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10Algebra 1Chapter 1 Resource Book
Evaluate algebraic expressions and use exponents.
VocabularyA variable is a letter used to represent one or more numbers.
An algebraic expression, or variable expression, consists of numbers, variables, and operations.
To evaluate an expression, substitute a number for the variable, perform the operation(s), and simplify the result if necessary.
A power is an expression that represents repeated multiplication of the same factor.
A power can be written in a form using two numbers, a base and an exponent. The exponent represents the number of times the base is used as a factor.
GOAL
Study GuideFor use with pages 227
LESSON
1.1
Evaluate algebraic expressions
Evaluate the expression when x 5 5.
a. 7x
b. 12 1 x
Solution
a. 7x 5 7(5) Substitute 5 for x.
5 35 Multiply.
b. 12 1 x 5 12 1 5 Substitute 5 for x.
5 17 Add.
Exercises for Example 1
Evaluate the expression for the given value of the variable.
1. 15 2 a when a 5 3 2. 3b when b 5 7
3. 11 1 c when c 5 10 4. 28}d
when d 5 4
5. 1}2
n when n 5 18 6. 0.4f when f 5 8
EXAMPLE 1
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11Algebra 1
Chapter 1 Resource Book
Evaluate an expression The cost of filling a car’s gas tank can be represented by the expression xy where x is the price per gallon of gasoline and y is the number of gallons purchased. You purchase 10 gallons of gasoline when the price per gallon is $2.35. Find the total cost.
Solution
Total Cost 5 xy Write expression.
5 2.35(10) Substitute 2.35 for x and 10 for y.
5 23.50 Multiply.
The total cost is $23.50.
Exercises for Example 2 7. You purchase 5 gallons of gasoline when the price of gasoline is $2.26 per
gallon. Find the total cost.
8. You purchase 8 gallons of gasoline when the price of gasoline is $2.20 per gallon. Find the total cost.
EXAMPLE 2
Read and write powers
Write the power in words and as a product.
a. 83
b. m6
Solution
a. eight to the third power, or eight cubed; 8 p 8 p 8b. m to the sixth power; m p m p m p m p m p m
Exercises for Example 3
Write the power in words and as a product.
9. 48 10. 11}3 24 11. x2
EXAMPLE 3
Study Guide continuedFor use with pages 227
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22Algebra 1Chapter 1 Resource Book
Use the order of operations to evaluate expressions.
VocabularyThe order of operations was established to evaluate an expression involving more than one operation.
Order of Operations
STEP 1 Evaluate expressions inside grouping symbols.
STEP 2 Evaluate powers.
STEP 3 Multiply and divide from left to right.
STEP 4 Add and subtract from left to right.
GOAL
Study GuideFor use with pages 8213
LESSON
1.2
Evaluate expressions
Evaluate the expression 42 p 5 2 62.
Solution
42 p 5 2 62 5 16 p 5 2 36 Evaluate powers.
5 80 2 36 Multiply.
5 44 Subtract.
Exercises for Example 1
Evaluate the expression.
1. 20 2 32 1 7 2. 5 p 23 4 6 3. 4 p 6 2 21 4 3
EXAMPLE 1
Evaluate expressions with grouping symbols
Evaluate the expression.
a. 47 2 2(9 1 12) b. 6[23 1 (13 2 8)]
Solution
a. 47 2 2(9 1 12) 5 47 2 2(21) Add within parentheses.
5 47 2 42 Multiply.
5 5 Subtract.
b. 6[23 1 (13 2 8)] 5 6[8 1 (13 2 8)] Evaluate power.
5 6[8 1 5] Subtract within the parentheses.
5 6[13] Add within the parentheses.
5 78 Multiply.
EXAMPLE 2
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23Algebra 1
Chapter 1 Resource Book
Evaluate an algebraic expression
Evaluate the expression4y 1 8}2 1 y
when y 5 3.
Solution4y 1 8}2 1 y
5 4(3) 1 8}
2 1 3Substitute 3 for y.
5 12 1 8}2 1 3
Multiply.
5 20}5 Add.
5 4 Divide.
Exercises for Example 3
Evaluate the expression when w 5 9.
10. 17 2 3w 11. w2 2 13
12. 5w
}w 1 6
13. 7(13 2 w)
14. 2w2 2 15 15. 5w 2 1}3
w
EXAMPLE 3
Exercises for Example 2
Evaluate the expression.
