]

Now divide both sides by $2$:

\[2x\div2 = 18 \div 2\] \[x=9\]

Here’s an example of a word problem using these basic rules.

Maria bought four concert tickets and spent $\$30$ dollars on souvenirs. If she spent a total of $\$270$, what was the price of a concert ticket?

Let $x$ be the price of a ticket. The amount she spent on tickets was $4x$. Therefore, she spent $4x$ and $\$30$. We know the total she spent was $\$270$, so we can write this equation (we’ll leave the dollar signs out for clarity):

\[4x+30=270\]

Subtract $30$ from both sides:

\[4x+30-30 = 270 -30\] \[4x = 240\]

Divide both sides by $4$:

\[4x \div 4 = 240 \div4\] \[x=60\]

So, the tickets cost $\$60$ each.

It’s always good practice to ask yourself if your answer is reasonable. Clearly, an answer of $\$6$ or $\$600$ per ticket would be pretty far off. Mentally trying $\$6$ for a ticket $(x)$ would give you only $\$24$ for the tickets. Add $\$30$ and you’ll get $\$54$, nowhere near $\$270$. Likewise, using $\$600$ per ticket would give you $\$2\text{,}400$ plus $\$30$, or $\$2\text{,}430$, way bigger than $\$270$.

If the numbers are not easy enough to mentally do the math, round them to simpler numbers and use those.

Distance and Time

There are a few ways to mathematically represent the motion of an object. Suppose a child on a tricycle pedals down the sidewalk and every second he goes four feet. We could show this by making a little table showing his total distance traveled in different amounts of time.

\[\begin {array}{|c|c|} \hline \text{Time (seconds)} & \text{Distance (feet)}\\ \hline 1 & 4\\ \hline 5 & 20\\ \hline 10 & 40\\ \hline \end{array}\]

Another thing we could do is use the three pairs of numbers in the table to make a graph.

72 Distance-Time Graph.png

A third way to represent the tricycle’s motion is to write an equation. If you look at the points in the table, you can likely see that the distance is always four times the time. That leads directly to this equation, where $d$ is distance and $t$ is time:

\[d=4\ t\]

So, those are the three different ways to represent a distance and time situation.

Example problem:

Given a table of values for Car 1 and an equation for the motion of Car 2, can you tell which car is going faster?

Car 1:

\[\begin {array}{|c|c|} \hline \text{Time (seconds)} & \text{Distance (feet)}\\ \hline 1 & 30\\ \hline 2 & 60\\ \hline 3 & 90\\ \hline 4 & 40\\ \hline \end{array}\]

Car 2:

\[d=32 \ t\]

We can make a table for Car 2 using the given equation. Try putting the same values into the equation for $t$ that were used in the first table and make a table for Car 2.

\[\begin {array}{|c|c|} \hline \text{Time (seconds)} & \text{Distance (feet)}\\ \hline 1 & 32\\ \hline 2 & 64\\ \hline 3 & 96\\ \hline 4 & 128\\ \hline \end{array}\]

We can see that Car 2 went farther each second than Car 1, which means Car 2 is faster.

Equivalent Expressions

Suppose you have two expressions and substitute the same number into the variables in both expressions. If both expressions then give you the same value, the expressions are equivalent. Showing this is easier than it may sound. See the examples below.

Consider these two expressions:

$3x +8 -x$ and $4+2x+4$

Substitute any number you want for $x$. How about $2$? It’s easy.

$3\cdot 2 + 8-2$ and $4 + 2\cdot 2 +4$

$6 +8-2$ and $4+4+4$

$12$ and $12$

We see the same value for each expression and we will no matter what number we choose to substitute. That makes the original expressions equivalent.

Another, maybe easier, way is to simplify both expressions by rearranging and rewriting each expression:

$3x +8 -x$ and $4+2x+4$

Rearranging:

$3x-x+8$ and $2x +4+4$

Combining like terms:

$2x+8$ and $2x+8$

Seeing that both expressions are now the same is another way to show that they are equivalent.

Applying Operation Properties

This section is just a little elaboration on the section above. The important point here is that if you correctly use basic number properties in an expression, the new expression will always be equivalent to the original expression. Again, let’s look at some examples.

The commutative property states that order doesn’t matter in addition and multiplication. That means the expressions $x+22$ and $22+x$ are equivalent. Likewise, $7 \times x$ and $x \times 7$ are equivalent expressions.

The distributive property states that $a(x+y) = ax+ay$. By that property, $4(2x -8)$ and $8x-32$ are equivalent expressions. Likewise, you can take an expression like $3x-9$ and factor it as $3(x-3)$, sort of a reverse distributive operation.

The standard operations of $+, -, \times, \text{and } \div$ also fit into this section. Remember that like terms can be combined to produce a new expression that is equivalent to the original. For example, use addition to rewrite $x+x+y+y+y$ as $2x + 3y$.

Substitution

After you have solved an equation or an inequality, it’s always a good idea to check your result. You do that by substitution. Take your answer and substitute it back into the original equation or inequality and see if that value makes it true.

Is $x=2$ a correct solution to the equation $x^2-4x=4$? Substitute $2$ in for $x$ and see:

\[\begin {array}{l|r} x^2-4x & 4\\ 2^2-4(2) &4\\ 4-8 & 4\\ -4&4\\ \end{array}\]

We know that $-4$ does not equal $4$, so, no, $2$ is not a solution to the equation.

Is $x=10$ a correct solution to the inequality $3x-1>x+11$? Substitute $10$ in for $x$ and see:

\[\begin {array}{l|r} 3x-1&x+11\\ 30-1 &10+11\\ 29 & 21\\ \end{array}\]

We know that $29$ is greater than $21$, so $10$ is a correct solution to the inequality.