Chapter 2 Equations and Inequalities

Worksheet 5 (2.1) Chapter 2 Equations and Inequalities

2.1 Solving First-Degree Equations

Important Terminology Used when Solving Equations

An equation is a statement that two symbols or groups of symbols are names for the same number. An algebraic equation contains one or more variables. It is called an open sentence because it is neither true nor false. Solving an equation is the process used to find the number or numbers that make an algebraic equation a true numerical statement. A solution or root of an equation is the number that satisfies a given equation. The solution set is the set of all solutions of an equation. First-degree equations in one variable are equations with one variable which has an exponent of one. Equivalent equations are equations that have the same solution.

Summary 1: Warm-up 1. a) Is 5 a solution of 2x + 4 = 19 - x? Verify this by showing that the numerical statement is true or false. 2( ) + 4 = 19 - + 4 = 14 = True or False?

Answer: ______yes or no

Problem

1. Is 0 a solution of 3(x - 1) = 2x + 5? Verify this by showing that the numerical statement is true or false.

Worksheet 5 (2.1)

17 Further Properties of Equality Used to Solve Equations

1. Addition Property of Equality

For all real numbers a, b, and c, a = b if and only if a + c = b + c.

Note: Subtraction can be rewritten as addition. Therefore, this property also allows us to subtract the same number on both sides.

2. Multiplication Property of Equality

For all real numbers a, b, and c, where c  0, a = b if and only if ac = bc. Note: Division can be rewritten as multiplication. Therefore, this property also allows us to divide the same number on both sides.

Summary 2: Warm-up 2. Solve: a) -11 = 3x + 1 -11 + = 3x + 1 + -12 = 3x ( )(-12 ) = ( ) ( 3 x) = x x = The solution set is .

b) 2(x - 1) = 5x + 7 2x - = 5x + 7 2x - 2 + = 5x + 7 + - 3x - 2 = 7 - 3x - 2 + = 7 + - 3x =

Worksheet 5 (2.1) ( )(-3 x) = ( )( 9 ) x = The solution set is .

Problems - Solve:

2. -5x - 3 = 12

18 3. 2(2x - 3) = 5x - 11

Solving First-Degree Equations in One Variable

1. Simplify each side of the equation. 2. Apply the addition property of equality to isolate the variable term on one side and constant on the other side. 3. Apply the multiplication property of equality to obtain the coefficient of 1 for the variable. 4. Write an appropriate solution set and check the solution to verify that the resulting numerical statement is true.

Summary 3: Warm-up 3. Solve and check:

a) 5n - 4 - 2n = 4n + 12 + 3n - 4 = + 12 3n - 4 + = 7n + 12 + -4n - 4 = 12 -4n - 4 + = 12 + -4n = ( )(-4n) = ( )(16) n =

The solution set is .

Worksheet 5 (2.1)

Check: 5( ) - 4 - 2( ) = 4( ) + 12 + 3( ) ____ - 4 - ____ = ____ + 12 + ______= ____ True or False?

b) 4(m - 1) - (m + 6) = 5(m + 2) 4m - 4 - m - ____ = 5m + 10 ____ - 10 = 5m + 10 3m - 10 + ____ = 5m + 10 + ______- 10 = 10 -2m - 10 + ____ = 10 + ____ -2m =

19 ( )(-2m) = ( )(20) m = ____ The solution set is ______.

Problems - Solve and check:

4. 3n - 1 - 4n = 6n - 7 - 4n

5. -(n - 5) + 3(n + 2) = 4(n - 3) - 1

Using Equations to Solve Word Problems

1. Carefully read the word problem. Reread to get an overview of the situation. Determine known facts and identify unknown quantities. 2. Declare the variable by choosing a letter to use as a variable and tell in words what it will represent. 3. Express other unknown quantities using this variable. 4. Translate the word problem or use a guideline from the word problem to write an algebraic equation to solve. 5. Check results to determine whether or not they satisfy the conditions stated in the original problem. 6. Express answer as a complete sentence.

