Mathematics Study Guide for the PRAXIS Test

Algebra

In the revised version of the PRAXIS Math Test, algebra and geometry form one of the three concept areas for testing. We’ll look at algebra skills first.

Algebra is the type of math that uses letters to represent unknown quantities and enables the user to solve all sorts of everyday problems. For a very simple example, if you have 16 sodas and there will be \(8\) people at a party, you could let \(s\) represent sodas and use this equation to find out how many sodas each person can have:

\[8 \cdot s = 16\]

When solving an equation, you try to isolate the letter on one side of the equation by performing needed operations to get rid of the numbers on that side. Be sure to remember to do the same thing to both sides. Here, divide both sides by \(8\) and you’ll have the answer: \(2\).

Properties

Basic operations in algebra are conducted based on generally agreed principles. Here are some of the important ones. You don’t need to remember the names of these, just be careful not to violate them.

Commutative Property

The commutative property has to do with order. Order matters in some operations, but not in others. If order doesn’t matter, we say that that operation has a commutative property.

In Addition—Does \(8 + 11 = 11 + 8\)? Yes.
This is the commutative property of addition: \(a + b = b + a\) (same results either way)

In Subtraction—Does \(25 - 13 = 13 - 25\)? No.
The commutative property doesn’t exist for subtraction. (different results)

In Multiplication—Does \(7 \times 9 = 9 \times 7\)? Yes.
This is the commutative property for multiplication: \(a \cdot b = b \cdot a\) (same results)

In Division—Does \(8\div 2 = 2\div 8\) ? No.
The commutative property doesn’t exist for division. (different results)

Summary: Addition and multiplication are commutative because you get the same results with either order.

Associative Property

The associative property has to do with grouping, meaning using parentheses. Does it matter where you put the parentheses? It can.

In Addition:

\[\begin{array} {c c} (6+2)+11 & 6+(2+11)\\ 8 + 11 & 6+13\\ 19 & 19 \end{array}\]

(same results)

In Subtraction:

\[\begin{array} {c c} (25-3)-13 & 25-(3-13)\\ 22-13 & 25-(-10)\\ 9 & 35 \end{array}\]

(different results)

In Multiplication:

\[\begin{array} {c c} (7 \cdot 9) \cdot 2 & 7\cdot( 9 \cdot 2)\\ 63 \cdot 2 & 7 \cdot 18\\ 126 &126 \end{array}\]

(same results)

In Division:

\[\begin{array} {c c} (24 \div 4) \div 2 & 24 \div (4 \div 2)\\ 6 \div 2 & 24 \div 2\\ 3 &12 \end{array}\]

(different results)

Addition and multiplication are associative. You get the same results no matter where you put the parentheses.

Distributive Property

The commutative and associative properties are pretty easy and unlikely to cause you trouble. In this section, we look at the distributive property, which can mess you up if you’re not careful. It has to do with multiplying a quantity times two or more quantities in parentheses. For example:

\[4(x+9)\] \[\require{enclose} \enclose{circle}4 \cdot x + \enclose{circle}4 \cdot 9\] \[4x + 36\]

The \(4\) has been distributed to the \(x\) and to the \(9\).
This is the distributive over addition property. This also works over subtraction.

The most common mistake people make is to do this:

\(4(x+9)\)
\(4x + 9\)

Don’t do that. Always multiply both quantities in the parentheses.

Also, never do this:

\(4(x \times 9)\)
\(4\times x \times 4 \times 9\)
\(144x\)

Wrong. There is no distributive over multiplication property or distributive over division property.

Arithmetic and Algebraic Procedures

Be able to follow or create procedures for solving algebraic problems. Most problems you run into will require you to perform several steps to arrive at an answer, especially equation solving. Be able to follow a simple flow chart or carry out a recurrence sequence.

Finding Equivalent Expressions

Equivalent expressions are expressions that express the same quantity but are not in the same form (they don’t look the same). You run into these all the time when you are simplifying algebraic expressions or using the commutative, associative, or distributive properties. For example:

\(3 \times x \times 5 \text{, }\; 3 \cdot 5 \cdot x \text{, and } 15x\) are all equivalent expressions.

Other examples are:

\(5 \times 8\) is equivalent to \(8+8+8+8+8\)

\((x-2)(x+2)\) is equivalent to \(x^{2}-4\)

Writing Equations and Expressions to Solve Problems

Be able to write expressions and equations based on given information to model a real-life situation or a purely mathematical problem.

