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

Algebra

Percentage of Test Level Specifically Assessing Algebra (— = Assumed)

L	E	M	D	A
not tested	not tested	not tested	not tested	28%

Algebra has been a big part of this study guide so far. In this section, we will revisit some of those topics and expand into a few new things.

Algebra is a very broad subject that involves exponents, radicals, logarithms, polynomials, simplifying and factoring, linear and quadratic equations, functions, word problems, and math concepts learned previously. Many real-life situations can be solved using algebra. Think of algebra as solving puzzles and finding values based on clues.

In algebra, we solve for unknown values with the use of other known values and relationships between these values. Unknown values are often assigned letters called variables. We don’t know their values yet, hence, we assign \(x\), \(y\), or any letter to represent them momentarily. The equal sign (=) is placed between expressions on the left and right to denote that the expression on the right is equal to the expression on the left.

In this equation, for instance:

\(x + 3 = 7\), we mean that if \(x\) is added to \(3\), we get \(7\).

Mathematical operators, such as subtraction (-), addition (+), multiplication (x), and division (\(\div\)), are used to define relationships between values. Instead of the equality sign (=), inequality signs such as less than or equal (\(\leq\)), greater than or equal (\(\geq\)), less than (<) and greater than (>) can also be used.

For example, a person travels by car and wants to know how fast he must drive to get from one city to another to catch a meeting in \(1.5\) hours. The distance between the two cities is \(60\) miles. The question can be rephrased as “What speed times the travel time of \(1.5\) hours will result in \(60\) miles?” In algebra, we assign variables to unknowns, instead of blanks or question marks. There are also rules followed and standard symbols used. That question is better written as:

\[x \cdot 1.5\text{ hr} = 60 \text{ mi}\]

That’s a very simple illustration, and there are a lot more complicated situations in algebra. The basic thing is to be able to figure out an equation that describes the question, designate symbols and variables, make use of known values and formulas, and follow algebraic rules.

Parts of an Expression

Algebraic expressions are combinations of terms and symbols. By symbols we mean \(+ \ \ - \ \ \times\) and \(\div\). Terms can be constants, like \(4, \ 19, \ \sqrt{7}, \ \frac{2}{3}, 5^2\), or they can be variables, like \(x\), \(y\), or \(z\). A term can also be a combination of a constant and a variable, such as \(9x\) or \(\frac{y^2}{4}\).

These are examples of algebraic expressions:

\[3x-4\] \[x^2 + 2x +1\] \[\sqrt{x^2 +y^2}\]

Notice that there is no equal sign in these. If one was there, you would have an equation, not an expression.

Other things to know:

In the terms \(6y\) and \(7x^2\), the \(6\) is a coefficient of the \(y\) and the \(7\) is a coefficient of the \(x^2\).

Factors is the term for quantities that are multiplied together. An easy example is \(2 \times 4=8\), where \(2\) and \(4\) are factors of \(8\). Also, \(2x\) and \(3x\) are factors of \(6x^2\) because \(2x \times 3x = 6x^2\). Sometimes factors will be expressions with two terms. For example, \((x+3)(x-2) = x^2+x-6\), where \((x+3)\) and \((x-2)\) are factors of \(x^2+x-6\).

Rewriting Expressions

Very often in math, expressions need to be rewritten to help solve a problem. Factoring can be used in a lot of problems and there are certain patterns that show up. One pattern that can show up is when two terms share a factor.

In \(7x^2 +4x\), both terms have an \(x\). Factor the \(x\) out of each term:

\[7x^2+4x\] \[x(7x)+x(4)\] \[\text{Rewrite as } x(7x+4)\]

Always check for this pattern first, as it’s the easiest one to factor.

Another pattern is called the difference of two squares:

In \(x^2-9^2\), both \(x^2\) and \(9\) are perfect squares, with one being subtracted from the other. Whenever you see that pattern, it always factors the same way. In this case, the factors are \((x+3)(x-3)\). We can describe the pattern this way: if you see two perfect squares with a negative sign between them, say \(x^2-a^2\), it can be factored as \((x+a)(x-a)\).

Another example: \(4x^2-y^2 = (2x+y)(2x-1)\)

Here’s a tricky one. Does \(x^4-y^4\) fit the pattern?? Yes, both terms are perfect squares. The first term is \((x^2)^2\) and the second term is \((y^2)^2\), so we just use the same factoring pattern \((x^2+y^2)(x^2-y^2)\). Another trick comes up here. The second factor, \((x^2-y^2)\), is also the difference of two squares, so we can factor it as \((x+y)(x-y)\). Now we have three factors, \((x^2+y^2)(x+y)(x-y)\), which can happen when the exponents are greater than two.

