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# Algebra: Praxis I Exam (page 2)

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Updated on Jul 5, 2011

#### Like Terms

If two or more terms have exactly the same variable(s), and these variables are raised to exactly the same exponents, they are said to be like terms. Like terms can be simplified when added and subtracted.

Examples
7x + 3x = 10x
6y2 – 4y2 = 2y2

However, 3cd2 + 5c2d cannot be simplified. Since the exponent of 2 is on d in 3cd2 and on c in 5c2d, they are not like terms.

The process of adding and subtracting like terms is called combining like terms. It is important to combine like terms carefully, making sure that the variables are exactly the same.

### Algebraic Expressions

An algebraic expression is a combination of monomials and operations. The difference between algebraic expressions and algebraic equations is that algebraic expressions are evaluated at different given values for variables, while algebraic equations are solved to determine the value of the variable that makes the equation a true statement.

There is very little difference between expressions and equations, because equations are nothing more than two expressions set equal to each other. Their usage is subtly different.

Example
A mobile phone company charges a \$39.99 a month flat fee for the first 600 minutes of calls, with a charge of \$.55 for each minute thereafter.
Write an algebraic expression for the cost of a month's mobile phone bill: \$39.99 + \$.55x, where x represents the number of additional minutes used.
Write an equation for the cost (C) of a month's mobile phone bill: C = \$39.99 + \$.55x, where x represents the number of additional minutes used.

In the preceding example, you might use the expression \$39.99 + \$.55x to determine the cost if you are given the value of x by substituting the value for x. You could also use the equation C = \$39.99 + \$.55x in the same way, but you can also use the equation to determine the value of x if you were given the cost.

#### Simplifying and Evaluating Algebraic Expressions

We can use the mobile phone company example above to illustrate how to simplify algebraic expressions. Algebraic expressions are evaluated by a two-step process: substituting the given value(s) into the expression, and then simplifying the expression by following the order of operations (PEMDAS).

Example
Using the cost expression \$39.99 + \$.55x, determine the total cost of a month's mobile phone bill if the owner made 700 minutes of calls.
Let x represent the number of minutes over 600 used, so in order to find out the difference, subtract 700 – 600; x = 100 minutes over 600 used.
Substitution: Replace x with its value, using parentheses around the value.
\$39.99 + \$.55x
\$39.99 + \$.55(100)

Evaluation: PEMDAS tells us to evaluate parentheses and exponents first. There is no operation to perform in the parentheses, and there are no exponents, so the next step is to multiply, and then add.

\$39.99 + \$.55(100)
\$39.99 + \$55 = \$94.99
The cost of the mobile phone bill for the month is \$94.99.

You can evaluate algebraic expressions that contain any number of variables, as long as you are given all of the values for all of the variables.

### Simple Rules for Working with Linear Equations

A linear equation is an equation whose variables' highest exponent is 1. It is also called a first-degree equation. An equation is solved by finding the value of an unknown variable.

1. The equal sign separates an equation into two sides.
2. Whenever an operation is performed on one side, the same operation must be performed on the other side.
3. The first goal is to get all of the variable terms on one side and all of the numbers (called constants) on the other side. This is accomplished by undoing the operations that are attaching numbers to the variable, thereby isolating the variable. The operations are always done in reverse PEMDAS order: start by adding/subtracting, then multiply/divide.
4. The final step often will be to divide each side by the coefficient, the number in front of the variable, leaving the variable alone and equal to a number.
Example
5m + 8 = 48
–8 = –8
m = 8

Undo the addition of 8 by subtracting 8 from both sides of the equation. Then undo the multiplication by 5 by dividing by 5 on both sides of the equation. The variable, m, is now isolated on the left side of the equation, and its value is 8.

### Checking Solutions to Equations

To check an equation, substitute the value of the variable into the original equation.

Example
To check the solution of the previous equation, substitute the number 8 for the variable m in 5m + 8 = 48.
5(8) + 8 = 48
40 + 8 = 48
48 = 48

Because this statement is true, the answer m = 8 must be correct.

#### Isolating Variables Using Fractions

Working with equations that contain fractions is almost exactly the same as working with equations that do not contain variables, except for the final step. The final step when an equation has no fractions is to divide each side by the coefficient. When the coefficient of the variable is a fraction, you will instead multiply both sides by the reciprocal of the coefficient. Technically, you could still divide both sides by the coefficient, but that involves division of fractions, which can be trickier.

Example

Undo the addition of by subtracting from both sides of the equation. Multiply both sides by the reciprocal of the coefficient. Convert the 11 to an improper fraction to facilitate multiplication. The variable m is now isolated on the left side of the equation, and its value is .

### Equations with More than One Variable

Equations can have more than one variable. Each variable represents a different value, although it is possible that the variables have the same value.