Solving Word Problems

Word problems can be both the hardest to solve, and the most important to deal with in mathematics. If there are situations in life that call for calculations, you can bet that no one will be there to hand you just exactly the numbers you need and no more. The problems won’t come along just after you finished a class unit where you learned how to solve that exact type of problem. Math educators are beginning to realize that word problems need to be given more attention, and this type of problem is becoming much more common on tests today. Refer often to the earlier section on Problem-Solving Techniques.

Linear Relationships

Word problems can be based on a linear relationship. You can tell that two quantities have a linear relationship if you see that when one of them keeps changing by a constant amount, the other one also changes by a constant, though different, amount. Here’s an example:

Suppose a corn plant is \(12\) cm high and grows \(2.5\) cm each day. (Each time the number of days increases by \(1\), the height increases by \(2.5\) cm.) A graph of this over a period of days would be a straight line.

An expression for the growth (not the height, just the growth) of the plant, in terms of \(d\), the number of days would be:

\[2.5 \ d\]

A word equation for the height would be:

height = beginning height + growth
\(h=12+2.5 \ d\)

How tall would the plant be after \(30\) days?
\(h=12+2.5\cdot 30= 87\text{ cm}\)

Venn Diagrams

Some word problems are about different groups of things that overlap in some way. They are often more easily understood with Venn diagrams. Here’s an example. Suppose \(20\) students are on the school soccer team and \(29\) are on the tennis team. Eight of these students are actually on both teams. How many students are there all together?

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The overlapping area represents the 8 students on both teams. If \(29\) students are on the tennis team, and 8 of them are on both teams, that leaves \(21\) tennis players not on the soccer team. Likewise, out of \(20\) soccer players, with \(8\) on both teams, that leaves \(12\) soccer players not on the tennis team. Adding all three groups, we get \(21 + 8 + 12 = 41\) students.

Solving Linear Equations

A linear equation has the form \(y=mx+b\). If you replace \(y\), \(m\), and \(b\) with numbers, you have an equation in \(x\) that can be solved. Here is a simple example:

Solve \(31=2x+5\) .

\[\begin {array}{rrrrrl} 31&=&2x&+&5&\\ -5&&&-&5& \text{Subtract 5 from both sides to isolate the variable.}\\ \hline 26&=&2x&&&\\ \frac{26}{2}&=&\frac{2x}{2}&&&\text{Divide both sides by 2}\\ \hline 13&=&x&&&\\ \end{array}\]

Notice that we did our subtracting before we divided. You want to collect like terms by adding and subtracting to isolate the variable before you divide by the coefficient of the variable.

Here is a second example with more terms:

Solve \(4x+9=-11+6x\)

\[\begin {array}{llllllll} 4x&+&9&=&-11&+&6x&\\ -6x&&&=&&-&6x\;\;\text{Subtract 6x from both sides.}&\\ \hline -2x&+&9&=&-11&&&\\ &-&9&=&-9&&\text{Subtract 9 from both sides.}\\ \hline -2x&&&=&-20&&&\\ \dfrac{-2x}{-2}&&&=&\dfrac{-20}{-2}&&\text{Divide both sides by -2}\\ \hline \;\;x&&&=&\;\;10&&&\\ \end{array}\]

Solving Quadratic Equations

If the highest power of \(x\) (or whatever variable) in an equation is \(2\), it is a quadratic equation. The simplest quadratics look like this: \(x^{2}=64\). To solve it, ask yourself what number, when squared, gives you \(64\)? It’s \(8\), of course, but don’t forget it could also be \(-8\). A more formal way to describe how to solve it is to say “take the square root of both sides of the equation.”

\[\sqrt{x^{2}}=\sqrt{64}\] \[x = \pm 8\]

Here is one that is just a bit more complicated:

\[\begin {array}{rrrrr} 2x^{2}&-6&=&44&\\ &+6&&+6&\\ \hline 2x^{2}&&=&50&\\ \dfrac{2x^{2}}{2}&&=&\dfrac{50}{2}&\\ \hline x^{2}&&=&25&\\ \sqrt{x^{2}}&&=&\sqrt{25}\\ \hline x&&=&\pm5&\\ \end {array}\]