Mathematics Study Guide for the TABE Test

Algebraic Concepts: Expressions and Equations—Part 1

Now that you have learned about the essential elements and tools of expressions and equations, you are prepared to learn how to manipulate them to make them easier to use or solve. This section is about how to write and organize equations and expressions to simplify or solve them.

Evaluating an Expression

Some questions will ask you to evaluate an expression, which simply means to reduce the expression to a single numeric value. This is done by substituting a number for the variable (or variables) in the expression and then simplifying the expression. So, for instance, suppose you are given this expression:

\[3(x^2 -11)\]

A question on the TABE might ask you to evaluate that expression assuming that \(x=3\). So, to answer the question, we substitute \(3\) for \(x\):

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

Notice how the steps followed the order of operations you learned earlier.

Solving an Equation

Solving an equation means finding the value of the variable, which generally will require isolating the variable on one side of the equal sign. As you’ve learned, \(x\) is the most common variable, so often your answer will state \(x=\) some number. If there is more than one variable in an equation, then you may be asked to find the value in terms of one or more of the variables. For example, the equation \(x = 5y + 3z\) has written \(x\) in terms of \(y\) and \(z\).

There are many steps you can take when solving an equation, but the important thing to remember is that any action on one side of the equation must also happen on the other. So, if you add \(6\) to one side, you must add \(6\) to the other. Likewise, if you divide one side by \(2x\), the other side must also be divided by \(2x\). Very often, you will need to do two or more of these steps to solve an equation.

Let’s do a couple examples.

Solve: \(9x=45\)

Solution

There is a \(9\) on the left side with the \(x\), so we have to get rid of it. Addition or subtraction won’t work in this case because the \(9\) is multiplied by the \(x\). To make the \(9\) go away, we need to divide both sides by \(9\):

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

Solve: \(2x+4 = 22\)

Solution

Here we have to complete two steps, because we have to get rid of both the \(2\) and the \(4\) on the side of the \(x\). There’s no rule that dictates which one you have to get rid of first, but, in this case, it will be simpler to start by subtracting \(4\) from both sides:

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

Now, we’ll divide both sides by \(2\):

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

Solving an Inequality

You learned about inequalities earlier when you learned the four different signs:

• \(<\) means less than
• \(>\) means greater than
• \(\le\) means less than or equal to
• \(\ge\) means greater than or equal to

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 \(x\). This can be a bit difficult to understand at first, so it’s helpful to visualize the answer using a number line.

The inequality \(x<-2\) means that \(x\) can be any value less than \(-2\). The number line below illustrates this:

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 \(x\).

This can also work with the greater/less than or equal to symbols. For instance, we can have the inequality \(x\ge-2\), which means that \(x\) can be any value greater than or equal to \(-2\). On the number line, that looks like this:

Again, we’ve illustrated this on the number line. The arrow going to the right means that \(x\) can be any value greater than \(-2\), but the filled-in dot at \(-2\) means that \(-2\) is also included.

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?

Solution

You have some practice converting word problems into equations. Now you’re going to convert this problem into an inequality. First, write an expression for the perimeter using variables. If the width is \(w\), then the length is \(2w\), because it’s twice the width. Therefore, the perimeter expression for the four sides is:

\[p = w + w + 2w + 2w\]

We can write that as \(p=6w\).

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. In fact, 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 \(10\).

Equivalent Expressions

Since expressions use variables, and variables can have any value, sometimes one expression can be the same as another, even if their constants and coefficients are completely different. When two expressions are the same, we say they are equivalent.

Consider these two expressions:

\[3x +8 -x\] \[4+2x+4\]

They don’t look the same, but if we pick a value for \(x\), you will see they actually are equivalent. Let’s try substituting \(2\) for the variable:

\[3\cdot 2 + 8-2 = 6 +8-2 = 12\] \[4 + 2\cdot 2 +4 = 4+4+4 = 12\]

We got the same value for each expression, and we would no matter what number we chose to substitute.

It can sometimes be easier to simplify both expressions by rearranging and rewriting each expression:

\[3x +8 -x\] \[4+2x+4\]

Rearranging these expressions, we get:

\[3x-x+8\] \[2x +4+4\]

Now, we combine like terms to get:

\[2x+8\] \[2x+8\]

Rearranging both expressions to make them the same is another way to show that they are equivalent.

