Algebra: Praxis I Exam
This section will help in mastering algebraic equations by reviewing variables, cross multiplication, algebraic fractions, reciprocal rules, and exponents. Algebra is arithmetic using letters, called variables, in place of numbers. By using variables, the general relationships among numbers can be easier to see and understand.
A term of a polynomial is an expression that is composed of variables and their exponents, and coefficients. A variable is a letter that represents an unknown number. Variables are frequently used in equations, in formulas, and in mathematical rules to help illustrate numerical relationships. When a number is placed next to a variable, indicating multiplication, the number is said to be the coefficient of the variable.
Examples 8c 8 is the coefficient to the variable c. 6ab 6 is the coefficient to both variables, a and b.
Three Kinds of Polynomials
- Monomials are single terms that are composed of variables and their exponents and a positive or negative coefficient. The following are examples of monomials: x, 5x, –6y3, 10x2y, 7, 0.
- Binomials are two nonlike monomial terms separated by + or – signs. The following are examples of binomials: x + 2, 3x2 – 5x, –3xy2 + 2xy.
- Trinomials are three nonlike monomial terms separated by + or – signs. The following are examples of trinomials: x2 + 2x – 1, 3x2 – 5x + 4, –3xy2 + 2xy – 6x.
- Monomials, binomials, and trinomials are all examples of polynomials, but we usually reserve the word polynomial for expressions formed by more than three terms.
- The degree of a polynomial is the largest sum of the terms' exponents.
- The degree of the trinomial x2 + 2x – 1 is 2, because the x2 term has the highest exponent of 2.
- The degree of the binomial x + 2 is 1, because the x term has the highest exponent of 1.
- The degree of the binomial –3x4y2 + 2xy is 6, because the x4y2 term has the highest exponent sum of 6.
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.
- 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.
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.
- 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).
- 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.
- The equal sign separates an equation into two sides.
- Whenever an operation is performed on one side, the same operation must be performed on the other side.
- 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.
- 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.
- 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.
- 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.
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.
- 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.
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.
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.
- 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.
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.
- 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.
- Ten less than a number: x – 10.
- The difference of two numbers: x – y.
- Phrases meaning multiplication: times; times the sum/difference; product; of.
- 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.
- 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.
- 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
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.
- 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?
- _____ is 40% of 50.
- x is .40 × 50.
- 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?
- 24 is 40% of _____.
- 24 = .40 × x.
- 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?
- 15 is _____% of 75.
- 15 = x × 75.
- .20 = x
- 20% = x
- 15 is 20% of 75. Therefore, 20% of the employees will receive raises.
Problems Involving Ratio
A ratio is a comparison of two quantities measured in the same units. It is symbolized by the use of a colon—x:y. Ratios can also be expressed as fractions or using words (x to y).
Ratio problems are solved using the concept of multiples.
- A bag contains 60 screws and nails. The ratio of the number of screws to nails is 7:8. How many of each kind are there in the bag?
- From the problem, it is known that 7 and 8 share a multiple and that the sum of their product is 60. Whenever you see the word ratio in a problem, place an "x" next to each of the numbers in the ratio, and those are your unknowns.
- Let 7x = the number of screws.
- Let 8x = the number of nails.
- Write and solve the following equation:
- Therefore, there are (7)(4) = 28 screws and (8)(4) = 32 nails.
- Check: 28 + 32 = 60 screws, .
Problems Involving Variation
Variation is a term referring to a constant ratio in the change of a quantity.
- Two quantities are said to vary directly if their ratios are constant. Both variables change in an equal direction. In other words, two quantities vary directly if an increase in one causes an increase in the other. This is also true if a decrease in one causes a decrease in the other.
- If it takes 300 new employees a total of 58.5 hours to train, how many hours of training will it take for 800 employees?
- Because each employee needs about the same amount of training, you know that they vary directly. Therefore, you can set the problem up the following way:
- Cross multiply to solve:
- Therefore, it would take 156 hours to train 800 employees.
- Two quantities are said to vary inversely if their products are constant. The variables change in opposite directions. This means that as one quantity increases, the other decreases, or as one decreases, the other increases.
- If two people plant a field in six days, how many days will it take six people to plant the same field? (Assume each person is working at the same rate.)
- As the number of people planting increases, the days needed to plant decreases. Therefore, the relationship between the number of people and days varies inversely. Because the field remains constant, the two products can be set equal to each other.
- 2 people × 6 days = 6 people × x days
- Thus, it would take six people two days to plant the same field.
