2.1 – Linear Equations in One VariableAlgebraic equation is a statement that two expressions have equal value.

Solving algebraic equations involves finding values for a variable that make the equation true.

Linear equation in one variable can be written in the form:ax + b = c, a 0.

Equivalent equations are equations with the same solutions in the form of:

variable = number, x = 3, or

number = variable, 3 = x.

2.1 – Linear Equations in One VariableA linear equation in one variable can be written in the form:

ax + b = c where a, b, and c are real numbers and a ≠ 0.

Addition Property of Equality

a = b and a + c = b + c

Multiplication Property of Equality

equivalent equations

a = b and ac = bc

equivalent equations

2.1 – Linear Equations in One Variable

Simplify both sides

8 + (–8) + z = –8 + –8

8 + z = – 8

Add –8 to each side

z = –16

Example

2.1 – Linear Equations in One Variable

Example

3z – 1 = 26

3z = 27

z = 9

3z – 1 + 1 = 26 + 1

3

27

3

3z

2.1 – Linear Equations in One Variable

Example

4p – 11 – p = 2 + 2p – 20

3p – 11 = 2p – 18

p = –7

p – 11 = –18

3p + (– 2p) – 11 = 2p + (– 2p) – 18

p – 11 + 11 = –18 + 11

2.1 – Linear Equations in One Variable

Example

5(3 + z) – (8z + 9) = – 4z

15 + 5z – 8z – 9 = – 4z

6 – 3z = – 4z

6 + z = 0

z = –6

6 – 3z + 4z = – 4z + 4z

6 + (–6) + z = 0 + (–6)

2.1 – Linear Equations in One Variable

Example

12x + 30 + 8x – 6 = 10

20x + 24 = 10

20x = – 14

20

14

20

20 x

10

7x

20x + 24 + (– 24) = 10 + (– 24)

2.1 – Linear Equations in One Variable

Example

301093 yy

7

7

7

21 y

y 3

3( 3) 2 65y y

6255

)3(35

y

y

30)3(109)3(3 yyyy

)30(307)30(9 y

LCD: 5

2.1 – Linear Equations in One Variable

Example

5x – 5 = 2(x + 1) + 3x – 7

5x – 5 = 2x + 2 + 3x – 7

5x – 5 = 5x – 5

Both sides of the equation are identical.

This equation will be true for every x that is substituted into the equation. The solution is “all real numbers.”

The equation is called an identity.

2.1 – Linear Equations in One Variable

Example

3x – 7 = 3(x + 1)

3x – 7 = 3x + 3

–7 = 3

3x + (– 3x) – 7 = 3x + (– 3x) + 3

No value for the variable x can be substituted into this equation that will make this a true statement.

There is “no solution.”

The equation is a contradiction.

2.2 - An Introduction to Problem SolvingGeneral Strategy for Problem Solving

UNDERSTAND the problem.

Read and reread the problem..Choose a variable to represent the unknown.

Construct a drawing.

Propose a solution and check.TRANSLATE the problem into an equation.

SOLVE the equation.

INTERPRET the result:Check proposed solution in problem.State your conclusion.

2.2 - An Introduction to Problem SolvingExample

Twice a number plus three is the same as the number minus six.

2 3 6

2 3 6

3 6

3 3 6 3

9

x x

x x x x

x

x

x

2x +3 = x – 6

2.2 - An Introduction to Problem Solving

The product of twice a number and three is the same as the difference of five times the number and ¾. Find the number.

Example

2x ·3 = 5x

2x · 3 = 5x – ¾

6x = 5x – ¾

x = –¾

6x + (–5x) = 5x + (–5x) – ¾

2.2 - An Introduction to Problem Solving

A car rental agency advertised renting a Buick Century for $24.95 per day and $0.29 per mile. If you rent this car for 2 days, how many whole miles can you drive on a $100 budget?

