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Review of Elementary Algebra Content

Review of Elementary Algebra Content

Review of Elementary Content

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1 Table of Contents

Fractions ...... 1 ...... 5 Order of Operations ...... 9 Exponents ...... 11 ...... 18 Factoring ...... 22 Solving Linear ...... 31 Solving Linear Inequalities ...... 34 Graphing Linear Equations ...... 38 Finding the of a Line ...... 43 Application of Linear Equations ...... 48 Functions ...... 50 Systems of Linear Equations ...... 53 Application of a System of Linear Equations ...... 62 Solving Linear Inequalities in Two Variables ...... 64 Solving Systems of Linear Inequalities ...... 67 Solving Equations ...... 72 Solving Absolute Value Inequalities ...... 76 Vocabulary Used in Application Problems...... 79 Solving Application Problems ...... 81 Comprehensive Review of ...... 88

Revised May 2013 2i

3 FRACTIONS

Lowest Terms

Example: 1 Simplify: 12 4⋅ 3 = 15 5⋅ 3 1 4 = 5

Addition/

Rule: To add and subtract fractions you need a Common Denominator.

Example: 2 3 23+ += 77 7 For this example, we already have the common denominator. 5 = 7

1 2 13 24 Example: + =⋅+⋅ 4 3 43 34 This example, we need to find a common denominator, or a denominator which is 38 = + DIVISIBLE by 3 and 4. One such denominator 12 12 is 12. 11 = 12

Example: 71713 −= −⋅ 12 4 12 4 3 73 = − 12 12

4 = 12 1 = 3

Note: Any over itself is equal to 1. When you multiply a number by 1, you are not changing the number, just renaming it.

43 9 =1 = 1 = 1 43 9

1 FRACTIONS

Multiplication/

Rule: and division of fractions do not require a common denominator.

3 Example: 7 12 7 12 ⋅=⋅ Cancel a common factor of 4 85 85 2

73⋅ = Multiply across 25⋅ 21 = 10

2 5 26 Example: ÷=⋅ 3 6 35 2 5 26 To divide by , multiply by the = ⋅ 6 35 6 1 reciprocal which is 22⋅ 5 = 15⋅ 4 = 5

2 FRACTIONS

Problems Answers

Perform the following operations.

75 1 1. + 12 12

21 13 2. + 35 15

31 5 3. − 43 12

55 5 4. + 6 12 4

51 23 5. − 76 42

71 1 6. − 12 2 12

3 11 7. + 2 4 4

4 11 8. 5 − 3 3

16 13 41 9. − 35 15

16 12 76 10. − 37 21

3 FRACTIONS

Problems (continued) Answers

Perform the following operations.

25 49 11. + 6 4 4

13 3 12. ⋅ 24 8

32 6 13. ⋅ 75 35

2 4 14. 2⋅ 15 15

32 9 15. ÷ 53 10

3 3 16. ÷ 2 8 16

53 35 17. ÷ 6 14 9

95 27 18. ÷ 43 20

4 INTEGERS

Definition: The absolute value of a number is the distance between 0 and the number on the . The symbol for “The absolute value of a” is a . The absolute value of a number can never be negative.

Examples: 1) −=44 2) 33=

Addition of Signed :

• Like signs: Add their absolute values and use the common .

Examples: 1) −+−3( 2) =−( 32 + ) = −5

2) 48+=( 48 +)

=12

• Unlike signs: Subtract their absolute values and use the sign of the number with the larger absolute value.

Examples: 1) −+34 =+( 43 −)

=1

2) 2+−( 6) =−( 62 − ) = − 4

5 INTEGERS

Subtraction of Signed Numbers

1) Change the subtraction symbol to addition. 2) Change the sign of the number being subtracted. 3) Add the numbers using like signs or unlike signs rules for addition.

Examples: 1) −−34 =−+− 3( 4)

= −7

2) 6−−( 262) = +( ) = 8

3) −−−2( 4) =−+ 24

= 2 Multiplication and Division of Two Signed Numbers:

• Like signs: The product or quotient of two numbers with like signs is positive.

Examples: 1) (−3)( −= 4) 12

8 2) = 2 4

−16 3) = 4 −4 = 2 • Unlike signs: The product or quotient of two numbers with unlike signs is negative.

Examples: 1) (−=−3)( 4) 12

−15 2) = −5 3 • Division by 0 is . 3 Example: is undefined. 0

6 INTEGERS

Addition and Subtraction of Signed Numbers

Problems Answers

Perform the following operations.

1. 53+−( ) 2

2. −+−62( ) –8

3. 7− 12 −5

4. −−11 4 −15

5. 6+ 2 +−( 13) −5 

6. 85−−( ) 13

7. −−3( 4 − 11) 4

8. 51 4 −− − 62 3 9. (−9) +−( 14) + 12 −11 

10. −−4.4 8.6 −13

11. 2+−−( 48) −10

12. 93 3 +− 10 5 10

13. −−8( −− 41) −( 92 −) 4 

14. −+−83( )  +−+− 76( )  –24   

15. −+4( − 12 + 1) −−−( 1 9) −5 

7 INTEGERS

Multiplication and Division of Signed Numbers

Problems Answers

Perform the following operations.