4. 3(14 2 5) 5. 6(9 2 14)
6. (7 1 5) 2 (8 1 4) 7. (33 2 6) 4 3
8. 42(2 1 8) 9. 9[15 4 (2 1 3)]
Study Guide continuedFor use with pages 8213
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19Algebra 1
Chapter 2 Resource Book
Add positive and negative numbers.
VocabularyThe number 0 is the additive identity.
The opposite of a is its additive inverse.
Rules of Addition
Same signs To add two numbers with the same sign, add their absolute values. The sum has the same sign as the numbers added.
Different signs To add two numbers with different signs, subtract the lesser absolute value from the greater absolute value. The sum has the same sign as the number with the greater absolute value.
Properties of Addition
Commutative Property The order in which you add two numbers does not change the sum.
Associative Property The way you group three numbers in a sum does not change the sum.
Identity Property The sum of a number and 0 is the number.
Inverse Property The sum of a number and its opposite is 0.
GOAL
Add real numbersa. 17 1 (224) 5 ⏐224⏐ 2 ⏐17⏐ Rule of different signs
5 24 2 17 Take absolute values.
5 27 Subtract. Use the sign of the number with the greater absolute value.
b. 21.3 1 (25.8) 5 ⏐21.3⏐ 1 ⏐25.8⏐ Rule of same sign
5 1.3 1 5.8 Take absolute values.
5 27.1 Add and use the sign of the numbers added.
Exercises for Example 1
Find the sum.
1. 23.6 1 7.1 2. 5.3 1 (29) 3. 20.2 1 (20.6)
4. 231}2 1 122
1}5 2 5. 11 1 (215) 1 8 6. 25 1 (28) 1 6
EXAMPLE 1
Study GuideFor use with pages 73 –79
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20Algebra 1Chapter 2 Resource Book
Identify the properties of addition
Identify the property being illustrated.
Statement Property Illustrated
a. 215 1 0 5 215 Identity property of addition
b. 12 1 (217) 5 217 1 12 Commutative property of addition
Exercises for Example 2
Identify the property being illustrated.
7. 9 1 (2x) 5 2x 1 9
8. 5.1 1 (25.1) 5 0
9. [3 1 (22)] 1 1 5 3 1[(22) 1 1]
EXAMPLE 2
Study Guide continuedFor use with pages 73 –79
LESSON
2.2
Solve a multi-step problem Gas Prices The table shows the changes in gas prices for two companies. Which company had the greater total change in gas prices for the three weeks?
Week Price change for company A Price change for company B
1 $.05 2$.06
2 2$.08 $.13
3 $.11 2$.04
Solution
STEP 1 Calculate the total change in gas prices for each company.
Company A: Company B:Total change 5 0.05 1 (20.08) 1 0.11 Total change 5 20.06 1 0.13 1 (20.04)
5 20.08 1 (0.05 1 0.11) 5 0.13 1 [2(0.06 1 0.04)]
5 20.08 1 0.16 5 0.13 1 (20.1)
5 0.08 5 0.03
STEP 2 Compare the total change in gas prices: 0.08 > 0.03.
Company A had the greater total change in gas prices.
Exercise for Example 310. In Example 3, suppose that the changes in gas prices for week 4 are 2$.06 for
company A and $.07 for company B and the changes in gas prices for week 5 are $.04 for company A and 2$.03 for company B. Which company has the greater total change in gas prices for the five weeks?
EXAMPLE 3
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42Algebra 1Chapter 2 Resource Book
Multiply real numbers.
VocabularyThe number 1 is called the multiplicative identity.
The Sign of a Product
p The product of two real numbers with the same sign is positive.
p The product of two real numbers with different signs is negative.
Properties of Multiplication
Commutative Property The order in which two numbers are multiplied does not change the product.
Associative Property The way you group three numbers when multiplying does not change the product.
Identity Property The product of a number and 1 is that number.
Property of Zero The product of a number and 0 is 0.
Property of 21 The product of a number and 21 is the opposite of the number.
GOAL
Multiply real numbers
Find the product.
a. (27)(23) b. 4(22)(26)
Solution
a. (27)(23) 5 21 Same signs; product is positive.
b. 4(22)(26) 5 (28)(26) Multiply 4 and 22; product is negative.
5 48 Same signs; product is positive.
Exercises for Example 1
Find the product.
1. 25(4) 2. 9(28)
3. 212(26) 4. (24)121}2 2
5. 4(28)(5) 6. 28(22)(28)
EXAMPLE 1
Study GuideFor use with pages 87–93
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43Algebra 1
Chapter 2 Resource Book
Identify the properties of multiplication
Identify the property being illustrated.
