At Colorado State University, the College Algebra sequence
extensively uses the graphing features of

the Texas Instruments TI-83 or TI-84 graphing calculator. Whenever possible, the
questions on the

Math Placement Exam and in this review are designed to accommodate the student
who has no

knowledge of these graphing features. However, students may find it helpful to
learn about these

features.

Polynomial expressions contain letters that symbolize
numbers and obey the usual laws of arithmetic.

In adding, subtracting or multiplying polynomials, we apply these laws and
combine like terms .

**Example:** Simplify the polynomial expression (x -
3)(2x + 5) - 6x(x - 1).

Solution:

**Problems:**

1. Simplify: (3x - 7)(2x - 5) - x(4x - 3)

2. Simplify: (7x + 2)^{2} - 5x[-3(x + 2)]

3. Simplify: (2x^{2} - 5x + 3)(3x^{2} + x - 2)

4. Simplify: 2(x + 3)^{3} - (x - 1)(x + 1)

An equation (a statement equating two expressions ) can be
manipulated by adding (or subtracting) the

same quantity from both sides, or by multiplying (or dividing) both sides by the
same non- zero

quantity. To solve a simple linear equation we manipulate the equation into the
form

Ax = B,

then divide by A to conclude that

x = B/A

**Example:** Solve for x in the fol lowing equation : 3(x
- 2) - x = -5x - 4(2 - x).

Solution:

First perform all indicated multiplications: 3x - 6 - x = -5x - 8 + 4x.

Next simplify each side separately: 2x - 6 = -x - 8.

Now isolate all x terms on one side of the equation and all constant terms on
the other

by adding x to both sides and adding 6 to both sides

3x = -2.

Divide by 3 to obtain the solution

x = -2/3

**Problems:**

5. Solve for x: 2(x + 4) = 5x - 12

6. Solve for u: -6[u - 4(u + 2)] = 5u - 4

7. Solve for R: 2R - 2(R + 3) = 4[2R - 3(1 + R)]

## MATHEMATICAL MODELS

A mathematical model is a formula or algebraic equation
that can be formed to model real-world

situations.

To model a real-world situation:

· Read the problem carefully. What information is given to you? What is the
problem looking

for? Let x (or any variable ) represent one of the quantities of the problem.

· Write an expression with the information given.

· Evaluate the expression.

· Reread the problem. Does your answer make sense with the context of the
problem?

**Example:** The length of the rectangular tennis court
is 6 feet longer than twice the width. If the

court’s perimeter is 228 feet, what are the courts dimensions?

Solution: Let x = the width of the court. The length must
be = 6 + 2x. We are given that the

perimeter = 228. The formula for perimeter is: 2W + 2L = P.

The dimensions of the tennis court are 36 ft. by 78 ft.

**Problems:**

8. For large parties, many restaurants automatically add an 18% gratuity to the
bill. If the total bill,

including gratuity is $163.43, what was the bill before the gratuity was added?

9. A job pays an annual salary of $44,550, which includes a year-end bonus of
$750. If paychecks

are issued twice a month, what is the gross amount of each paycheck?

## INEQUALITIES/ABSOLUTE VALUES

Inequalities are manipulated in the same way as equalities
with one notable exception. The direction

of the inequality changes when an inequality is multiplied or divided by a
negative number . Such is

the case in the following:

8 > 3 so (-2)(8) < (-2)(3), since -16 < -6.

**Example:** Simplify the following: 2x + 5 ≥- x + 4.

Solution: Simplify the following: 2x + 5 ≥ -x + 4 .

Add x to both sides: 3x + 5 ≥4.

Subtract 5 from both sides, 3x ≥ -1, so
that x ≥ -1/3.

Note that we divided by a positive number, therefore, no
change in the inequality sign

occurs. The solution of an inequality is a set of numbers and can be graphed on
the

number line. In the above example, the solutions are graphed as the (closed)
half- line to

the right of -1/3 (inclusive).