1. (34)(− ) –12

24 2. − 4 −6

3. (−−10)( 12) 120

0 4. 0 −2

3  10  5 5. −−   89   12

6. (−9.8) ÷−( 7) 1.4

−30 5 7. 28−

8. (−5.1)( .02) –.102

−40 − 4 9. 82−−( )

10. (−−91)( − 2) −−( 6) 26

8 ORDER OF OPERATIONS

Please Excuse My Dear Aunt Sally

Parentheses ( ) Exponents Multiply Divide Add Subtract   "as you see it, left to right" Brackets [ ] "as you see it, left to right"

Square Roots

Grouping Symbols Absolute Value

Examples:

Note: When one pair of grouping symbols is inside another pair, perform the operations within the innermost pair of grouping symbols first.

2 36÷( 5 − 2) +⋅ 6 7 54÷−⋅ 3( 2 3) −+ 12 3

2 36÷( 3) +⋅ 6 7 54÷− 3( 6) −+ 12 3 36÷+⋅ 9 6 7 54÷−[ 3] − 12 + 3 4+ 42 −−+18 12 3 46 −+30 3 − 27

Examples:

Note: For a problem with a fraction bar, perform the operations in the numerator and denominator separately.

49− 7 +− 7 4⋅+ 2 7 8 + 7 15 = = = = 3 322−− 2 94 5 5

4 (6− 5) −−( 21) 14 −−( 21) 1 −−( 21) 1+ 21 22 = = = = = 2 −9 −− 3 422 − 9 −− 3 4 − 9 −− 3 16 27 − 16 11 ( )( ) ( )( ) ( )( )

9 ORDER OF OPERATIONS

Problems Answers

Perform the following operations.

1. 5+− 35 2 18

2. 12÷+ 2 6 1 37

3. 43−+2 6 1

4. 62 −÷+ 8 2 4 12

5. −−+12( 4 3) −19

6. 5−− 2( 43 ) 5 −24

2 − 7. 92652 ÷−( ) 17

24− 1 8. 2 − (−+31) 5

+−2 − 19 9. 4 22( 5) 3

2 3 21 7 10. −  4 32 12

10 EXPONENTS

Laws of Exponents

Definition: an = aaa... a  n times

a = the base

n = the exponent

Properties:

1. Product rule: aam n= a mn+ Example: aa2 3= aa  aaa = a23+ = a 5   2 times 3 times

m 7 1 1 1 1 1 a − a aaaaaaa − = mn =  =75 = 2 2. Quotient rule: n a Example: 5 1 1 1 1 1 aa a a aaaaa

m 2 3. Power rule: aan= nm Example: (a3) = aa 32 = 6 ( )

Raising a product to a power: Raising a quotient to a power: n n n aa = nn = (ab) a b  n bb

00 4. Zero power rule: a0 =1 Example: ( x) =1;( 3 xx) = 300 = 1 1 = 1

Examples:

(12ab2)( 3 a 3 b 5 c) = 36 a13++ b 25 c = 36 a 4 b 7 c

4 2x2 2 4 xx 24 16 8 = = 3 4 3 4 12 3y 3 yy 81

++ + 3xy22( 2 xy+= 5 y2) ( 3 xy 22)( 2 xy) +( 3 xy22)( 5 y 2) = 6 x 2121 y + 15 xy 222 = 6 xy 33 + 15 xy 24

11 EXPONENTS

Problems Answers

Perform the following operations and simplify.

10 8 1. ( xy25)( xy 83) xy

99 2. (57ab6)(− ab 38) −35ab

3. xy10 4 xy43

x6 yz z

4. 10ab37 −5ab3

−2ab24

5. 423 3 12 6 (ab c ) ab c

6. 2 532 4 10 6 (−4xyz) 16xy z

7. 3 3 93 (−mn) −mn

8. 0 2 2 (3nx) 9x

9. 2303 4 3 12 (53x y) (− xy ) −27xy

3 10. 3 −8x6 −4xy  3 2xyz z

12 EXPONENTS

Negative Exponents

Definitions:

a−m 1 a−m =  1 am Note: The act of moving a factor from the 1 am  = am numerator to the denominator (or from the a−m 1 denominator to the numerator) changes a negative exponent to a positive exponent. a−m bn −n m b a

Examples:

10−4 1 1 −4 = = 10  4 1 10 10,000

−− ab2 a 12 b a 1 a −2 ab−2 = =  = Since b has a negative exponent, move 22 2 11bb it to the denominator to become b

xy−−52 x 5 y 2 yz23 = Only the factors with negative exponents −3 = z z−3 x5 are moved.