Solution
Statement Property Illustrated
a. x p 5 5 5 p x Commutative property of multiplication
b. 26 p (21) 5 6 Multiplicative property of 21
c. (23 p x) p 2 5 23 p (x p 2) Associative property of multiplication
Exercises for Example 2
Identify the property being illustrated.
7. 25 p 1 5 25
8. [5 p (22)] p (23) 5 5 p [(22) p (23)]
9. 0 p 11 5 0
10. 23 p (212) 5 212 p (23)
EXAMPLE 2
Use properties of multiplication
Find the product (23x) p (22). Justify your steps.
Solution
(23x) p (22) 5 (22) p (23x) Commutative property of multiplication
5 [22 p (23)]x Associative Property of multiplication
5 6x Product of 22 and 23 is 6.
Exercises for Example 3
Find the product. Justify your steps.
11. 29(2x)
12. w(23)(12)
13. (28)(5)(2z)
EXAMPLE 3
Study Guide continuedFor use with pages 87–93
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43Algebra 1
Chapter 1 Resource Book
Translate verbal sentences into equations or inequalities.
VocabularyAn open sentence is a mathematical statement that contains two expressions and a symbol that compares them.
An equation is an open sentence that contains the symbol 5.
An inequality is an open sentence that contains one of the symbols <, ≤, >, or ≥.
When you substitute a number for the variable in an open sentence, the resulting statement is either true or false. If the statement is true, the number is a solution of the equation, or a solution of the inequality.
GOAL
Write equations and inequalities
Write an equation or an inequality.
a. 8 times the quantity of 11 plus a number x is 112.
b. The product of 7 and a number y is no more than 31.
c. A number z is more than 8 and at most 15.
Solution
Verbal phrase Equation or inequality
a. 8 times the quantity of 11 plus a number x is 112. 8(11 1 x) 5 112
b. The product of 7 and a number y is no more than 31. 7y ≤ 31
c. A number z is more than 8 and at most 15. 8 < z ≤ 15
Exercises for Example 1
Write an equation or an inequality.
1. The difference of 73 and a number x is 17.
2. The product of 8 and the quantity of a number y plus 6 is less than 21.
3. The quotient of a number w and 5 is at most 4.
4. The sum of a number z and 2 is greater than 15 and less than 23.
EXAMPLE 1
Study GuideFor use with pages 21226
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44Algebra 1Chapter 1 Resource Book
Check possible solutions
Check whether 5 is a solution of the equation or inequality.
Equation/inequality Substitute Conclusion
a. 3x 2 7 5 12 3(5) 2 7 0 12 8 Þ 12 ✗
5 is not a solution.
b. 9 1 2x ≤ 23 9 1 2(5) ≤? 23 19 ≤ 23 ✓
5 is a solution.
Exercises for Example 2
Check whether the given number is a solution of the equation or inequality.
5. 13 1 a 5 17; 4 6. 7b 2 3 5 10; 2 7. 4c < 15; 3
8. 21 2 3d ≥ 11; 2 9. 4g 1 6 ≤ 14; 3 10. 7 < m 1 8 < 15; 6
EXAMPLE 2
Solve a multi-step problem A soccer team is selling pizzas for $6 each. Each pizza costs $4 to make. The team has 10 players and wants to raise $900 for equipment and uniforms. How many pizzas does the team need to sell? How many pizzas will each player sell if every player sells the same number of pizzas?
Solution
STEP 1 Write a verbal model. Let p be the number of pizzas sold. Write an equation.
1Price ofpizza 2 Cost to make
each pizza 2 3 1Number ofpizzas sold 2 5 Profi t
(6 2 4) 3 p 5 900
STEP 2 Use mental math to solve the equation (6 2 4)p 5 900, or 2p 5 900. Think: 2 times what number is 900? Because 2(450) 5 900, the solution is 450.
The team needs to sell 450 pizzas.
STEP 3 Find the number of pizzas each player sells: 450 pizzas}10 players
5 45 pizzas per player
Each player will sell 45 pizzas.
Exercise for Example 3 11. Your family is driving 188 miles to visit a relative. Your father drives 63 miles
then stops for a break. How many more miles are left in the trip? Your father drives 50 miles per hour. How long will the remainder of the trip take? Write a verbal model for the situation, then solve.
EXAMPLE 3
Study Guide continuedFor use with pages 21226
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32Algebra 1Chapter 3 Resource Book
Solve multi-step equations.GOAL
Solve an equation by combining like terms
Solve 17x 2 11x 1 8 5 20.