Applying Operation Properties

This section elaborates on what you just learned. 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.

For example, the commutative property that you learned earlier when reading about operation properties 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.

Furthermore, 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 \(3(x-3)\), which is sort of a reverse distributive operation.

Finally, remember that like terms can be combined to produce a new expression that is equivalent to the original. For example, the basic rules of addition mean we can rewrite \(x+x+y+y+y\) as \(2x + 3y\).

Substitution

We use the term substitution to mean replacing a variable with a specific value. Many questions will ask you to substitute a value into an equation or inequality to solve it. Another way of saying this is that the given value is the solution for the equation or inequality.

For example, is \(x=2\) a correct solution to the equation \(x^2-4x=4\)? To find out, let’s substitute \(2\) in for \(x\):

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

We know that \(-4\) does not equal \(4\), so \(2\) is not a solution to the equation.

Let’s try another. 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.

Substitution is also used to check your answer after you have solved an equation or an inequality to find the value of a variable. Just take your answer and substitute it back into the original equation or inequality and see if that value makes it true.

Expressions with Exponents and Roots

As you move further into algebra, you will often work with expressions that include exponents and roots. These concepts allow us to represent repeated multiplication and undo that multiplication in a structured way. You learned about exponents and roots earlier, but now you’ll be doing them with variables. That can make these processes seem more confusing, but they work exactly the same.

To review, an exponent tells you how many times a number is multiplied by itself. For example:

\[3^4 = 3 \times 3 \times 3 \times 3 = 81\]

A root does the opposite. It asks, “What number multiplied by itself a certain number of times gives this result?” For example:

\[\sqrt{25} = 5\]

Properties of Exponents

There are several important properties that help us rewrite and simplify expressions involving exponents. We discussed these previously, but it will be good to review them as they are vital rules to know when the math gets more complex:

1. product of powers (same base)—When multiplying powers with the same base, add the exponents:

\[x^3 \cdot x^5 = x^{3+5} = x^8\]

2. quotient of powers (same base)—When dividing powers with the same base, subtract the exponents:

\[\frac{x^7}{x^2} = x^{7-2} = x^5\]

3. power of a power—When raising a power to another power, multiply the exponents:

\[(x^3)^4 = x^{3 \cdot 4} = x^{12}\]

4. power of a product—Apply the exponent to each factor inside parentheses:

\[(2x)^3 = 2^3 \cdot x^3 = 8x^3\]

5. zero exponent rule—Any nonzero number raised to the power of zero equals \(1\):

\[x^0 = 1\]

6. Negative Exponents—A negative exponent means take the reciprocal:

\[x^{-3} = \frac{1}{x^3}\]

Working with Roots and Radicals

Radicals (like square roots and cube roots) can also be written using exponents. This helps connect roots and exponents into one system. For example:

\[\sqrt{x} = x^{\frac{1}{2}}\] \[\sqrt[3]{x} = x^{\frac{1}{3}}\]

More generally:

\[\sqrt[n]{x^m} = x^{\frac{m}{n}}\]

These are called rational exponents because the exponent is a fraction.

Rewriting Expressions using Exponents

Being able to switch between radicals and exponents is an important skill to master. For example, this is how we can rewrite \(\sqrt{y^5}\) using exponents:

\[\sqrt{y^5} = (y^5)^{\frac{1}{2}} = y^{\frac{5}{2}}\]

Likewise, this is how we write \(x^{\frac{3}{2}}\) using radicals:

\[x^{\frac{3}{2}} = \sqrt{x^3}\]

We can also simplify using the exponent rules:

\[x^{\frac{1}{2}} \cdot x^{\frac{3}{2}} = x^{\frac{1}{2} + \frac{3}{2}} = x^2\]

Rewriting expressions using exponent properties allows you to simplify complex expressions, combine like terms more easily, and solve equations involving powers and roots.

Scientific Notation

Scientific notation is a way of writing very large and very small numbers that doesn’t require as much space. Instead of writing numbers like \(3\text{,}000\text{,}000\text{,}000\) or \(0.00000007\) with all those zeros, we can use powers of ten, which are specific types of exponential expressions in which the base is \(10\).