13 EXPONENTS

Example:

−−22 Within the parenthesis, move the factors with −−12xy−23 12y3  = negative exponents. 42 242  33xy  xxy 

−2 −4y =  Simplify within the parenthesis. x6

−2 (−4) y−2  =  −2 ( x6 )    Use the power rule for exponents. −2  (−4) y−2 =  x−12  

x12 = 2 Move all factors with negative exponents. (−4) y2

x12 = 16y2 Simplify.

14 EXPONENTS

Problems Answers

Perform the following operations and simply.

2 1. (−3) 9

2. −32 –9

3. 330⋅⋅− 12 3 8 11 = 34 81

−2 −1 52 25 45⋅ = 4. −3 26 2 4 2 1024

5. (33a04 b)( ab 32 c ) 9ab72 c

2 − 2 −13 18ac42 e 6. ( 32ebc) ( ab ) b11

− 32− 3 y6 7. ( xy ) x9

6ab02 c− −18ac2 8. −1 (−3a) bc−3 b

15 EXPONENTS

Problems (continued) Answers

9. x−8 1

x8

10. −3 1 2 8

11. aaa−−4 10 8 1

a2

12. a−−6( aa 23) 1 7 a

13. x8 x12 −4 x 14. −3 1 b 5 8 b b 15. a−4 a2 −6 a

16. p p6 −5 p 17. −5 1 4 −2 4 64 18. 28 8 5 2 19. −1 1 2 2 2 8 20. −4 1 ab 53− 4 ab ab

21. xy−52 y5 −−23 3 xy x

16 EXPONENTS

Problems (continued) Answers

22. −−28 1 bc 42− 66 bc bc

23. 3−4 9 −3 9

24. 2−−53a a3 −6 2a 64

5 20 25. (2ab43− ) 32a 15 b

−3 12 26. 3x−4 x ( ) 27

−2 6 27. 32a−−45( ab) 3a 2 4b

−−12 92 28. 2ab34−− a 6 b ab ( ) ( ) 2

−2 4 29. (−4ab52− ) b 10 16a

−3 19 30. (ab−45) a 17 −72 b ab

−6 24 31. −5a−−1( ab 34) −5b 19 a

2 2ab54− 32a 32. ( ) 2 −3 b (2ab−32)

− −−254 27a22 33. 32aa( ) −2 16 (3a2 )

17 POLYNOMIALS

n Definition: A sum of a finite number of terms of the form: axn where an is a and n is a non-negative . (No negative exponents, no fractional exponents.)

43 Examples: 3xxx+ 2 −+ 81 is a −52 23xx+ is not a polynomial

Types of Polynomials:

. Monomial: A polynomial with 1 term

4 Example: 3x

. Binomials: A polynomial with 2 terms

3 Example: 24x +

. : A polynomial with 3 terms

Example: 3xx2 +− 42

Degree:

• Degree of a term – the sum of the exponents on the variables.

43 Example: 3xy has degree 437+=

• Degree of a polynomial – the largest degree of any of the terms. In a polynomial with one it is the largest exponent.

Example:

43 2 5328xxx−+− has degree 4 2 +− 2 += 2x y 34 xy has degree 3 since 2xy has degree 213

18 POLYNOMIALS

Addition of Polynomials

Adding Polynomials: Combine like terms.

Example:

(3xx2+−+ 45) ( 2 xx 2 −+= 823) xx 222 +−+−+= 45282543 xx xx −−

Subtraction of Polynomials

Subtracting Polynomials: Distribute the negative sign and combine like terms.

Example:

(3xx2+−− 45) ( 2 xx 2 −+= 823) xx 2 +−−+−=+ 45282 xx 22 x 127 x −

Multiplication of Polynomials

Multiplying Polynomials: Use the .

Example:

(32x+)( xx2 −+= 343) xxx( 22 −++ 342) ( xx −+ 34) =3x32 − 9 x + 12 xx + 2 2 −+ 6 x 8 32 =3xxx − 7 ++ 68

Example:

(2x+− 3)( 3 x 4) =( 2 xx)( 3) +−++−( 2 x)( 4) 33( x) 3( 4) =6x2 −+− 8 xx 9 12 2 =6xx +− 12

Example:

(3yy+− 43)( 4) The multiplication of the sum and difference of two terms. =3yy( 3) + 3 y( −+ 4) 43( y) + 4( − 4) =−+−9y2 12 yy 12 16

=+−9y2 0 16 =9y2 − 16 The answer is “the difference of two squares.”

19

POLYNOMIALS

Example:

2 (43y + ) The of a . (4343y++=)( y) 44433433 yy( ) +++ y( ) ( y) ( ) =16y2 +++ 12 yy 12 9

=++16yy2 2( 12) 9 =16yy2 ++ 24 9 The answer is a “perfect square trinomial.”