Remember that like terms have the same variable and exponent. All of the rules for working with variables apply in equations that contain more than one variable, but you must remember not to combine terms that are not alike.

Equations with more than one variable cannot be solved, because if there is more than one variable in an equation there is usually an infinite number of values for the variables that would make the equation true. Instead, we are often required to "solve for a variable," which instead means to isolate that variable on one side of the equation.

Example
Solve for y in the equation 2x + 3y = 5.
There are an infinite number of values for x and y that that satisfy the equation. Instead, we are asked to isolate y on one side of the equation.

### Cross Multiplying

Because algebra uses percents and proportions, it is necessary to learn how to cross multiply. You can solve an equation that sets one fraction equal to another by cross multiplication. Cross multiplication involves setting the cross products of opposite pairs of terms equal to each other.

Example

### Algebraic Fractions

Working with algebraic fractions is very similar to working with fractions in arithmetic. The difference is that algebraic fractions contain algebraic expressions in the numerator and/or denominator.

Example
A hotel currently has only one-fifth of its rooms available. If x represents the total number of rooms in the hotel, find an expression for the number of rooms that will be available if another tenth of the total rooms are reserved.

Because x represents the total number of rooms, represents the number of available rooms. One-tenth of the total rooms in the hotel would be represented by the fraction . To find the new number of available rooms, find the difference: .

Just like in arithmetic, the first step is to find the LCD of 5 and 10, which is 10. Then change each fraction into an equivalent fraction that has 10 as a denominator.
Therefore, rooms will be available after another tenth of the rooms are reserved.

### Reciprocal Rules

There are special rules for the sum and difference of reciprocals. The reciprocal of 3 is and the reciprocal of x is .

• If x and y are not 0, then .
• If x and y are not 0, then .

### Translating Words into Numbers

The most important skill needed for word problems is being able to translate words into mathematical operations. The following will be helpful in achieving this goal by providing common examples of English phrases and their mathematical equivalents.

Phrases meaning addition: increased by; sum of; more than; exceeds by.

Examples
A number increased by five: x + 5.
The sum of two numbers: x + y.
Ten more than a number: x + 10.

Phrases meaning subtraction: decreased by; difference of; less/fewer than; diminished by.

Examples
Ten less than a number: x – 10.
The difference of two numbers: xy.

Phrases meaning multiplication: times; times the sum/difference; product; of.

Examples
Three times a number: 3x.
Twenty percent of 50: 20% × 50.
Five times the sum of a number and three: 5(x + 3).

Phrases meaning "equals": is; result is.

Examples
Fifteen is 14 plus 1: 15 = 14 + 1.
Ten more than two times a number is 15: 2x + 10 = 15.

### Assigning Variables in Word Problems

It may be necessary to create and assign variables in a word problem. To do this, first identify any knowns and unknowns. The known may not be a specific numerical value, but the problem should indicate something about its value. Then let x represent the unknown you know the least about.

Examples
Max has worked for three more years than Ricky.
Unknown: Ricky's work experience = x
Known: Max's experience is three more years = x + 3

Heidi made twice as many sales as Rebecca.
Unknown: number of sales Rebecca made = x
Known: number of sales Heidi made is twice Rebecca's amount = 2x

There are six less than four times the number of pens than pencils.
Unknown: the number of pencils = x
Known: the number of pens = 4x – 6

Todd has assembled five more than three times the number of cabinets that Andrew has.
Unknown: the number of cabinets Andrew has assembled = x
Known: the number of cabinets Todd has assembled is five more than three times the number
Andrew has assembled = 3x + 5

### Percentage Problems

To solve percentage problems, determine what information has been given in the problem and fill this information into the following template:

_____ is _____% of _____

Then translate this information into a one-step equation and solve. In translating, remember that is translates to = and of translates to ×. Use a variable to represent the unknown quantity.

##### Examples
Finding a percentage of a given number:
In a new housing development there will be 50 houses. 40% of the houses must be completed in the first stage. How many houses are in the first stage?
1. Translate.
_____ is 40% of 50.
x is .40 × 50.
2. Solve.
x = .40 × 50
x = 20
20 is 40% of 50. There are 20 houses in the first stage.
Finding a number when a percentage is given:
40% of the cars on the lot have been sold. If 24 were sold, how many total cars were there on the lot originally?
1. Translate.
24 is 40% of _____.
24 = .40 × x.
2. Solve.
60 = x
24 is 40% of 60. There were 60 total cars on the lot.
Finding what percentage one number is of another:
Matt has 75 employees. He is going to give 15 of them raises. What percent of the employees will receive raises?
1. Translate.
15 is _____% of 75.
15 = x × 75.
2. Solve.
.20 = x
20% = x
• 15 is 20% of 75. Therefore, 20% of the employees will receive raises.