Finding the Zeros

If you have an expression such as \(6x-18\), finding the zero of the expression just means to set it equal to zero and solve for \(x\). The zero of an expression means the value of \(x\) that makes the expression zero:

\[6x-18=0\] \[6x=18\] \[x=3\]

We say that \(3\) is a root of the equation \(6x-18=0\) and a zero of the expression \(6x-18\).

This comes up more often when the expression is a quadratic, where it means the same thing.

Find the zeros of \(x^2 +8x+12\).

Set the expression equal to zero and solve for \(x\):

\[x^2 +8x+12 = 0\]

Since the equation we now have is a quadratic, we need to factor it, if we can:

\[(x+2)(x+6)=0\]

Set each factor equal to zero and solve each equation:

\[x+2=0 \ \ \ \ \ \ \ \text{ and } \ \ \ \ \ \ x+6=0\] \[x= -2 \ \ \ \ \ \ \ \text{ and } \ \ \ \ \ \ \ x=-6\]

We can say that \(-2 \text{ and }-6\) are the roots of the equation \(x^2 +8x+12 = 0\), and \(-2 \text{ and }-6\) are the zeros of the expression \(x^2 +8x+12\).

Polynomials as a System

Combinations of constants and variables can have special names, depending on how many terms there are.

Monomials have one term, such as:

\(3x\)
\(x^2\)
\(26z\)

Binomials have two terms, such as:

\(81x^2+ 14\)
\(11-5x\)
\(9y-2y^3\)

Trinomials have three terms, such as:

\(y^2-3y+74\)
\(x^2 +xz + z^2\)

There is a term, polynomial, that includes all of these and more, possibly with many terms, though four or five are the most you’re likely to see.

In math, we think in terms of different systems of numbers and expressions. If a system of numbers is closed, that means that for certain operations, the result of the operation is always in that system.

For example, think of the system of integers, \(\{... -3,\, -2, \,-1,\, 0, \ ,1, \ ,2, \, 3, …\}\). If you add any two of them you always get an integer for an answer. The same is true for subtraction and multiplication, so we say that the system of integers is closed under addition, subtraction, and multiplication. Why not division too? If you divide integers, you can get fractions, like \(\frac{3}{5}\), which are not part of the system of integers.

Polynomials are in a system that is also closed under addition, subtraction, and multiplication. Let’s take a little look at adding polynomials. It’s important to do two things when you do this. First, make sure both polynomials have terms in the same order, usually in order of decreasing exponents. Second, line them up so that like terms are in the same column. Here’s an example:

Add \(5x^2 -2x +1\) and \(x^2+2x-2\).

\[\begin{array}{rrr} 5x^2 & -2x & +1\\ +x^2 &+2x & -2 \\ \hline 6x^2 &+ 0 &-1\\ \end{array}\]

To subtract would require the same setup, but of course you would be subtracting.

Here is one way to multiply two polynomials. It uses the distributive property:

\[(x-5)(x^2 -3x+9)\]

First, take the \(x\) from the left and multiply it by every term in \((x^2 -3x+9)\):

$$x(x^2) +x(-3x) +x(9)

Now do the same with the \(-5\): \((-5)(x^2) +(-5)(-3x)+(-5)(9)\)

Put all six terms together and simplify:

\[x^3 -3x^2 + 9x -5x^2+15x-45\] \[x^3 -8x^2 +24x -45\]

Notice that in both cases we ended up with another polynomial, which is the point of this section. The set of polynomials is closed under addition, subtraction, and multiplication. No matter how you add, subtract, or multiply polynomials, you will always get another polynomial.

Creating Equations and Inequalities

You need to be able to create one or two variable equations and inequalities to solve problems. These could include linear, quadratic, rational, and exponential equations.