Solution
17x 2 11x 1 8 5 20 Write original equation.
6x 1 8 5 20 Combine like terms.
6x 1 8 2 8 5 20 2 8 Subtract 8 from each side.
6x 5 12 Simplify.
6x}6 5
12}6 Divide each side by 6.
x 5 2 Simplify.
Exercises for Example 1
Solve the equation. Check your solution.
1. 9x 2 13x 1 7 5 31
2. 13 2 5x 1 8x 5 22
3. 15x 2 9 2 8x 5 12
4. 18 2 2x 2 4x 5 224
EXAMPLE 1
Study GuideFor use with pages 1482153
LESSON
3.3
Solve an equation using the distributive property
Solve 4x 1 3(2x 2 1) 5 17.
Solution
METHOD 1 Show All Steps METHOD 2 Do Some Steps Mentally
4x 1 3(2x 2 1) 5 17 4x 1 3(2x 2 1) 5 17
4x 1 6x 2 3 5 17 4x 1 6x 2 3 5 17
10x 2 3 5 17 10x 2 3 5 17
10x 2 3 1 3 5 17 1 3 10x 5 20
10x 5 20 x 5 2
10x}10
5 20}10
x 5 2
EXAMPLE 2
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33Algebra 1
Chapter 3 Resource Book
Exercises for Example 2
Solve the equation. Check your solution.
5. 3(x 2 4) 1 4x 5 16
6. 9x 2 6(3x 2 3) 5 9
7. 22x 1 7(3x 21) 5 31
8. 5(2x 1 8) 2 6x 5 16
Multiply by a reciprocal to solve an equation
Solve 3}4
(5x 2 4) 5 12.
Solution
3}4 (5x 2 4) 5 12 Write original equation.
4}3 p 3}
4 (5x 2 4) 5
4}3 p 12 Multiply each side by
4}3, the reciprocal of
3}4 .
5x 2 4 5 16 Simplify.
5x 5 20 Subtract 4 from each side.
x 5 4 Simplify.
Exercises for Example 3
Solve the equation. Check your solution.
9. 1}2 (x 2 11) 5 9
10. 23}2 (2y 1 6) 5 15
11. 215 5 5}7 (4z 2 1)
12. 36 5 23}4 (5m 1 12)
EXAMPLE 3
Study Guide continuedFor use with pages 1482153
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44Algebra 1Chapter 3 Resource Book
Solve equations with variables on both sides.
VocabularyAn equation that is true for all values of the variable is an identity.
GOAL
Solve an equation with variables on both sides
Solve 13 2 6x 5 3x 2 14.
Solution
13 2 6x 5 3x 2 14 Write original equation.
13 2 6x 1 6x 5 3x 2 14 1 6x Add 6x to each side.
13 5 9x 2 14 Simplify.
27 5 9x Add 14 to each side.
3 5 x Divide each side by 9.
The solution is 3. Check by substituting 3 for x in the original equation.
CHECK 13 2 6x 5 3x 2 14 Write original equation.
13 2 6(3) 5 3(3) 2 14 Substitute 3 for x.
25 5 3(3) 2 14 Simplify left side.
25 5 25 ✓ Simplify right side. Solution checks.
Exercises for Example 1
Solve the equation. Check your solution.
1. 9a 5 7a 2 8 2. 17 2 8b 5 3b 2 5 3. 25c 1 6 5 9 2 4c
EXAMPLE 1
Study GuideFor use with pages 154 –160
LESSON
3.4
Solve an equation with grouping symbols
Solve 4x 2 7 5 1}3
(9x 2 15).
Solution
4x 2 7 5 1}3 (9x 2 15) Write original equation.
4x 2 7 5 3x 2 5 Distributive property
x 2 7 5 25 Subtract 3x from each side.
x 5 2 Add 7 to each side.
The solution is 2.
EXAMPLE 2
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45Algebra 1
Chapter 3 Resource Book
Identify the number of solutions of an equation
Solve the equation, if possible.
a. 4(3x 2 2) 5 2(6x 1 1) b. 4(4x 2 5) 5 2(8x 2 10)
Solution
a. 4(3x 2 2) 5 2(6x 1 1) Write original equation.
12x 2 8 5 12x 1 2 Distributive property
12x 5 12x 1 10 Add 8 to each side.
The equation 12x 5 12x 1 10 is not true because the number 12x cannot be equal to 10 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.