Let’s look at the process using the number \(3\text{,}000\text{,}000\text{,}000\). To write this number in terms of a power of ten, move the decimal point left until it’s just to the right of \(3\), giving you \(3.000000000\). You’ll be able to drop all those zeros, but first, count how many there are. In this case, there are nine, so that becomes the exponent of your exponential expression with base \(10\). Finally, multiply the \(3\) by the \(10^9\) to get your original number in scientific notation:

\[3 \times 10^9\]

Scientific notation is always written in that form. The first part, known as the significand, must be between \(0\) and \(10\), but it may be a decimal number. For example, the distance from here to the Sun is \(9.3 \times 10^7\) miles.

But what if the number you start with is very small? The process is almost the same, but you move the decimal point to the right and make the power of ten negative.

So, say you have the number \(0.00000007\). First, count the total number of zeros, including the one to the left of the decimal place. There are eight. That is going to be your exponent (but negative). Now, move the decimal point to the right until it’s just to the right of the \(7\) and drop all the zeros. You’re going to create the scientific notation like you did before, with \(7\) multiplied by a power of ten. This time, though, the exponent is negative, so you have \(10^{-8}\), and your result is:

\[7 \times 10^{-8}\]

Comparing Numbers in Scientific Notation

You should be able to make comparisons between two different quantities written in scientific notation. Here is how that’s done.

First, understand that each power of ten is \(10\) times greater than a power that is one number smaller:

\[\begin{array}{|c|c|} \hline 10^2 & 10 \\ \hline 10^3 & 100 \\ \hline 10^4 & 1\text{,}000 \\ \hline 10^5 & 10\text{,}000 \\ \hline 10^6 & 100\text{,}000 \\ \hline \end{array}\]

Also, each power of ten is \(100\) times the one that is two numbers smaller, \(1{,}000\) times the one that is three numbers smaller, and so on. Likewise, \(3 \times 10^8\) is \(10\) times bigger than \(3 \times 10^7\).

Now, if we have \(6 \times 10^8\) and \(3 \times 10^7\), we know that \(10^8\) is \(10\) times bigger than \(10^7\). Also, we know that \(6\) is two times bigger than \(3\). Therefore, we can conclude that \(6 \times 10^8\) is \(10 \times 2 = 20\) times bigger than \(3 \times 10^7\).

So, to make a comparison between numbers written in scientific notation, you have to take into account the power of \(10\) and the significand.

Pairs of Equations

You now have experience working with equations with a single variable. But what happens when there are two unknowns and thus two different variables? A single equation like \(x+y=12\) shows two numbers adding up to \(12\). It can’t be solved except to say that there are an infinite number of solutions that will make it true:

\[(3+9), (4+8), (10+2), (11+1), (6\frac{1}{2} + 5\frac{1}{2}) …\]

However, if you are given a pair of distinct (not equivalent) equations in terms of two variables (e.g., \(x\) and \(y\)), you can find a solution. These pairs of equations in two unknowns are called simultaneous equations. Say you are given these two equations:

\[x+y=12\] \[x-y=6\]

The simultaneous equations show two numbers adding up to \(12\) and the same two numbers having a difference of \(6\). What two numbers would do both of these things? After thinking for a while, you may come up with \(x=9\) and \(y=3\):

\[9+3=12\] \[9-3=6\]

This wasn’t hard to get because the numbers were simple. If the numbers are not so simple, though, you will need a method for solving the problem. There are actually multiple methods.

One method, known as the elimination method, involves adding or subtracting the two equations to get rid of one of the variables. In the simple example from above, we can add the two equations and the \(y\) variable is eliminated, leaving an equation we can solve:

\[\begin{array}{llclcr} & &x&+&y&=&12\\ &+&x&-&y&=&6\\ \hline & &2x& & & =&18\\ & &x& & & =&9\\ \end{array}\]

Now that we know \(x = 9\), we can substitute \(9\) for \(x\) in either of the original equations. Let’s choose the second equation:

\[x-y=6\] \[9-y=6\] \[y=9-6\] \[y=3\]

We see again that \(x=9\) and \(y=3\).

Another method for solving two equations is the substitution method. This involves isolating one of the variables in one of the equations. Then we substitute that variable’s equivalent value into the other equation. For example, here are two new equations:

\[2x+y=18\] \[x+y=11\]

We need to get one of the variables in one of the equations on its own. Let’s use the second equation and isolate the \(y\):

\[y=11 - x\]

Now, we substitute that value for \(y\) into the first equation and do basic math operations:

\[2x+(11-x)=18\] \[x + 11 = 18\] \[x = 7\]

We have the value for \(x\), so we’ll input that into one of the equations and get the value of \(y\):

\[7+y=11\] \[y=4\]

So, our answer is \(x=7\) and \(y=4\).