Example:

3 (ab− ) The of a binomial. (ababab−−−=−−−)( )( ) ( abab)( ) ( ab) =+aa a( − b) − b( a) − b( − b) ( a − b) 22 =a −−+ ab ab b( a − b) 22 =−+−a2 ab b( a b) 22 =aab( −−) 2 ababbab( −+) ( −) Use the distributive property. =−−a32 a b22 a 2 b + ab 2 +− ab 23 b =−+−a333 a 2 b ab 23 b

Division of Polynomials

Dividing a polynomial by a monomial.

Example:

9xy43−+ 24 xy 25 xy 49 xy 43 24 xy 25 xy 4 =−+ 3xy22 333xy22 xy 22 xy 22 1 =−+38xy42−− 32 xy 22 −− 52 xy 12 −− 42 3 1 =−+38xy2 xy 03 x− 12 y 3 y2 =38xy23 −+ y 3x

20 POLYNOMIALS

Problems Answers

Find the degree of the following and determine what type of polynomial is given.

1. 3x32 y+− 28 xy Trinomial, degree 5

2. 52− x3 Binomial, degree 3

3. x5 Monomial, degree 5

4. 8 Monomial, degree 0

Add: 5. (3xxx32+ 2 −++ 54) ( 6 xx3 − 28 −) 9274xxx32+ −−

Subtract: 6. (3xxx32+ 2 −+− 54) ( 6 xx3 − 28 −) −3xxx32 + 2 −+ 3 12

Multiply: 7. ( x+22)( xx2 −+ 5 3) 2xx32−−+ 76 x

8. (3xx+− 22)( 3) 6xx2 −− 56

9. (4xx+− 34)( 3) 16x2 − 9

2 2 10. (57x + ) 25xx++ 70 49

2 2 11. (92y − ) 81yy−+ 36 4

3 32 12. (34x − ) 27xxx− 108 +− 144 64

Divide: 15 12 10 8 13. 12x− 48 xx +− 18 x 5213 28xx− +− 6x10 6 x2

21 FACTORING

Factoring Out the Greatest Common Factor

1. Identify the TERMS of the polynomial. 2. Factor each term to its prime factors. 3. Look for common factors in all terms. 4. Factor out the common factor. 5. Check by multiplying.

Example: Factor 68x22+− xy 4 x Terms: 6x22 , 8 xy , and − 4x

23⋅⋅⋅+⋅⋅⋅⋅⋅−⋅⋅xx 222 xyy 22 x Factor into primes

23⋅⋅⋅+⋅⋅⋅⋅⋅−⋅⋅xx 222 xyy 22 x Look for common factors

Answer: 23xx( +− 4 y2 2) Factor out the common factor

Check: 23xx( ) +− 24 xy( 2 ) 22 x( ) Check by multiplying

68x22+− xy 4 x

Factoring of the Form: x2 ++ bx c

To factor xx2 ++5 6, look for two numbers whose product = 6, and whose sum = 5 List factors of 6 Choose the pair which adds to 5 and multiplies to 6. 16 2 + 3 = 5 −−16 2 • 3 = 6 2•3 −−23 Since the numbers 2 and 3 work for both, we will use them.

So, x22+56 x += x + 236 xx + + Substitute 23xx+ in for 5x.

=xx( ++23) ( x + 2) Factor by grouping. =++( xx23)( )

Answer: ( xx++23)( )

Check by multiplying: ( x+2)( x +=+ 3) x2 236 xx ++ =++xx2 56

22 FACTORING

Factoring Trinomials of the Form:ax2 ++ bx c

Example: Factor 6xx2 ++ 14 4

1. Factor out the GCF 6xx2 ++ 14 4 GCF is 2 23( xx2 ++ 7 2)

2. Identify the values of ab=3; = 7; and c = 2 ab, , and c for the 3xx2 ++ 72

3. Find the product of ac ac=23 = 6

4. List all possible factor pairs 1 6 1+6=7 1 & 6 and 2 & 3 are the that equal ac⋅ and identify 23 2+= 3 5 only factor pairs. the pair whose sum is b . 1 & 6 is the pair that adds up to 7 and multiplies up to 6. 5. Substitute boxed values 372xx2 ++ Substitute from Step 4 in for the bx ++1xx 6 for + 7 x 2 ++ + term. 3x 16 xx 2

6. the two pairs 3x2 ++ 162 xx + First step in factor by grouping. 7. Factor the GCF out of each xx(31231++) ( x +) The resulting common pair. factor is (31x + )

8. Factor out (31x + ) (31xx++)( 2) These are the two factors of 3xx2 ++ 72

9. Recall the original GCF 23( xx2 ++= 7 2) 23( x + 1)( x + 2) There are 3 factors of from Step 1. 6xx2 ++ 14 4

Answer: 23xx++ 1 2 ( )( )

10. Check by multiplying. 2( 3xx+ 1)( += 2) 6 xx2 + 14 + 4

23 FACTORING

2 −− Example: Factor 3yy 44

Note: If ac⋅ is negative, the factor pairs have opposite signs.