12x 2 12x 5 12x 1 10 212x Subtract 12x from each side.
0 5 10 Simplify.
The statement 0 5 10 is not true, so the equation has no solution.
b. 4(4x 2 5) 5 2(8x 2 10) Write original equation.
16x 2 20 5 16x 2 20 Distributive property
Notice that the statement 16x 2 20 5 16x 2 20 is true for all values of x. So, the equation is an identity.
Exercises for Example 3
Solve the equation, if possible.
7. 11x 1 7 5 10x 2 8
8. 5(3x 2 2) 5 3(5x 2 1)
9. 1}2 (6x 1 18) 5 3(x 1 3)
EXAMPLE 3
Exercises for Example 2
Solve the equation. Check your solution.
4. 2m 2 7 5 3(m 1 8)
5. 1}5 (15n 1 5) 5 8n 2 9
6. 7p 2 3 5 3}4 (8p 2 12)
Study Guide continuedFor use with pages 154 –160
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38Algebra 1Chapter 6 Resource Book
Solve multi-step inequalities.GOAL
Solve a two-step inequality
Solve 24x 1 3 > 15. Graph your solution.
Solution
24x 1 3 > 15 Write original inequality.
24x > 12 Subtract 3 from each side.
x < 23 Divide each side by 24.Reverse inequality symbol.
The solutions are all real numbers less than 23.0 1 2212223242526Check by substituting a number less than 23 in
the original inequality.
CHECK 24x 1 3 > 15 Write original inequality.
24(25) 1 3 >? 15 Substitute 25 for x.
23 > 15 ✓ Solution checks.
Exercises for Example 1
Solve the inequality. Graph your solution.
1. 7x 1 8 > 22 2. 27 ≥ 22x 1 9 3. 2.3x 2 6.9 < 7.13
EXAMPLE 1
Study GuideFor use with pages 369–374
LESSON
6.3
Solve a multi-step inequality
Solve the inequality.
a. 21}3 (x 1 12) < 5 b. 9x 1 2 < 5x 2 18
Solution
a. 21}3 (x 1 12) < 5 Write original inequality.
2x}3 2 4 < 5 Distributive property
2x}3 < 9 Add 4 to each side.
x > 227 Multiply each side by 23. Reverse the inequality symbol.
The solutions are all real numbers greater than 227.
b. 9x 1 2 < 5x 2 18 Write original inequality.
9x < 5x 2 20 Subtract 2 from each side.
4x < 220 Subtract 5x from each side.
x < 25 Divide each side by 4.
The solutions are all real numbers less than 25.
EXAMPLE 2
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39Algebra 1
Chapter 6 Resource Book
Identify the number of solutions of an inequality
Solve the inequality, if possible.
a. 5(3x 2 2) < 15x 1 7 b. 9 2 28x > 4(5 2 7x)
Solution
a. 5(3x 2 2) < 15x 1 7 Write original inequality.
15x 2 10 < 15x 1 7 Distributive property
210 < 7 Subtract 15x from each side.
All real numbers are solutions because 210 < 7 is true.
b. 9 2 28x > 4(5 2 7x) Write original inequality.
9 2 28x > 20 2 28x Distributive property
9 > 20 Add 28x to each side.
There are no solutions because 9 > 20 is false.
Exercises for Example 3
Solve the inequality, if possible.
7. 2m 2 7m 2 4 > 1 2 5m
8. 3n 2 13 < 3(n 2 2)
9. 11p 2 3p 1 6 ≥ 4(2p 2 1)
EXAMPLE 3
Exercises for Example 2
Solve the inequality.
4. 3(2x 2 7) > 15
5. 10 2 3x ≤ 5x 2 14
6. 1}2 (8x 1 6) <
1}3 (9x 2 15)
Study Guide continuedFor use with pages 369–374
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52Algebra 1Chapter 6 Resource Book
Solve and graph compound inequalities.
VocabularyA compound inequality consists of two separate inequalities joined by and or or.
GOAL
Write and graph a compound inequality
Translate the verbal phrases into an inequality. Then graph the inequality.
a. All real numbers that are less than or equal to 7 or greater than or equal to 10.
Inequality: x ≤ 7 or x ≥ 10
3 4 5 7 8 9 10 116
b. All real numbers that are greater than 21 and less than or equal to 1.
Inequality: 21 < x < 1
0 1 2 3 421222324
Exercises for Example 1
Translate the verbal phrases into an inequality. Then graph the inequality.