Summary 4:

Worksheet 5 (2.1)

Warm-up 4. Set up and write an algebraic equation, then solve:

a) The sum of two numbers is 55. The larger number is eleven more than the smaller number. Find the two numbers.

Declare the variable:

Let n = the smaller number = the larger number

20 Write an algebraic equation and solve:

n + ______= 55 2n + ____ = 55 2n + 11 + ____ = 55 + ____ 2n = ____ ( )(2n) = ( )(44) n = ____ ; n + 11 = ____

The numbers are ______and ______.

Problems - Set up and write an algebraic equation, then solve:

6. In a class containing 58 students, the number of women students is one more than twice the number of men students. How many women students are in the class?

7. An air-conditioner repair bill is $87. This included $30 for parts and an amount for 3 hours of labor. What was the hourly rate that was charged for labor?

Worksheet 6 (2.2)

2.2 Equations Involving Fractional Forms

Summary 1:

21 Using the LCD to Clear Fractions in an Equation The least common denominator, LCD, refers to the least common multiple of a set of denominators.

Clearing the equation of all fractions is accomplished by applying the multiplication property of equality. Both sides of the equation are to be multiplied by the LCD in the equation to produce an equivalent equation.

When an equation has more than one term on one or both sides of the equation, the distributive property is used to multiply each term by the LCD.

Warm-up 1. Find the LCD and multiply both sides of the equation by the LCD to clear the fractions. Give the resulting equivalent equation. 3 1 2 a) - + x = The LCD is _____. 4 2 3  3 1   2  - + x =    4 2   3   3   1   2  -  +  x =    4   2   3  -9 + _____ = _____

The resulting equivalent equation is ______. 2 1 b) (4x - 1) + 1 = - (5x + 2) The LCD is _____. 5 4 2   1  4x - 1 + 1 = - 5x + 2 5   4  2   1  (4x - 1) + 1 = - (5x + 2) 5   4  (4x - 1) + = (5x + 2) The resulting equivalent equation is ______.

Worksheet 6 (2.2)

Problems - Find the LCD and multiply both sides of the equation by the LCD to clear the fractions. Give the resulting equivalent equation. 4x - 1 3x + 1 1. = - x + 4 9 3

1 1 2. (6x - 1) + = - 2x - 5 2 2

22 Solving Equations Involving Fractional Forms

1. Find the LCD in the equation. 2. Apply the multiplication property of equality to clear the equation of all fractions. This is done by multiplying each term in the equation by the LCD. (See summary 1 above.) 3. Solve the resulting equivalent equation. 4. Check when directed to do so.

Summary 2:

3 1 2 Warm-up 2. a) Solve and check: - + x = 4 2 3  3 1   2  - + x =    4 2   3   3   1   2  -  +  x =    4   2   3  -9 + _____ = _____

_____ = 17

x = _____

The solution set is ______. Worksheet 6 (2.2) 3 1 2 Check: - + ( ) = 4 2 3

23 3 2 - + = 4 3 2 = True or False? 3

4x - 1 3x + 1 b) Solve: = - x + 4 9 3  4x - 1  3x + 1    =  - x + 4  9   3   4x - 1  3x + 1   =   - (x) + (4)  9   3  4x - 1 = (3x + 1) - 9x + 36 4x - 1 = 9x + - 9x + 36 4x - 1 = 4x = x = The solution set is ______.

Problems - Solve: 3a - 1 a - 2 21 a - 1 3. + = + 4 3 20 5

1 1 4. (2x - 1) = 3 + (5x + 2) 2 3

Worksheet 6 (2.2)

Summary 3:

24 Suggestions for Solving Word Problems 1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Find a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer in a complete sentence.

a) Find a number such that one-half of the number is three more than one-fourth of the number.

Let ____ = a number

Write an algebraic equation and solve:

= + 3 4( ) = 4( ) + 4(3) = + 12 = 12

The number is ______.