1. Factor out the GCF. There is no GCF.

2. Identify the values of a, ab=3; =−=− 4; c 4 b, and c.

3. Find the product of ac . ac=−=−3( 4) 12

4. List all possible factor +1 ( − 12) ++− 1( 12) =− 11 Include all possible factor pairs that equal ac and combinations, including identify the pair whose −1 ( + 12) −++ 1( 12) =+ 11 the +−/ combinations sum is b. since ac is negative. +26 ( −) + 264 +−( ) =− −26 ( +) − 264 ++( ) =+ +34 ( −) + 34 +−( ) =− 1 −34 ( +) − 34 ++( ) =+ 1

5. Substitute the boxed 3y22− 443264 y −= y + yy − − Substitute values from Step 4 in for 2yy−− 6 in for 4 y the bx term.

6. Group the two pairs. =2 + −− First step in factor by 32yy 64y grouping.

7. Factor the GCF out of =yy(32232 +−) ( y +) The resulting common each pair factor is (32y + ) .

8. Factor out (32y + ) (32yy+−)( 2) These are the factors.

Answer: (32yy+−)( 2)

9. Check: (32y+)( y −= 23) yy2 − 44 −

24 FACTORING

Factoring Perfect Square Trinomials

Example: Factor xx2 ++18 81

STEP NOTES 1. Determine if the trinomial is a Perfect Square. 2 2 • Is the first term a perfect square? x Yes xx = ( x) 2 • Is the third term a perfect square? +81 Yes 9 9=( 9) = 81 • Is the second term twice the product of +18x Yes 2( xx 9) = 18 the square roots of the first and third terms?

2. To factor a Perfect Square Trinomial: • Find the of the first term x xx2 = • Identify the sign of the second term + • Find the square root of the third term 9 81= 9 • Write the factored form 2 + +=+ +=+2 xx18 81( xx 9)( 9) ( x 9)

Answer: + 2 ( x 9) • Check your work ( xx+9)( +=+ 9) xx2 18 + 81

Example: Factor 25x22++ 20 xy 4 y

STEP NOTES 1. Is the trinomial a Perfect Square Trinomial? 2 2 • Is the first term a perfect square? 25x Yes (5xx) = 25 2 2 2 • Is the third term a perfect square? 4y Yes (24yy) = 2 • Is the second term twice the product of 20xy Yes 2( 5x 2 y) = 20 xy the square roots of the first and third terms?

2. Factor the trinomial • Square root of the first term 25xx2 = 5 • Sign of the second term +20xy →+ • Square root of the third term 42yy2 = 2 • Factored form 25x22+ 20 xyy +=+ 4( 5 xyxy 2 )( 5 += 2 ) (52xy+ )

2 Answer: (52xy+ ) • Check your work (5xyxy+ 2)( 5 += 2) 25 x22 + 20 xyy + 4

25 FACTORING

Factoring the Difference of Two Squares

Example: Factor x2 − 36

STEP NOTES 1. Are there any common factors? No 2. Determine if the binomial is the Difference of Two Squares • Is the first term a perfect square? x2 Yes xx = x2 • Is the second term a perfect square? 36 Yes 6 6= 36

• Is this the difference of the two terms? x2 − 36 Yes, “–” means the difference of the two terms.

3. To factor the Difference of Two Squares: • Find the square root of the first term. x xx2 = • Find the square root of the second term. 6 36= 6

• Write the factored form. x2 −=+36( xx 6)( − 6) The factors are the product of the sum +− Answer: ( xx66)( ) (+) and difference (–) of the terms’ square roots. • Check your work. ( xx+6)( −=− 6) x2 36

Example: Factor 36ab22− 121

STEP NOTES 1. Are there any common factors? No

2. Is the binomial a Difference of Two Squares? • Is the first term a perfect square? 36a2 Yes 6aa 6= 36 a2

• Is the second term a perfect square? 121b2 Yes 11bb 11= 121 b2

• Is this the difference of the two terms? 36ab22− 121 Yes, “–” means the difference of the two terms.

3. To factor the Difference of Two Squares: • Find the square root of the first term. 6a 36aa2 = 6

• Find the square root of the second term. 11b 121bb2 = 11

• Write the factored form. 22=+−The factors are the product of 36a 121 b( 6 a 11 ba)( 6 11 b) the sum (+) and difference (–) Answer: (6a+− 11 ba)( 6 11 b) of the terms’ square roots. • Check your work. (6a+ 11 ba)( 6 −= 11 b) 36 a22 − 121 b

26 FACTORING

FACTORING STRATEGY

1. Is there a common factor? If so, factor out the GCF, or the opposite of the GCF so that the leading is positive.

2. How many terms does the polynomial have?

If it has two terms, look for the following problem type:

a. The difference of two squares

If it has three terms, look for the following problem types:

a. A perfect-square trinomial b. If the trinomial is not a perfect square, use the grouping method.

