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College Algebra K/DCFriday, 23 October 2015

• OBJECTIVE TSW review terminology of equations, solve linear equations, and identify equations as identities, conditionals, or contradictions.

• TESTS are graded.

• TODAY’S ASSIGNMENT (due on Monday, 10/26/15)

– Sec. 1.1: pp. 84-85 (9-35 odd)

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Write the problem and solve each equation. Use solution sets.

9) 11)

13) 15)

17) 19)

21) 23)

25) 27)

Write the problem. Decide whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set.

29) 31)

33) 35)

If you finish today, keep it until Monday.

Assignment: Sec. 1.1: pp. 84-85 (9–35 odd) Due Monday, 10/26/15

5 4 3 4x x 6 3 1 8 10 14x x

5 4 5

26 3 3

x x

2 4 2 3 2 2x x x

0.2 0.5 0.1 7x x

4

0.5 103

x x x

4 2 7 2 22 3 2 2x x x

0.3 2 0.5 2 0.2 4x x x

3 5 5 1 6 7x x x

1 10

3 214 10

xx

4 2 6 8 5 24x x x x

0.08 0.06 12 7.72x x

2 8 3 16x x

4 7 2 12 2 1x x x

Solving Linear Equations1.1

Terminology

• An equation is a statement that two expressions are equal.

• To solve an equation means to find all numbers that make the equation true.

• These numbers are the solutions or roots of the equation.

• A number that is a solution satisfies the equation.

• The solution(s) of an equation make up its solution set.

• Equations with the same solution set are equivalent equations.

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Solve .

Solving a Linear Equation

Solution set: {6}

Distributive property

Combine terms.

Add 4 to both sides.Add 12x to both sides.Combine terms.Divide both sides by 4.

NOTE: You are solving an equation; Use solution sets!

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Solve .

Clearing Fractions Before Solving a Linear Equation

Solution set: {–10}

Multiply by 10, the LCD of all the fractions.

Distributive property

Combine terms.

Add –4s and –6 to both sides. Combine terms.

Divide both sides by –6.

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Solve

Clearing Decimals Before Solving a Linear Equation

Solution set: {12}

0.4 3.6 0.2 4 6 .x x

0.4 3.610 10 0.2 4 6x x

4 36 2 4 6x x

4 36 8 12x x 48 4x12 x

Multiply both sides by 10(one decimal place).

Distribute on the right side.

Add 12 and subtract 4x from both sides.

Divide both sides by 4.

Types of Equations

• An identity is an equation that leads to a true statement.– It can be simplified to 0 = 0.– The solution set is {all real numbers}, or ℜ.

• A conditional equation is an equation whose solution is a single number.– Ex: x = 3– Braces are used to indicate solutions { }.

• A contradiction is an equation that leads to a false statement.– Ex: –3 = 7– The solution set is the empty set, or Ø but it is NOT

{Ø}.

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Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set.

Identifying Types of Equations

This is a conditional equation.Solution set: {11}

Add –4x and 9 to both sides. Combine terms.

Divide both sides by 2.

1-10

Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set.

Identifying Types of Equations

This is a contradiction.Solution set: ø

Distributive property

Subtract 14x from both sides.

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Decide whether the equation is an identity, a conditional equation, or a contradiction. Give the solution set.

Identifying Types of Equations

This is an identity.Solution set: {all real numbers}

Distributive property

Combine terms.

Add x and –3 to both sides.

The most common mistake on this type of problem is to forget to state the solution set !

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Write the problem and solve each equation. Use solution sets.

9) 11)

13) 15)

17) 19)

21) 23)

25) 27)

Write the problem. Decide whether each equation is an identity, a conditional equation, or a contradiction. Give the solution set.

29) 31)

33) 35)

If you finish today, keep it until Monday.

Assignment: Sec. 1.1: pp. 84-85 (9–35 odd) Due Monday, 10/26/15

5 4 3 4x x 6 3 1 8 10 14x x

5 4 5

26 3 3

x x

2 4 2 3 2 2x x x

0.2 0.5 0.1 7x x

4

0.5 103

x x x

4 2 7 2 22 3 2 2x x x

0.3 2 0.5 2 0.2 4x x x

3 5 5 1 6 7x x x

1 10

3 214 10

xx

4 2 6 8 5 24x x x x

0.08 0.06 12 7.72x x

2 8 3 16x x

4 7 2 12 2 1x x x

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College Algebra K/DCMonday, 26 October 2015

• OBJECTIVE TSW solve equations for a specified variable.

• ASSIGNMENT DUE (wire basket)

– Sec. 1.1: pp. 84-85 (9-35 odd)

• TODAY’S ASSIGNMENT (due Wed/Thur)

– Sec. 1.1: p. 85 (39-47 all, 49-58 all)

Assignment: Sec. 1.1: p. 85 (39-47 all, 49-58 all) Due Wednesday/Thursday, 28/29 October 2015.

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Write each problem. Then, solve for the indicated variable.

, for 39) V lwh l , for 40) P r t PI

, for 41) P a b c c 2 2 ,42) for P l w w

1 , for

244) A h B b h 1

, for 2

43) A h B b B

21 , for

246) s gt g2 2 2 , fo 4 ) r5 S rh r h

2 2 2 , f r 47) oS lw wh hl h Write each problem. Then, solve for x.

49) 2 3x a b x a 5 250) 1x a c a x

51) 3ax b x a 52) 4 3a ax b bx

53) 31

xax

a

1

22

54)x

x aa

2 2 3 255) a x x a 2 2 56) ax b bx a

7) 3 1 45 2x x m 5 38 1) 35 x x k

15

Solving For a Specified Variable1.1

2-16

Solve for the specified variable.

Solving for a Specified Variable

(a) d = rt, for t

(b) , for k

Divide both sides by r.

Factor out k.

Divide both sides by

2-17

Solve for the specified variable.

Solving for a Specified Variable

(c) , for y

Distributive property

Subtract 8y and 8 from both sides.

Divide both sides by 3.

16 62 3 42 , for

210) t s t s t

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Write each problem and solve. Leave answers as fractions when necessary. Use solution sets

Class Problems – Sec. 1.1 (10/26/2015) (Due today – show work) Bring to me, please.

9 11) 1 7 1x x 2) 8 3 4 6 4 8 4x x x x

8 3 83) 1 5x x x

0.6 5 0.8 6 0.2 . 1 84) x x x

6 2 1 35) 4 15 1x x x

Write the equation and solve. Idenitfy as an identity, conditional, or contradiction. Give the solution set.

Write the equation and solve for the indicated variable. Use solution sets.

7) 12 31 8 , for g h g h 56 for

28)

xy ny

, for9) PV nRT n

2 6 81 5 126) 7 3 27x x x

Assignment: Sec. 1.1: p. 85 (39-47 all, 49-58 all) Due Wednesday/Thursday, 28/29 October 2015.

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Write each problem. Then, solve for the indicated variable.

, for 39) V lwh l , for 40) P r t PI

, for 41) P a b c c 2 2 ,42) for P l w w

1 , for

244) A h B b h 1

, for 2

43) A h B b B

21 , for

246) s gt g2 2 2 , fo 4 ) r5 S rh r h

2 2 2 , f r 47) oS lw wh hl h Write each problem. Then, solve for x.

49) 2 3x a b x a 5 250) 1x a c a x

51) 3ax b x a 52) 4 3a ax b bx

53) 31

xax

a

1

22

54)x

x aa

2 2 3 255) a x x a 2 2 56) ax b bx a

7) 3 1 45 2x x m 5 38 1) 35 x x k