Linear Equations in One variable

Nonlinear Equations

4x = 8

3x – = –9

2x – 5 = 0.1x +2

Notice that the variable in a linear equation is not under a radical sign and is not raised to a power other than 1. The variable is also not an exponent and is not in a denominator.

Solving a linear equation requires isolating the variable on one side of the equation by using the properties of equality.

+ 1 = 32

+ 1 = 41

3 – 2x = –5

An equation is a mathematical statement that two expressions are equivalent. The solution set of an equation is the value or values of the variable that make the equation true. A linear equation in one variable can be written in the form ax = b, where a and b are constants and a ≠ 0.

To isolate the variable, perform the inverse or opposite of every operation in the equation on both sides of the equation. Do inverse operations in the reverse order of operations.

The local phone company charges $12.95 a month for the first 200 of air time, plus $0.07 for each additional minute. If Nina’s bill for the month was $14.56, how many additional minutes did she use?

Ex 1:

monthly charge plus

additional minute charge

times

12.95 0.07

number of additional minutes

total charge

+

=

m 14.56* =

12.95 + 0.07m = 14.56

0.07m = 1.61

0.07 0.07

m = 23

Nina used 23 additional minutes.

–12.95 –12.95

Ex 2: Solve 4(m + 12) = –36

Method 1

4(m + 12) = –364 4

m + 12 = –9

m = –21

–12 –12

4m + 48 = –36

4m = –84

–48 –48

=4m –84 4 4

m = –21

Method 2

If there are variables on both sides of the equation, (1) simplify each side. (2) collect all variable terms on one side and all constants terms on the other side. (3) isolate the variables as you did in the previous problems.

Ex 3:

Simplify each side by combining like terms.

–11k + 25 = –6k – 10

Collect variables on the right side.

Add.

Collect constants on the left side.

Isolate the variable.

+11k +11k

25 = 5k – 10

35 = 5k

5 5

7 = k

+10 + 10

Solve 3k– 14k + 25 = 2 – 6k – 12.

You have solved equations that have a single solution. Equations may also have infinitely many solutions or no solution.

An equation that is true for all values of the variable, such as x = x, is an identity. An equation that has no solutions, such as 3 = 5, is a contradiction because there are no values that make it true.

Solve 3v – 9 – 4v = –(5 + v).

Ex 4:

3v – 9 – 4v = –(5 + v) –9 – v = –5 – v + v + v

–9 ≠ –5 x

The equation has no solution. The solution set is the empty set, which is represented by the symbol .

Solve 2(x – 6) = –5x – 12 + 7x.

2(x – 6) = –5x – 12 + 7x

2x – 12 = 2x – 12

–2x –2x

–12 = –12

The solutions set is all real number, or .

An inequality is a statement that compares two expressions by using the symbols <, >, ≤, ≥, or ≠. The graph of an inequality is the solution set, the set of all points on the number line that satisfy the inequality.

The properties of equality are true for inequalities, with one important difference. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

These properties also apply to inequalities expressed with >, ≥, and ≤.

Solve and graph 8a –2 ≥ 13a + 8.

Example 5: Solving Inequalities

Subtract 13a from both sides.8a – 2 ≥ 13a + 8

–13a –13a

–5a – 2 ≥ 8Add 2 to both sides. +2 +2

–5a ≥ 10Divide both sides by –5 and reverse the inequality.

–5 –5

–5a ≤ 10

a ≤ –2

Example 5 Continued

Check Test values in the original inequality. –10 –9 –8 –7 –6 –5 –4 –3 –2 –1

•

Test x = –4 Test x = –2 Test x = –1

8(–4) – 2 ≥ 13(–4) + 8 8(–2) – 2 ≥ 13(–2) + 8 8(–1) – 2 ≥ 13(–1) + 8

–34 ≥ –44

So –4 is a solution.

So –1 is not a solution.

So –2 is a solution.

–18 ≥ –18 –10 ≥ –5 x

Solve and graph 8a – 2 ≥ 13a + 8.