Creating Equations and Inequalities in One Variable

We’ve dealt with linear, quadratic, and exponential functions in this guide, but not much with rational functions. A rational function contains ratios with polynomial expressions in the top and bottom (numerator and denominator). Here are a few examples of rational expressions:

\[\frac{z-1}{z+1}\] \[\frac{x^2-4}{x+2}\] \[\frac{t^3-4t}{t^2-4t+4}\]

What do you need to be able to do with these expressions? The main thing is to simplify them by factoring and canceling. Simplify the second expression above:

\[\frac{x^2-4}{x+2}\] \[\frac{(x+2)(x-2)}{x+2}\] \[\require{cancel} \frac{\cancel{{(x+2)}}(x-2)}{\cancel{(x+2)}}\] \[x-2\]

If you want it to be a rational function, write \(f(x)=\) before the expression:

\(f(x) = \frac{x^3-4x}{x^2-4x+4}\).

Creating Equations in Two or More Variables

Be able to create two equations with two variables to represent a certain situation. For example:

There are cows and pigs on a farm, \(17\) in all. There are five fewer pigs than cows. How many cows are there?

Let the number of cows be \(c\) and the number of pigs be \(p\). Together they add up to \(17\):

\[c+p=17\]

The difference between their numbers is \(5\):

\[c-p=5\]

Add the two equations:

\[\begin{array}{rcrcr} c &+&p&=&17\\ c &-&p&=&5\\ \hline & &2c&=&22\\ \hline & &c& =&11\\ \end{array}\]

The number of cows is \(11\).

What else do you need to be able to do with these equations? Graph them. There’s nothing really new about it. Just make up a data table and plot the points you get. There’s an example below.

\[\text{Graph } y = \frac{x^2-4}{2}\]

Pick some values for \(x\) and calculate corresponding \(y\) values:

\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline x&0&1&2&3&-1&-2&-3\\ \hline y&-2&\frac{-3}{2}&0&\frac{5}{2}&\frac{-3}{2}&0&\frac{5}{2}\\ \hline \end{array}\]

Constraints

Given a description of a situation, be able to write equations and/or inequalities to describe the limitations (called constraints) on the quantities. It’s not as hard as it may sound.

Suppose you have \(18\) marbles in a bag, red and blue ones. There are more red than blue marbles. Letting \(b\) equal the number of blue marbles and \(r\) the number of red marbles, we know two things:

\(b +r =18\)
\(r>b\)

Now, solve the equation for \(b\) and substitute that into the inequality:

\[b=18-r\] \[r>18-r\] \[2r>18\] \[r>9\]

This doesn’t tell us what the value of \(r\) is, but it does narrow it down. The number of red marbles is constrained to be any number from \(10\) to \(18\).

Another example:

A rectangle’s length is three times its width. If its perimeter is less than \(24\), what must be true of its length and width?

We know two things.

\[l=3w\]

And if the perimeter is less than \(24\), we can use the formula for perimeter and write:

\[2(l+w)<24\]

Substitute \(3w\) in place of \(l\):

\[2(3w+w)<24\]

Solve for \(w\):

\[8w<24\] \[w<3\]

That tells us that the width must be less than \(3\). Now, multiply both sides by \(3\) and substitute \(l\) for \(3w\):

\[3w<9\] \[l<9\]

Now we know that \(l\) must be less than \(9\).

Those are the constraints on the length and width.

Explanations

You should be able to explain the steps in solving a simple equation the way it’s done in the example below.

\[\begin{array}{rl} 3x+7-x=19& \text{ Original equation}\\ 2x+7=19& \text{ Simplified by collecting like terms.}\\ 2x+7-7=19-7& \text{ Subtraction property for equality}\\ 2x =12 & \text{ Simplified}\\ \frac{2x}{2} = \frac{12}{2} &\text{ Division property for equality.} \\ x =6 & \text{ Simplified}\\ \end {array}\]

Linear Equations and Inequalities

This section is about solving linear equations and inequalities that will have only one variable. Actually, there is a good example of this exact thing directly above this section. It shows the steps it takes to get the variable alone on one side of the equation. Because that is there, let’s start with an inequality.

The properties of equality, like addition, subtraction, multiplication, and division, all work for inequalities too. Except, if you multiply or divide both sides by a negative number, you have to change the direction of the inequality sign. You can see this is done in the last step of our example below.

\[\begin{array}{ll} \text{Start with} & 5-2x <9 \\ \text{Subtract 5 from both sides.} & 5-2x-5<9-5\\ \text{Simplify.}& -2x< 4\\ \text{Divide both sides by -2 and switch < to >}& \frac{-x}{-2} >\frac{4}{-2}\\ \text{Simplify} & x>-2\\ \end{array}\]

A problem you may come across is solving an equation in terms of a certain constant. What’s different is that, rather than a number for a coefficient, there will be a letter variable. Something like this:

Solve \(5x-a=-3\) in terms of \(a\).