Example

0.29x + 2(24.95) = 100

x = the number of whole miles driven

0.29x = the cost for mileage driven

cost for mileage driven + daily rental cost = $100

0.29x + 49.90 = 100

2.2 - An Introduction to Problem Solving

0.29x = 50.10

29.0

10.50

29.0

29.0x

x 172.75

0.29x + 49.90 – 49.90 = 100 – 49.90

0.29x + 49.90 = 100

miles

2.2 - An Introduction to Problem Solving

A pennant in the shape of an isosceles triangle is to be constructed an Athletic Club and sold at a fund-raiser. The company manufacturing the pennant charges according to perimeter, and the athletic club has determined that a perimeter of 149 centimeters should make a nice profit. If each equal side of the triangle is twice the length of the third side, increased by 12 centimeters, find the lengths of the sides of the triangular pennant.

Example

Let x = the third side of the triangle

2x + 12 = the first side

2x + 12 = the second side

2x + 122x + 12

x

2.2 - An Introduction to Problem Solving

2x + 122x + 12

x

Perimeter is 149 cm.

2x + 12 x+ = 149

2 2 12 12 149x x x 5 24 149x

5 125x 25x

+ 2x + 12

cm

2(25) + 12 = 62 cm

2.2 - An Introduction to Problem SolvingExample

The measure of the second angle of a triangle is twice the measure of the smallest angle. The measure of the third angle of the triangle is three times the measure of the smallest angle. Find the measures of the angles.

Let x = degree measure of smallest angle

xº 2xº

3xº

2x = degree measure of second angle

3x = degree measure of third angle

2.2 - An Introduction to Problem SolvingThe sum of the measures of the angles of a triangle equals 180.

x 2x+ 3x+ = 180

x = 30º

6x = 180

6 180

6 6

x

2x = 60º

3x = 90º

2.2 - An Introduction to Problem SolvingExample

The sum of three consecutive even integers is 252. Find the integers.

x = the first even integer

x + 2 = next even integer

x + 4 = next even integer

x + x + 2 + x + 4 = 252

3x + 6 = 252

3x = 2463 246

3 3

x 82x The integers are:

82, 84 and 86.

2.3 – Formulas and Problem SolvingA formula:An equation that states a known relationship among multiple quantities (has more than one variable in it).

A = lw (Area of a rectangle = length · width)I = PRT (Simple Interest = Principal · Rate · Time)d = rt (distance = rate · time)V = lwh (Volume of a rectangular solid = length·width·height)

2.3 – Formulas and Problem SolvingSolving Equations for a Specified Variable

Step 1: Multiply on both sides to clear the equation of fractions if they appear.

Step 2: Use the distributive property to remove parentheses if they appear.

Step 3: Simplify each side of the equation by combining like terms.

Step 4: Get all terms containing the specified variable on one side and all other terms on the other side by using the addition property of equality.

Step 5: Get the specified variable alone by using the multiplication property of equality.

2.3 – Formulas and Problem SolvingExample

mr

mnr

mr

T

nmr

T

Solve: T = mnr for n.

2.3 – Formulas and Problem Solving

23 22 7xy x x

Solve: 3y – 2x = 7 for y.

3 7 2y x

3 7

3 3

2y x

7 2 7 2 or

3 3 3

xy y x

Example

2.3 – Formulas and Problem Solving

Solve: for A = P + PRT for T.

PRTPPPA

PRTPA

PR

PRT

PR

PA

TPR

PA

Example

2.3 – Formulas and Problem Solving

)1( RTPA

RT

RTP

RT

A

1

)1(

1

PRT

A

1

Solve: A P PRT for P.

Example

Factor out P on the right side.

Divide both sides by 1 + RT.

Simplify.

2.3 – Formulas and Problem SolvingExample

A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet.

x = the length of the shortest side

2x = the length of the second side

x + 30 = the length of the third side

x 2x

x + 30

2.3 – Formulas and Problem Solving

x 2x

x + 30

Perimeter is 102 feet

P = a + b + c

102 = x + 2x + x + 30

102 = 4x + 30

102 – 30 = 4x + 30 – 30

72 = 4x

4

4

4

72 x x = 18 feet 2(18) = 36 feet

18 + 30 = 48 feet