Worksheet 6 (2.2)

b) In triangle ABC, angle A is one-third angle C and angle B is 40 degrees more than two-thirds of angle A. Find the measures of the three angles in the triangle.

Draw a figure:

25 Let x = measure of angle C

______= measure of angle A

______= measure of angle B

x + + = 180 9( ) + 9( ) + 9( ) = 9(180) 9x + 3x + = 1620 14x + 360 = 1620 14x = 1260 x =

x ( ) = = 3 3 2x 2( ) + 40 = + 40 = 9 9

The angle measures are _____, _____, and _____.

Worksheet 6 (2.2)

5. John is paid 2½ times his normal hourly rate for each hour he works on a holiday. Last week he worked 45 hours, five hours were on a holiday. Find his normal hourly rate if he earned $420.

26 6. Maria took three college algebra exams and had an average score of 84. Her second exam was eight points better than her first and her third exam was eleven points better than her second exam. Find the three exam scores.

Worksheet 7 (2.3)

2.3 Equations Involving Decimals and Problem Solving

Summary 1:

27 Using 10n, a Power of 10, to Clear Decimals in an Equation Multiplying a decimal by 10n will result in moving the decimal point n places to the right.

Clearing the equation of all decimals is accomplished by applying the multiplication property of equality. Both sides of the equation are to be multiplied by an appropriate power of 10 to produce an equivalent equation. This power of 10 is equal to the most number of decimal places for any given decimal in the equation.

When an equation has more than one term on one or both sides of the equation, the distributive property is used to multiply each term by the power of 10.

Warm-up 1. Find an appropriate power of 10 to clear all decimals. Give the resulting equivalent equation:

a) 2y = .6y + .21 The power of 10 is ______.

_____(2y) = _____(.6y + .21)

_____(2y) = _____(.6y) + _____(.21)

The resulting equivalent equation is ______.

Problem - Find an appropriate power of 10 to clear all decimals. Give the resulting equivalent equation:

1. 0.05x + 0.08(10000 - x) = 620

Worksheet 7 (2.3)

Summary 2:

28 Solving Equations Involving Decimals 1. Find an appropriate power of 10 to clear all decimals using the multiplication property of equality. (See summary 1 above.) 2. Solve the resulting equivalent equation. 3. Check when directed to do so.

Warm-up 2. a) Solve and check: .07x + 160 = 152 + .08x _____(.07x + 160) = _____(152 + .08x) _____(.07x) + _____(160) = _____(152) + _____(.08x) _____ + 16000 = ______+ 8x 7x + 16000 + _____ = 15200 + 8x + ______+ 16000 = 15200 -1x + 16000 + ______= 15200 + ______-1x = ______(-1x) = _____(-800) x = _____

The solution set is ______.

Check: .07( ) + 160 = 152 + .08( ) _____ + 160 = 152 + ______= _____ True or False?

b) Solve:

.11(7000 - x) - .12x = 310 _____[.11(7000 - x) - .12x] = _____[310] _____[.11(7000 - x)] - _____[.12x] = _____[310] 11(7000 - x) - _____ = ______77000 - _____ - 12x = 31000 77000 - _____ = 31000 77000 - 23x + ______= 31000 + ______-23x = ______(-23x) = _____(-46000) x = ______

The solution set is ______.

Worksheet 7 (2.3)

Problems - Solve:

2. 0.222x = 0.2 - 0.22x , round to nearest hundredth.

29 3. 0.06x + 0.05(8000 - x) = 450

Suggestions for Solving Word Problems

1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Find a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer as a complete sentence.

Summary 3: Warm-up 3. Set up and write an algebraic equation, then solve:

a) Kayla bought an opal necklace at a 15% discount sale for $1150. Find the original price of the necklace, rounded to the nearest cent.