If it has four or more terms, try to factor by grouping.

3. Can any factors be factored further? If so, factor them completely.

4. Does the check? Check by multiplying.

27 FACTORING STRATEGY Always check for

Greatest Common Factor First Then continue:

2 – Terms 3 – Terms 4 – Terms

2 Difference of Form: x++ bx c 2 Factor by Form: ax++ bx c 2 Two Squares xx−+65 Grouping 2 x2 − 4 5xx−− 76 List factors of 5 2 22 7x+ 14 xx −− 6 12 ( x) − (2) 51⋅ 5⋅− 6 =− 30 7xx( +− 26) ( x + 2) ( xx+−22)( ) −51 ⋅− List factors of –30

Choose the pair which −1 ⋅ 30 1 ⋅− 30 ( xx+−27)( 6) adds to – 6 −2 ⋅ 15 2 ⋅− 15

−516 +− =− −3 ⋅ 10 3 ⋅− 10 −56 ⋅ 5 ⋅− 6 −− ( xx51)( ) Choose the pair which adds to –7 Perfect Square Trinomials 3x+− 10 xx =− 7 2 4xx++ 20 25 2 Substitute this pair in for the 2 (2xx) ++ 22( )( 5) 5 middle term. 2 +− − 5xxx 3 10 6 2525xx++ ( )( ) Factor by Grouping. 2 (25x+ ) xx(53253+−) ( x +)

(53xx+−)( 2)

28 FACTORING

Problems Answers

Factor completely.

1. xx2 ++68 ( xx++24)( )

2. uu2 ++15 56 (uu++87)( )

3. xx2 +−8 20 ( xx+−10)( 2)

4. yy2 −−6 40 ( yy−+10)( 4)

5. mm2 −+15 54 (mm−−96)( )

6. x22+−5 xy 14 y ( x+−72 yx)( y)

7. a22++16 ab 28 b (a++14 ba)( 2 b)

8. xx32−−3 18 x xx( −+63)( x )

9. 14xx2 −− 3 (7xx+− 32)( 1)

10. 3xx2 −+ 19 20 ( xx−−53)( 4)

29

FACTORING

Problems (continued) Answers

11. 2xx2 −− 16 96 2( xx−+ 12)( 4)

12. 54xx54−− 54 108 x 3 54xx3 ( −+ 2)( x 1)

13. 6xx2 ++ 2 12 23( xx2 ++ 6)

14. 36xx2 +− 18 54 18( 2xx+− 3)( 1)

15. x2 −1 ( xx−+11)( )

16. 16x4 − 81 (4x2 ++− 92)( xx 32)( 3)

17. 27xx42− 48 3xx2 ( 3+− 43)( x 4)

18. 6xx2 ++ 33 45 32( xx++ 5)( 3)

2 2 19. xx−+18 81 ( x − 9)

2 2 20. 9xx++ 24 16 (34x + )

2 2 21. 20xx++ 60 45 52( x + 3)

30 SOLVING LINEAR EQUATIONS

To Solve a

1. Remove grouping symbols by using the distributive property. 2. Combine like terms to simplify each side. 3. Clear fractions by multiplying both sides of the equation by the Least Common Denominator. 4. Move the variable terms to one side and the constants to the other side. Do this by adding or subtracting terms. 5. Solve for the variable by multiplying by the inverse or dividing by the coefficient of x. 6. Check by substituting the result into the original equation.

Example Solve 3( x−− 4) ( 2844 xx +=) +

3x− 12 − 2 xx −= 8 4 + 4 Use the distributive property. xx−=+20 4 4 Combine terms −=3x 24 Move variable to one side, numbers to the other −3x 24 Divide to solve for the variable. = −−33 Solution: x = −8

Check: 384(−−−) ( 28( −+) 8) = 48( −+) 4

3(− 12) −−+=−+( 16 8) 32 4 −36 −−( 8) =− 28 −=−28 28

2 13 Example: Solve x −= 3 24

213 Clear fractions by multiplying by the LCD 12( 3x) −= 12( 24) 12( ) 8x −= 69

8x 15 = 88 15 x = Solution: 8 Check: 2 15 1 3 −= 38 2 4 215 15 1 3 −= 314 8 24 523 −= 444 33 = 44 31 SOLVING LINEAR EQUATIONS

Special Cases:

1. When the variable terms drop out and the result is a true statement, (i.e., 22= or 00= ), there are an infinite number of solutions. The equation is called an .

Example: Solve 2+ 9xx = 33( +− 1) 1 2+ 9xx = 33( +− 1) 1 29+xx = 9 +− 31

29+=+xx 9 2

00= Infinite number of solutions

Solution: All real numbers

2. When the variable terms drop out and the result is a false statement (i.e., 10 = 2 or 8 = 0), there is no solution. The equation is called a contradiction.