It’s not possible to get rid of the \(a\), so it will appear in your answer. Just follow the normal steps to get \(x\) alone on the left side. Whatever ends up on the right side is your answer. The phrase “in terms of \(a\)” means \(a\) will appear in your answer.

\[5x-a=-3\] \[5x = a- 3\] \[x = \frac{a-3}{5}\]

Let’s do one more example. A problem may have more than one constant term that is a letter. Solve the following equation in terms of \(a\) and \(b\):

\[2ax+b=10+2b\]

\[2ax=10 +2b -b\] \[2ax = 10 +b\] \[x = \frac{10+b}{2a}\]

Quadratic Equations

Be able to solve quadratic equations. There are two different ways to do this and we’ll look at them both.

The first way is to set up the quadratic expression so it’s equal to zero, if it isn’t already, and factor it. After that, set the two factors equal to zero and solve for \(x\) in each. You will get two answers.

\[x^2-2x-15=0\] \[(x+3)(x-5)=0\] \[x+3 = 0 \ \ \text{ or } \ \ \ x-5=0\] \[x=-3\ \ \text{ or } \ \ \ x=5\]

This method is a good one to use if you can factor the expression fairly quickly, but there are quadratics that are essentially impossible to factor. What then?

Then we need the quadratic formula.

When \(ax^2 + bx +c =0\), we use the following formula:

\[x= \frac{-b \pm\sqrt{b^2-4ac}}{2a}\]

Let’s try an example.

Solve \(5x^2+3x-2=0\).

Looking at the equation, we see that \(a=5\), \(b=3\), and \(c=-2\).

Substitute those values into the quadratic formula above:

\[x= \frac{-3 \pm\sqrt{3^2-4 \cdot 5 \cdot (-2)}}{2 \cdot 5}\]

Now, simplify everything you can:

\[x= \frac{-3 \pm\sqrt{9+40 }}{10}\] \[x= \frac{-3 \pm\sqrt{49}}{10}\] \[x= \frac{-3 \pm 7}{10}\] \[x=\frac{-3+7}{10}\ \ \ \text{ or }\ \ \ x=\frac{-3-7}{10}\] \[x = \frac{2}{5} \ \ \ \text{ or } \ \ \ x = -1\]

Systems of Linear Equations

We have solved systems of two linear equations earlier in this guide by adding or subtracting the equations to eliminate one of the variables. Another way to solve these pairs of equations is by graphing both equations and seeing where they intersect. That is because \(x\) and \(y\) are the only ones that work for both lines.

At what point do these lines intersect?

It’s a little bit hard to read, but the point is \((7.5, 4.5)\).

The equations for the two lines are \(d-.6t=0\) and \(-d+t=3\)

Now let’s solve these for \(d\) and \(t\) in the usual way of solving a pair of linear equations. First we add the two equations and the \(d\) disappears:

\[\begin{array}{rcrcr} d &-&0.6t&=&0\\ -d &+&t&=&3\\ \hline & &0.4t&=&3\\ & &t& =&\frac{3}{0.4}\\ & &t& =&7.5\\ \end{array}\]

Knowing \(t=7.5\), we can substitute it into \(-d+t=3\):

\[-d+t=3\] \[-d+7.5=3\] \[-d=-4.5\] \[d=4.5\]

It’s comforting to see that we get the same solution for \(t\) and \(d\) that we got from the graph.

Understanding Graphs

Suppose you have an equation in two variables, maybe something simple like \(y=2x-1\). How many \((x, y)\) pairs will make the equation true? Does \((1,1)\) work? Yes. How about \((2, 3)\)? It does. And \((-1, -3)\)? Yes, again. Try these for yourself to see that they all work.

Question 1: We saw three pairs that work, but how many pairs exist that will work? If you are thinking it’s an infinite number, you are right.

Question 2: How many of these points would we need to plot a graph of the equation? Luckily, we can see that this is a linear equation, so all we need is two points.

And here it is:

Last Question:

The line graph above consists of how many points? Are you thinking it’s an infinite number again? You’re correct.

We only needed two points to locate it, but once it’s drawn it’s assumed to have an infinite number of points, just like in geometry. This is true of any graph.

The only other thing to note here is that if the equation made a curved graph, it would have taken more points to show its shape.