Declare the variable: Let x = original price ______= amount of discount

Write an algebraic equation and solve: x - ______= ______(x - .15x) = _____(1150) _____(x) - _____(.15x) = _____(1150) 100x - _____ = 115000 _____ = 115000 x = ______The original price was______.

Worksheet 7 (2.3)

b) A total of $12,000 was invested, part at 6% and the remainder at 8%. Find how much was invested at each rate, if the total yearly interest earned was $820.

30 Declare the variable:

Let x = amount invested at 6% ______= amount invested at 8% ______= interest earned at 6% ______= interest earned at 8%

.06x + .08(______) = ______[.06x + .08(12000 - x)] = _____(820) _____[.06x] + _____[.08(12000 - x)] = _____(820) 6x + _____(12000 - x) = 82000 6x + ______- 8x = 82000 _____ + 96000 = 82000 -2x = ______x = ______12000 - x = ______

There was ______invested at 6% and ______invested at 8%.

Problems - Set and write an algebraic equation, then solve:

4. Higinio bought a car with 6.5% sales tax included for $17,944. Find the price of the car, rounded to the nearest cent, before tax.

5. Xurry invests a certain amount of money at 7% interest and $1500 less than that amount at 4.5%. His yearly interest is $392.50, find how much he invested at each rate.

Worksheet 8 (2.4)

2.4 Formulas

Summary 1:

31 Literal equations are equations that contain more than one variable. Formulas are usually literal equations that state a rule in symbolic form.

1 Warm-up 1. a) Solve i = Prt for i, given that P = $800, r =8 2 % , and t = 3 years.

i = Prt i = (800)( )(3) i = ______

b) Solve 3(x - 2y) = 4 for y, given that x = -1.

3(x - 2y) = 4 3( - 2y) = 4 3( ) - 3( ) = 4 -3 - _____ = 4 -3 - 6y + _____ = 4 + _____ -6y = ______(-6y) = _____(7)

y = _____

Problems

1. Solve A = P + Prt for r, given that A = $1080, P = $800, and t = 7 years. Express r as a percent.

2. Solve 2x - 5y = -10 for x, given that y = 0.

Worksheet 8 (2.4)

Summary 2:

32 Changing the Form of a Literal Equation or Formula 1. If necessary, clear all fractions using the LCD and multiplication property of equality. 2. Apply distributive property to clear parentheses when appropriate. 3. Use the addition property of equality to collect terms containing the desired variable on one side and all other terms on the opposite side. 4. If two or more terms contain the desired variable on one side, use the distributive property to rewrite the expression. 5. Apply the multiplication property of equality to obtain a coefficient of 1 for the desired variable. 6. If needed, apply the symmetric property of equality so the desired variable is on the left side of the equation.

1 Warm-up 2. a) Solve A = 2 h( b1 + b2 ) for b2. 1 _____[A] = _____2 h( b1 + b2 )

_____ = h(b1 + b2) 2A = _____ + _____ 2A + _____ = hb1 + hb2 + _____ 2A - hb1 = ______(2A - hb1) = _____(hb2)

= b2 h

b2 = h

b) Solve -3x - y = 5 + 4y for y.

-3x - y + _____ = 5 + 4y + _____ -3x - _____ = 5 -3x - 5y + _____ = 5 + _____ -5y = _____ + ______(-5y) = _____(3x + 5)

y = or y = - 5 5 Worksheet 8 (2.4)

Problems - Solve: 3. A = 2lh + 2hw + 2lw , for w.

x - c y - c 4. = , for x. b a

33 Suggestions for Solving Word Problems Using Formulas

1. Carefully read the word problem. Reread to get overview of given situation. Determine known facts and identify unknown quantities. 2. Utilize a chart, figure, or diagram to clarify the problem. 3. Declare the variable and express unknown quantities in terms of the variable. 4. Use a formula as a guideline in the word problem. 5. Translate into an equation using the declared variable. 6. Solve the equation and use the solution to determine all facts asked for in the problem. 7. Check results in the original problem. Express answer as a complete sentence.