In general, there are three different types of rate problems likely to be encountered in the workplace: cost per unit, movement, and work output. Rate is defined as a comparison of two quantities with different units of measure.
Cost per Unit
Some problems will require the calculation of unit cost.
- If 100 square feet cost $1,000, how much does 1 square foot cost?
In working with movement problems, it is important to use the following formula:
- (rate)(time) = distance
- A courier traveling at 15 mph traveled from his base to a company in of an hour less than it took when the courier traveled at 12 mph. How far away was his drop-off?
- First, write what is known and unknown.
- Unknown: time for courier traveling 12 mph = x
- Known: time for courier traveling 15 mph = x –
- Then, use the formula (rate)(time) = distance to find expressions for the distance traveled at each rate:
- 12 mph for x hours = a distance of 12x miles
- 15 mph for x – mph = a distance of 15x – miles.
- The distance traveled is the same; therefore, make the two expressions equal to each other:
- Be careful: 1.25 is not the distance; it is the time. Now you must plug the time into the formula (rate)(time) = distance. Either rate can be used.
- 12x = distance
- 12(1.25) = distance
- 15 miles = distance
Work-output problems are word problems that deal with the rate of work. The following formula can be used on these problems:
- (rate of work)(time worked) = job or part of job completed
- Danette can wash and wax two cars in six hours, and Judy can wash and wax the same two cars in four hours. If Danette and Judy work together, how long will it take to wash and wax one car?
- Because Danette can wash and wax two cars in six hours, her rate of work is , or one car every three hours. Judy's rate of work is , or one car every two hours. In this problem, making a chart will help:
- Because they are both working on only one car, you can set the equation equal to one: Danette's part + Judy's part = 1 car:
- Solve by using 6 as the LCD for 3 and 2 and clear the fractions by multiplying by the LCD:
Thus, it will take Judy and Danette hours to wash and wax one car.
Patterns and Functions
The ability to detect patterns in numbers is a very important mathematical skill. Patterns exist everywhere in nature, business, and finance.
When you are asked to find a pattern in a series of numbers, look to see if there is some common number you can add, subtract, multiply, or divide each number in the pattern by to give you the next number in the series.
For example, in the sequence 5, 8, 11, 14…you can add 3 to each number in the sequence to get the next number in the sequence. The next number in the sequence is 17.
- What is the next number in the sequence , 3, 12, 48?
- Each number in the sequence can be multiplied by the number 4 to get the next number in the sequence: × 4 = 3, 3 × 4 = 12, 12 × 4 = 48, so the next number in the sequence is 48 × 4 = 192.
Sometimes it is not that simple. You may need to look for a combination of multiplying and adding, dividing and subtracting, or some combination of other operations.
- What is the next number in the sequence 0, 1, 2, 5, 26?
- Keep trying various operations until you find one that works. In this case, the correct procedure is to square the term and add 1: 02 + 1 = 1, 12 + 1 = 2, 22 + 1 = 5, 52 + 1 = 26, so the next number in the sequence is 262 + 1 = 677.
Properties of Functions
A function is a relationship between two variables x and y where for each value of x, there is one and only one value of y. Functions can be represented in four ways:
- a table or chart
- an equation
- a word problem
- a graph
For example, the following four representations are equivalent to the same function:
Helpful hints for determining if a relationship is a function:
- If you can isolate y in terms of x using only one equation, it is a function.
- If the equation contains y2, it will not be a function.
- If you can draw a vertical line anywhere on a graph such that it touches the graph in more than one place, it is not a function.
- If there is a value for x that has more than one y-value assigned to it, it is not a function.
x = 5 Contains no variable y, so you cannot isolate y. This is not a function. 2x + 3y = 5 x2 + y2 = 36 Contains y2, so it is not a function. |y| = 5 There is no way to isolate y with a single equation; therefore, it is not a function.
Instead of using the variable y, often you will see the variable f(x). This is shorthand for "function of x" to automatically indicate that an equation is a function. This can be confusing; f(x) does not indicate two variables f and x multiplied together; it is a notation that means the single variable y.
Although it may seem that f(x) is not an efficient shorthand (it has more characters than y), it is very eloquent way to indicate that you are being given expressions to evaluate. For example, if you are given the equation f(x) = 5x – 2, and you are being asked to determine the value of the equation at x = 2, you need to write "evaluate the equation f(x) = 5x – 2 when x = 2." This is very wordy. With function notation, you only need to write "determine f(2)." The x in f(x) is replaced with a 2, indicating that the value of x is 2. This means that f(2) = 5(2) – 2 = 10 – 2 = 8.