1. All real numbers that are less than 23 or greater than 0.
2. All real numbers that are less than 9 and greater than or equal to 7.
3. All real numbers that are greater than or equal to 14 or less than or equal to 10.
EXAMPLE 1
Study GuideFor use with pages 379–388
LESSON
6.4
Solve a compound inequality with and
Solve 7 ≤ x 2 4 ≤ 12. Graph your solution.
Solution
7 ≤ x 2 4 ≤ 12 Write original inequality.
7 1 4 ≤ x 2 4 1 4 ≤ 12 1 4 Add 4 to each expression.
11 ≤ x ≤ 16 Simplify.
The solutions are all real numbers greater than or equal to 11 and less than or equal to 16.
9 10 11 13 14 15 16 1712
EXAMPLE 2
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53Algebra 1
Chapter 6 Resource Book
Solve a compound inequality with or
Solve 3x 1 4 < 16 or 5x 2 12 > 13. Graph your solution.
Solution
Solve the two inequalities separately.
3x 1 4 < 16 or 5x 2 12 > 13 Write original inequality.
3x 1 4 2 4 < 16 2 4 or 5x 2 12 1 12 > 13 1 12 Use addition or subtraction property of inequality.
3x < 12 or 5x > 25 Simplify.
3x}3 <
12}3 or
5x}5 >
25}5 Use division property of
inequality.
x < 4 or x > 5 Simplify.
The solutions are all real numbers less than 4 or greater than 5.
0 1 2 3 4 5 7621
Exercises for Examples 2 and 3
Solve the inequality. Graph your solution.
4. 9 < 2x 1 3 < 15
5. 30 ≥ 27x 2 12 > 16
6. 28 ≤ 4(2x 2 3) ≤ 68
7. 3x 2 7< 11 or 2x 1 4 < 15
8. 1}2 (x 1 18) > 6 or 7x 1 5 < 251
9. 3x 1 8 > 7x 2 12 or 9(x 2 2) > 8x 2 9
EXAMPLE 3
Study Guide continuedFor use with pages 379–388
LESSON
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85Algebra 1
Chapter 3 Resource Book
Write equations in function form and rewrite formulas.
VocabularyAn equation in x and y is written in function form when the dependent variable y is isolated on one side of the equation.
A literal equation is an equation that contains two or more variables.
GOAL
Rewrite an equation in function form
Write 9x 2 4y 5 8 in function form.
Solution
To write an equation in function form, solve the equation for y.
9x 2 4y 5 8 Write original equation.
24y 5 8 2 9x Subtract 9x from each side.
y 5 22 1 9}4x Divide each side by 24.
The equation y 5 22 1 9}4 x is written in function form.
EXAMPLE 1
Study GuideFor use with pages 184 –189
LESSON
3.8
Solve a literal equation
The formula for the volume of a rectangular prism is V 5 lwh. Solve the formula for l.
Solution
V 5 lwh Write original equation.
V}wh
5 lwh}wh
Assume w Þ 0 and h Þ 0. Divide each side by wh.
V}wh
5 l Simplify.
The rewritten equation is V}wh
5 l.
Exercises for Examples 1 and 2
Write the equation in function form.
1. 7x 1 y 5 12 2. 3y 2 9x 5 21 3. 5y 2 2x 5 15
Solve the literal equation.
4. I 5 Prt for P 5. A 5 1}2 (b1 1 b2)h for b2
EXAMPLE 2
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86Algebra 1Chapter 3 Resource Book
Solve and use a geometric formula
The area A of a triangle is given by the formula A 5 1}2
bh where b is the base and h is the height.
12 in.
b
a. Solve the formula for the base b.
b. Use the rewritten formula to find the base of the triangle shown, which has an area of 106.8 square inches.
Solution
a. Solve the formula for b.
A 5 1}2bh Write original formula.
2A 5 bh Multiply each side by 2.
2A}h
5 b Divide each side by h.
b. Substitute 106.8 for A and 12 for h in the rewritten formula.
b 5 2A}h
Write rewritten formula.
b 5 2(106.8)}
12Substitute 106.8 for A and 12 for h.
b 5 17.8 Simplify.
The base of the triangle is 17.8 inches.
Exercises for Example 3
The surface area S of a sphere is given by the formula S 5 4πr2 where ris the radius of the sphere.
6. Solve the formula for r.
7. Use the rewritten formula from Exercise 6 to find r when S 5 314 square meters. Use 3.14 for π.
EXAMPLE 3
Study Guide continuedFor use with pages 184 –189
LESSON
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