Summary 3:

a) The length of a rectangular lot is 16 feet. If the total distance around the lot is 56 feet, find the width of the lot.

Draw a figure:

Worksheet 8 (2.4)

Let x = width

P = 2l + 2w

_____ = 2( ) + 2( ) _____ = 32 + ______= 2x _____ = x

34 The width is ______.

b) After 3 hours, two cars traveling in opposite directions on I-24 are 360 miles apart. One car is traveling 10 mph faster than the other car. Find the rate of speed for both cars.

Let x = speed of slower car ______= speed of faster car

Chart: Diagram:

d=rt r t ------fast x + 10 3 3x 3( ) ------<------+------> slow x 3 distance traveled ------360 miles

3x + ______= _____ 3x + 3x + _____ = 360 _____ + 30 = 360 6x = _____ x = _____ ; x + 10 = _____

The speed of the slower car is ______and the speed of the faster car is ______.

Worksheet 8 (2.4)

5. How many milliliters of 15%-salt solution must be added to 200 ml of 25%-salt solution to obtain a 20%-salt solution?

35 6. In a given triangle, the largest angle is three times the smallest angle. The third angle is ten more than the smallest angle. Find the measures of all three angles.

Worksheet 9 (2.5)

2.5 Inequalities

Summary 1:

Statements1. a < b meansof inequality a is less express than b .one of the following: 2. a  b means a is less than or equal to b. 3. a > b means a is greater than b. 4. a  b means a is greater than or equal to b.

Algebraic inequalities contain one or more variables. They are open sentences which are neither true nor false.

Solving an inequality is the process used to find the number or numbers that make an algebraic inequality a true numerical statement. These numbers are solutions of the inequality.

36 Verify this by showing that the numerical statement is true or false. 2( ) + 5  2 - ( ) + 5  2 +  5 True or False? Answer: ______yes or no Problem

1. Is 2 a solution of 2x + 5  2 - x ? Verify this by showing that the numerical statement is true or false.

Worksheet 9 (2.5)

Summary 2:

37 Properties of Inequalities 1. Addition Property of Inequality

For all real numbers a, b, and c, a > b if and only if a + c > b + c.

Note: When adding the same number on both sides of the inequality, the sense of the inequality remains the same.

2. Multiplication Property of Inequality

a) For all real numbers a, b, and c with c > 0, a > b if and only if ac > bc.

Note: When multiplying the same positive number on both sides of the inequality, the sense of the inequality remains the same.

b) For all real numbers a, b, and c with c < 0, a > b if and only if ac < bc.

Note: When multiplying the same negative number on both sides of the inequality, the sense of the inequality changes; therefore, the inequality sign must be reversed.

In general, the process for solving inequalities closely parallels that for solving equations.

Warm-up 2. Solve:

a) 2x - 5 > 7 2x - 5 + _____ > 7 + _____ 2x ___ 12 _____(2x) > _____(12) x ___ 6

b) x + 2  5x - 14 x + 2 +  5x - 14 + x  5x - x +  5x - 16 + Worksheet 9 (2.5)  -16 (-4x)  (-16) x 4

Problems - Solve: 2. - 3x + 8  2

38 3. 7x + 8 > 2(x - 6)

Summary 3:

Appropriate Ways to Express the Solution of an Inequality

SET GRAPH INTERVAL NOTATION { x | x > a } < (===> ( a, infinity ) a { x | x  a } < [===> [ a, infinity ) a { x | x < b } <====) > ( - infinity, b ) b { x | x  b } <====] > ( - infinity, b ] b

Notes: 1. Set notation like{ x | x < - 4 }is translated as "the set of all x such that x is less than -4." 2. The symbol ( or ) is used to exclude an endpoint or with the infinity symbol, . 3. The symbol [ or ] is used to include an endpoint.

Worksheet 9 (2.5)

Warm-up 3. a) Sketch a graph of x < 3 and express in interval notation.