Feel free to verify the answer by substituting the values back into the original equations.

Graphing Linear Equations

We often try to find the relationship between two numbers or variables, which can be represented with linear equations. As you’ve learned, tables and graphs give us visual assistance in doing so.

For example, suppose a crop scientist is experimenting with fertilizing corn to see how it affects the amount of corn that can be grown in a season. In one garden plot, she puts five pounds of fertilizer. In the next three plots, she puts \(10\) pounds, \(15\) pounds, and \(20\) pounds. Then, in the fall, she measures the number of bushels of corn in each plot.

As you learned earlier, one method for displaying this data is a table, with the dependent variable in the first column and the independent variable in the second column. Let’s suppose the scientist’s table looks like this:

Fertilizer (lb)	Corn (bu)
\(5\)	\(6\)
\(10\)	\(12\)
\(15\)	\(18\)
\(20\)	\(24\)

With that data, we can create a linear equation. In this case, we will say that “Yield” is equal to “Fertilizer weight” times \(1.2\), which can be written as:

\[Y=1.2F\]

It might not be obvious that this is the right equation, but if you try putting the numbers from the data table in it, you’ll see that it works.

Now, we have another method for displaying this data: graphing the linear equation. The numbers in the table can be graphed, with the independent variable on the horizontal axis. Or you can input numbers into the equation, with the \(F\) variable being the \(x\)-value and the \(Y\) variable being the \(y\)-value:

The graph shows clearly that increasing the fertilizer increases the amount of corn grown.

Quick reality check: It’s incredibly unlikely that any real experiment done with plants and fertilizer would ever have a perfectly linear graph like that. It’s just in math problems that we see that kind of thing.

Slope and Intercept

Linear equations can all be written in the form \(y=mx +b\), where \(m\) is the slope of the line and \(b\) is the \(y\)-intercept, the point on the \(y\)-axis where the line crosses. Another way to think of slope is as rise over run, with rise referring to the movement along the \(y\)-axis and run referring to the movement along the \(x\)-axis. In other words, you are dividing the distance moved vertically by the distance moved horizontally.

Remember when you learned about distance-time graphs earlier in this guide. Here is an example:

In the above graph, we are shown that the rise is \(20\) feet and the run is five seconds. The slope of the line is therefore \(\frac{20}{5}\) or \(4\). Since the line crosses the \(y\)-axis at \(0\), the formula for this line is \(y = 4x\).

With any graph of a line, even if we don’t know the actual values of the rise and run, we can simply pick any two points on the line and find its slope using this formula:

\[m= \frac{y_2-y_1}{x_2-x_1}\]

In this case, we’ll use the points \((5, 20)\) and \((10, 40)\):

\[m = \frac{40-20}{10-5} = \frac{20}{5} = 4\]

If this was a typical algebra problem with \(x, y\) variables, we’d be done, but this graph is showing the motion of an object, and we can squeeze a little more information out of it. Namely, we can determine the speed of the object.

The \(y\)-axis has the units in feet, and the \(x\)-axis has the units in seconds. If we put the units into our calculation, we will have feet per second, abbreviated as ft/s. That’s a speed. Including the \(4\) that we have for a slope, our final conclusion is that the object is moving at a speed of \(4\) ft/s.

Solving Linear Equations in One Variable

As you have learned, solving an equation involves finding the value of one or more variables. Many of the equations you’ve already solved in this guide have been linear equations, meaning their variables don’t have a power greater than \(1\). That makes them fairly simple to solve and to graph.

When a linear equation only has one variable, we say we have a linear equation in one variable, such as:

\[6 = 2x + 5\] \[12y = 6 - 6y + 3\]

Both of these are linear equations. Furthermore, they are both easily solved based on what you’ve already learned. Let’s quickly solve them:

\[6 = 2x + 5\]

\[6 -5 = 2x + 5 - 5\] \[1 = 2x\] \[\frac{1}{2} = x\]

\[12y = 6 - 6y + 3\]

\[12y = 9 - 6y\] \[12y + 6y = 9 - 6y + 6y\] \[18y = 9\] \[y = \frac{9}{18} = \frac{1}{2}\]