Example: Solve 3xx−= 53( − 2) + 4 3xx−= 53 −+ 64 3xx−= 53 − 2 −=−52

Solution: No solution

32

SOLVING LINEAR EQUATIONS

Problems Answers

Solve.

1. 3−( 2x +=− 5) xx 32( − 5) 17 x = 3

11 = 2. ( 23)( xx−=52) ( )( +) x 19

3. 2x+= 33 xx +−+ 6 4 No solution

4. 2x+= 33 xx +−− 6 3 Infinite number of solutions: All real numbers

5. 3xx+− 4 5 += 2 23( + x) x = 0

33 SOLVING LINEAR INEQUALITIES

Solving Linear Inequalities

Linear inequalities are solved almost exactly like linear equations. There is only one exception: if it is necessary to divide or multiply by a , the inequality sign must be reversed.

The solution can be written as a statement of inequality or in notation. It can also be shown as a graph. In interval notation and graphing: • Use a , [ ], or closed , if the endpoint is included in the solution • Use parenthesis, ( ), or open circle, if the endpoint is not included in the solution

Example: x <1 Interval Notation: (−∞,1)

or 0 1 2 0 1 2

Example: −32( xx −) ≤− 5 Distribute −36xx +≤− 5 Move variable term to one side, −4x ≤− 11 constants to the other

−−4x 11 Divide by − 4 and reverse the ≥ −−44 inequality because of division − 4

11 Simplify x ≥ 4

 11 11 x ≥∞Interval Notation: ,   44   [  11 Solution:  0 1 2 3 4 5  4  or    11  0 1 2 4 3 4 5 

34 SOLVING LINEAR INEQUALITIES

Problems Answers

Solve.

1. 4aa−> 7 32( + 5) − 2 a <−10 or(∞− , 10)

2. 2( 5xx+ 3) −≤ 3 6( 2 − 3) + 15 x ≥ 3 or[ 3, ∞)

3. 3xx+− 4 5 +< 2 23( + x) x > 0 or( 0,∞)

4. 10−+≥−+ 4xx 8 6 2 10 x ≤1 or(−∞ ,1]

5. 5( x+− 3) 6 xx ≤ 32( +− 1) 4 x x ≥ 4 or[ 4,∞)

6. 2( x−+ 53) xx < 4( −+ 63) x <−11 or(−∞ , − 11)

35 SOLVING LINEAR INEQUALITIES

Compound Inequalities

Compound inequalities are solved the same way as simple inequalities. Any (addition, subtraction, multiplication, division) must be performed on all three pieces of the inequality. Never remove the variable from the middle piece of the inequality. Always remember that when dividing or multiplying by a NEGATIVE number, the inequality sign must be reversed.

Example: Solve −< 8 3x −≤ 2 4 +2 ++ 22 −<63x ≤ 6 Add 2 to all three parts of the inequality −63x 6 <≤ Divide all three parts of the inequality by 3 333 −<22x ≤ Simplify

In interval notation, this is (−2, 2], because x is between –2 and 2, including the endpoint 2, but not including the endpoint –2.

−<2x ≤ 2 Interval Notation: (−2,2] Solution:

-3 ( - 2 -1 0 1 2 3 or ○ -3 -2 -1 0 1 2 3 ∙ Example: Solve −5 <− 3x < 12

5− 3x 12 When dividing an inequality by a negative −> > −−−333 number, reverse the inequality sign.

5 Keep the variable in the middle piece of the >x >−4 3 inequality.

5 −<4 x < 3 Arrange the inequality so that the lesser 5 value is on the left and the greater value on −4, the right, and change the inequality symbols 3 to preserve the relationship.

5 5 −<4x < Interval Notation: −4, 3 3

Solution:

( ) or

-5 -4 0 2 -5 -4 0 2

36

SOLVING LINEAR INEQUALITIES

Problems Answers

Solve

1. −<23x +< 14 −<11x < or (−1,1)

2. −≤54x + 55 ≤ −≤5 x ≤0 or − 5 2 [ 2 ,0]

3. −<44x + 28 ≤ −<33x ≤ or − 33 22 ( 22, ]

4. 352<−x < 7 −<11x < or (1,1)

5. 21≤−x < 6 −51

−< − < 4 << 4 6. 243x 0 3 x 2 or ( 3 ,2)

7. 014<−x < 7 −<3 x <1 or − 3 1 24 ( 24, )

8. −432 ≤ −x ≤− 1 2 ≤≤x 7 or 7 2 [2, 2 ]

9. −742 < −x <− 4 4 <

10. 443<−x < 6 −<2 x <0 or − 2 ,0 3 ( 3 )

37 GRAPHING LINEAR EQUATIONS

Standard Form: ax + by = c

A line is made up of an infinite number of points in the form (x,y). The coordinates for each of these points will satisfy the equation for the line (make it true).

Two special points on the line are its x and y intercepts. These are the points on the line where the line crosses the x axis and the y axis. The x intercept is in the form (x,0), and the y intercept is in the form (0,y).