Graph:

<───┬──────┬──────┬──────┬──────┬──────┬──────┬─ > -1 0 1 2 3 4 5

Interval notation:

b) Express (-, 8] in set-builder notation.

39 Set: { x | }

Problems

4. Express the graph below both in set notation and interval notation:

< [======> -6 -5 -4

5. Sketch a graph for the solution expressed as (3 , ).

Summary 4:

Key Factors to Consider when Solving Inequalities

1. The solving process for inequalities closely parallels that of solving equations.

2. The inequality sign reverses when multiplying both sides of the inequality by a negative number.

3. Use an appropriate format to express the solution of an inequality - set, graph, or interval notation.

4. One solution can be checked to possibly catch a mistake. It is not possible to check all solutions of an inequality.

Worksheet 9 (2.5)

Warm-up 4. a) Solve and graph the solution set on a number line:

2(3 - x) < 14 6 - _____ < 14 6 - 2x + _____ < 14 + _____ -2x < ______(-2x) < _____(8) x ___ -4

Graph:

40 b) Solve and express the solution set using interval notation: 2(x + 4) - (x - 2)  - 3(x - 1) -3(x - 1) 2x + 8 - _____ + _____  -3x + _____ x + _____  -3x + 3 x + 10 + _____  -3x + 3 + ______+ 10  3 4x + 10 + _____  3 + _____ 4x  ______(4x)  _____(-7) x  _____

Interval notation:

Problems

6. Solve and graph the solution set on a number line: -3 + 6x > -15

7. Solve and express the solution set using interval notation: -3(2x + 1) - 2 -2(x + 5)

Worksheet 10 (2.6)

2.6 More on Inequalities and Problem Solving

Summary 1:

41 Compound Inequalities Compound statements are formed when mathematical statements are joined by the words and or or.

Conjunctions are compound statements that use and. The solution set for a given conjunction is the intersection of the solution sets for the inequalities joined by the word and.

Expressing the Solution of a Conjunction:

SET GRAPH INTERVAL NOTATION { x | a < x < b } < (===) > (a, b) a b { x | a  x < b } < [===) > [a, b) a b { x | a < x  b} < (===] > (a, b] a b { x | a  x  b} < [===] > [a, b] a b

Note: a < x < b is the compact form for a < x and x < b. The compact form can only be used for conjunctions where the solution is between two endpoints.

Disjunctions are compound statements that use or. The solution set for a given disjunction is the union of the solution sets for the inequalities joined by the word or.

Expressing the Solution of a Disjunction:

SET GRAPH INTERVAL NOTATION { x | x < a or x > b } <===) (===> ( - infinity, a) union ( b, infinity ) a b Note: The endpoints a and b will be included in the solution for , . In this case, use [ or ].

Worksheet 10 (2.6)

Warm-up 1. Graph the compound inequality and express in interval notation:

a) x > -2 and x < 3 Give the compact form of this statement: ______

Graph:

42 <─┬──────┬──────┬──────┬──────┬──────┬──────┬─ ─────┬─>

Interval notation:

b) x  0 or x > 4

Graph:

<─┬──────┬──────┬──────┬──────┬──────┬──────┬─ ─────┬─>

Interval notation:

c) x > -2 or x > 3

Graph:

<─┬──────┬──────┬──────┬──────┬──────┬──────┬──────┬─>

Interval notation:

Problems - Graph the compound inequality and express in interval notation:

1. x > 1 and x < 3 (What is the compact form of this statement?)

2. x < -5 or x > 1

3. x  -3 or x < 2 Worksheet 10 (2.6)

Summary 2:

43 Key Factors to Consider when Solving Compound Statements 1. For disjunctions, solve each inequality separately in the compound sentence. The solution set is the union of these solutions.

2. For conjunctions, solve each inequality separately in the compound sentence. The solution set is the intersection of these solutions.

Note: If the conjunction is in compact form, it can be solved by isolating the variable in the middle. The properties of inequality will be applied simultaneously on the middle, left, and right of the compact form.