All you need to do when given an equation f(x) and told to evaluate f(value) is to replace the value for every occurrence of x in the equation.
- Given the equation f(x) = 2x2 + 3x + 1, determine f(0) and f(–1).
- f(0) means replace the value 0 for every occurrence of x in the equation and evaluate.
- f(0) = 2(0)2 + 3(0) + 1
- = 0 + 0 + 1
- = 1
- f(–1) means replace the value –1 for every occurrence of x in the equation and evaluate.
- f(0) = 2(–1)2 + 3(–1) + 1
- = 2(1) + –3 + 1
- = 2 – 3 + 1
- = 0
Families of Functions
There are a number of different types, or families, of functions. Each function family has a certain equation and its graph takes on a certain appearance. You can tell what type of function an equation is by just looking at the equation or its graph.
These are the shapes that various functions have. They can appear thinner or wider, higher or lower, or upside down.
Systems of Equations
A system of equations is a set of two or more equations with the same solution. Two methods for solving a system of equations are substitution and elimination.
Substitution involves solving for one variable in terms of another and then substituting that expression into the second equation.
- 2p + q = 11 and p + 2q = 13
- First, choose an equation and rewrite it, isolating one variable in terms of the other. It does not matter which variable you choose.
- 2p + q = 11 becomes q = 11 – 2p.
- Second, substitute 11 – 2p for q in the other equation and solve:
- p + 2(11 – 2p) = 13
- p + 22 – 4p = 13
- 22 – 3p = 13
- 22 = 13 + 3p
- 9 = 3p
- p = 3
- Now substitute this answer into either original equation for p to find q.
- 2p + q = 11
- 2(3) + q = 11
- 6 + q = 11
- Thus, p = 3 and q = 5.
- q = 5
The elimination method involves writing one equation over another and then adding or subtracting the like terms on the same sides of the equal sign so that one letter is eliminated.
- x – 9 = 2y and x – 3 = 5y
- Rewrite each equation in the same form.
- x – 9 = 2y becomes x – 2y = 9 and x – 3 = 5y becomes x – 5y = 3.
- If you subtract the two equations, the "x" terms will be eliminated, leaving only one variable: Subtract:
- x – 2y = 9
- –(x – 5y = 3)
- Substitute 2 for y in one of the original equations and solve for x.
- x – 9 = 2y
- x – 9 = 2(2)
- x – 9 = 4
- The answer to the system of equations is y = 2 and x = 13.
- y = 2 is the answer.
- x – 9 + 9 = 4 + 9
- x = 13
If the variables do not have the same or opposite coefficients as in the preceding example, adding or subtracting will not eliminate a variable. In this situation, it is first necessary to multiply one or both of the equations by some constant or constants so that the coefficients of one of the variables are the same or opposite. There are many different ways you can choose to do this.
- 3x + y = 13
- x + 6y = –7
We need to multiply one or both of the equations by some constant that will give equal or opposite coefficients of one of the variable. One way to do this is to multiply every term in the second equation by –3.
- 3x + y = 13
- –3(x + 6y = –7) → –3x – 18y = 21
Now if you add the two equations, the "x" terms will be eliminated, leaving only one variable. Continue as in the preceding example.
- 3x + y = 13
- –3x – 18y = 21
- y = –2 is the answer.
- Substitute –2 for y in one of the original equations and solve for x.
- 3x + y = 13
- 3x + (–2) = 13
- 3x + (–2) + –2= 13 + –2
- The answer to the system of equations is y = –2 and x = .
- 3x = 11
Linear inequalities are solved in much the same way as simple equations. The most important difference is that when an inequality is multiplied or divided by a negative number, the inequality symbol changes direction.
10 > 5 so (10)(–3) < (5)(–3) –30 < –15
Solving Linear Inequalities
To solve a linear inequality, isolate the letter and solve the same way as you would in a linear equation. Remember to reverse the direction of the inequality sign if you divide or multiply both sides of the equation by a negative number.
- If 7 – 2x > 21, find x.
- Isolate the variable.
- 7 – 2x > 21
- Because you are dividing by a negative number, the inequality symbol changes direction.
- x < –7
- The answer consists of all real numbers less than –7.
Solving Compound Inequalities
To solve an inequality that has the form c < ax + b < d, isolate the letter by performing the same operation on each part of the equation.
- If –10 < –5y – 5 < 15, find y.
- Add five to each member of the inequality.
- –10 + 5 < –5y – 5 + 5 < 15 + 5 – 5 < –5y < 20
- Divide each term by –5, changing the direction of both inequality symbols:
The solution consists of all real numbers less than 1 and greater than –4.
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