**Example:** Simplify the following: -2x + 5 > x + 4.

Solution: Subtract x from both sides: -3x + 5 > 4.

Subtract 5 from both sides: -3x > -1.

Then x <1/3.

Note that we divided by a negative number; therefore, a
sign change occurs in our

inequality. In this example, all x < 1/3 is graphed as the (open) half-line to
the left of 1/3

excluding 1/3.

**Problems:**

10. Find the solution set of 5(x - 3) + 1 > -4(x - 1).

11. Find the solution set of -3x + 2 ≤ 5(2x + 3).

12. Find the solution set of

13. Find the solution set of (x + 2)(2x - 3) < 2(x + 1)(x + 5).

The absolute value of a number x is defined as

Hence, |5| = 5 and |-3| = 3. The absolute value of x can be visualized on a
number line as the distance

the point x lies from the origin. Graphically we see that -3 is three units from
the origin and 5 is five

units from the origin.

**Example:** Find the set of x-values satisfying |x| <
2.

Solution:

When x ≥ 0, |x| = x and the inequality becomes x < 2. Thus, the non-negative
x-values

that satisfy |x| < 2 are given by 0 ≤ x < 2.

When x < 0, |x| = -x and the inequality becomes -x < 2, or -2 < x. Thus, the
negative xvalues

that satisfy x < 2 are given by - 2 < x < 0 .

It follows that the set of all x-values satisfying |x| < 2 consists of all
numbers x strictly

between -2 and 2 (excluding -2 and 2). The set of x-values satisfying |x| < 2,
is the

collection of numbers within two units of the origin on the number line.

**Example:** Find the solution set of |2x - 1| < 3.

Solution:

Any solution of |2x - 1| < 3 must satisfy both of the inequalities 2x - 1 < 3
and

-3 < 2x - 1. These two inequalities can be written together as

-3 < 2x - 1 < 3

and then solved at the same time. Add 1 to all members of the inequality to get

-2 < 2x < 4. Divide by 2 to find that the solution set is the interval -1 < x <
2 of all

x-values strictly between -1 and 2.

**Problems:**

14. Find the solution set of |6 - 2x| ≤ 3.

15. Find the solution set of

**Example:** Find the set of x-values satisfying |x| ≥
2.

Solution:

When x ≥ 0, |x| = x and the inequality becomes x ≥ 2. Thus, the non-negative x-
values

that satisfy |x| ≥2 are given by x ≥ 2.

When x < 0, |x| = -x and the inequality becomes -x ≥ 2, or x ≤ -2. Thus, the
negative xvalues

that satisfy |x| ≥ 2 are given by x ≤ -2.

Thus, the set of all x-values satisfying |x| ≥2 is comprised of two disjoint
sets. One set

is the real numbers x greater than or equal to 2. The other set is the real
numbers x less

than or equal to -2. The set of x-values satisfying |x| ≥ 2 consists of the
numbers two or

more units from the origin on the number line.

**Example:** Find the solution set of |2x - 1| ≥3.

Solution:

Any solution of |2x - 1| ≥ 3 must satisfy either the inequality 2x - 1 ≥ 3 or
the inequality

2x - 1 ≤ -3. These two inequalities cannot be written together and should not be
solvedat the same time.

When 2x – 1 ≥ 3,

2x ≥ 4,

so x ≥ 2

When 2x – 1 ≤ -3,

2x ≤ -2,

so x ≤ -1.

The solution set of |2x - 1| ≥ 3 is comprised of two
pieces. One piece is the real

numbers x greater than or equal to 2. The other is the real numbers x less than
or equal
to -1.

Notice that the solution set for |2x - 1| < 3 is an
interval and the solution set for

|2x - 1| ≥ 3 is the complement of the same interval.

**Problems:**

16. Find the solution set of |5 - 2x| ≥ 3.

17. Find the solution set of