Warm-up 2. Solve, then express the solution set as a graph and in interval notation:

a) x - 5 < -3 or x - 3 > 3

x - 5 + _____ < -3 + _____ or x - 3 + _____ > 3 + _____ x < _____ x > _____

The solution set is {x______or______}.

Graph:

<─┬──────┬──────┬──────┬──────┬──────┬──────┬─ >

Interval notation: ______union ______

b)- 8  3x - 2  7 - 8 +  3x - 2 +  7 +  3x  (-6 )  (3 x)  (9 )

Worksheet 10 (2.6) - 2   3

The solution set is {x______}.

Graph:

<─┬──────┬──────┬──────┬──────┬──────┬──────┬─ ─────┬─>

44 Interval notation:

Problems - Solve, then express the solution set as a graph and in interval notation:

4. x - 3 > -2 or x + 1 < 3

5. 0  4x + 2 < 8

Summary 3:

Solving Word Problems Involving Inequalities

1. The strategies previously outlined for equations hold true for inequalities as well.

2. Analyze the situation to determine which inequality symbol is appropriate.

Worksheet 10 (2.6)

Warm-up 3. Set up and write an algebraic inequality, then solve:

a) In general chemistry, Kate scored 95, 83, 89, and 85 on four exams this semester. If she is to earn an A, 90 or higher, for the semester, what must she score on the fifth and final test for the semester?

45 Let x = score on the fifth test

Write an algebraic inequality and solve:

95 + 83 + 89 + 85 + ( )  90 ( )  95 + 83 + 89 + 85 + x  _____   _____(90)  5  95 + 83 + 89 + 85 + x ______+ x 450 352 + x + _____ 450 + _____ x _____

She must score at least ______.

Problem - Set up and write an algebraic inequality, then solve:

6. Darla has $2000 to invest. If she invests $1000 at 9% interest, at what rate must she invest the remaining amount so that the two investments earn more than $130 of combined yearly interest?

Worksheet 11 (2.7)

2.7 Equations and Inequalities Involving Absolute Value

Summary 1:

46 Solving Absolute Value Statements when k > 0 Absolute value equations and inequalities must be converted to their corresponding equivalent compound statements when k > 0. The compound statement can then be solved and the solution set can be graphed or expressed in interval notation.

1. | ax + b | = k is equivalent to ax + b = - k or ax + b = k

2. | ax + b | < k is equivalent to ax + b > - k and ax + b < k or- k < ax + b < k 3. | ax + b | > k is equivalent to ax + b < - k or ax + b > k

Warm-up 1. a) Solve and give the solution set in set notation: | 2 - m | = 4 2 - m = or 2 - m = 2 - m + = - 4 + 2 - m + = 4 + - m = - m = (-m) = (-6) (-m) = (2) m = 6 or m =

The solution set is { , }

b) Solve and graph the solution set: | 2x + 1 |  11  2x + 1  -11 +  2x + 1 +  11 + Worksheet 11 (2.7)

-12  2x  (-12)  (2x)  (10)  x 

Graph:

Problem - Solve and write solution in interval notation: 1. | 6 - 7x | > 22

47 Solving Absolute Value Statements when k  0

Absolute value equations and inequalities are solved by inspection when k < 0.

1. For| ax + b | = k , the solution set is .

2. For| ax + b | < k , the solution set is .

3. For| ax + b | > k , the solution set is (-, ).

Absolute value equations have exactly one solution when k = 0.

1.| ax + b | = 0 is equivalent to ax + b = 0.

Summary 2:

Worksheet 11 (2.7)

Warm-up 2. Solve:

a)| 2x + 1 | < - 5

The solution set is _____.

b)| 2x + 1 | > - 5

c)| 2x + 1 | = 0 2x + 1 = 0 x = _____

Problems - Solve: 2.| y - 5 | > - 2

48 3.| 3m + 4 | < -10