Find the x intercept of the line by substituting 0 for y, and solving for x, (x,0).

Find the y intercept of the line by substituting 0 for x, and solving for y, (0,y).

Example: Graph the line using the intercept method 3xy−=− 4 12

Find the x intercept: (4,6) y Substitute 0 for y and solve for x:

3xy−=− 4 12 3x −=− 4( 0) 12 3x −=− 0 12 3x =− 12 −12 x 3x = 0 3 x =− 4 The x intercept is (– 4,0).

Find the y intercept:

Substitute 0 for x and solve for y:

3xy−=− 4 12 3( 0) −=− 4y 12 0−=− 4y 12 −=−4y 12 −−4y 12 = −−44 y = 3

The y intercept is (0,3).

To graph the line, graph these two points, and connect them. As a check, find one more point on the line. Let x = 4 and find y = 6. The point (4,6) is on the line and collinear with (− 4,0) and ( 0,3) .

38 GRAPHING LINEAR EQUATIONS

Slope—Intercept Form: y = mx + b

If the equation of the line is in slope-intercept form, we use the slope m, and the y intercept, (0,b), to graph the line.

The slope of a line passing through points ( xy, ) and ( xy, ) is 11 22 The slope calculation is called a − vertical change rise change in y yy21 Rate of Change. The rate of change m= = = = if xx21≠ horizontal change run change in xxx21− describes how one quantity (the numerator) changes with respect to Example: Graph the line: yx=23 + . another (the denominator).

Identify the slope as a fraction:

In the equation y= mx + b , m is the slope. 2 In the equation yx=23 + , 2 is the slope, so m = . 1

Identify the y intercept as a point:

In the equation, y= mx + b,( 0, b) is the y intercept. In the equation, yx=23 + , (0,3) is the y intercept

Graph the y intercept: Graph the slope:

2 rise of 2 (0,3) is the first point to be graphed m = = 1 run of 1 y 2 is positive; move up 2 (rise) from the y 8 intercept, (0,3), to the point (0,5).

6 (1,5) 4 1 is positive; then move 1 to the right (run) 2 (0,3) from (0,5) to (1,5). x 0 2 4 6 8

(1,5) is the second point of the line.

Graph the line through the two points:

Graph the line through the points (0,3) and (1,5).

39 GRAPHING LINEAR EQUATIONS

Special Cases:

The graph of the linear equation yk= , where k is a real number, is the horizontal line going through the(0, k ) y

Example: y = 2

4

4 x

The graph of the linear equation xk= , where k is a real number, is the vertical line going through the point (k,0) . y

Example: x = 4

3

3 x

40 GRAPHING LINEAR EQUATIONS

Problems Graph the following equations:

1. xy−=55 y 2. xy+=35 y

x x

3. yx= + 4 4. xy= 2 y y

x x

5. 6y = 12 6. x = −2 y y

x x

41 GRAPHING LINEAR EQUATIONS

Answers

1. xy−=55 y 2. xy+=35 y

4 3

3 x 2 x

= + = 3. yx4 y 4. xy2 y

4 4

3 x 4 x

5. 6y = 12 6. x = −2 y y

5

2 3 x 2 x

42 FINDING THE EQUATION OF A LINE

The standard form of the equation of a line is written as

ax+= by c

where a and b are not both 0.

Definition of slope: Let L be a line passing through the points ( xy11, ) and ( xy22, ) . Then the slope of L is given by the formula:

yy− m = 21 xx21−

The slope-intercept form of the equation of a line is written as

y= mx + b

where m is the slope and the point (0,b) is the y-intercept.

Point-Slope Form: The equation of the straight line passing through ( xy11, ) and having slope m is given by y−= y11 mx( − x) .

Parallel Property: Parallel lines have the same slope.

Perpendicular Property: When two lines are perpendicular, their slopes are negative (opposite) reciprocals of one another. The product of their slopes is –1.

43 FINDING THE EQUATION OF A LINE

To find the equation of a line, use: Slope-Intercept form: y= mx + b , slope m, y intercept (0,b)

or Point-Slope form: y−= y11 mx( − x) , slope m, point ( xy11, ) Slope-Intercept Form

When given the slope and y intercept: use the slope-intercept form y= mx + b

3 Example: Find the equation of the line with slope and y intercept (0,− 2) 4

Use the slope-intercept form, y= mx + b , and fill in the values:

Solution: The equation of the line is: Make the substitutions of 3 33 yx= − 2 m = slope is and 4 44 by=−−2( intercept is ( 0, 2))

Point-Slope Form

When given the slope and a point on the line: use the point-slope form y−= y11 mx( − x)

2 Example: Find the equation of the line with slope , through the point (6,1) 3

Use the point-slope form, y−= y1 mx( − x1 ), and fill in the values.

2 yx−=16( −) 3 2 26 22 yx−=1 − ⋅ Make the substitutions of m = slope is and 3 31 33 2 xy= 6 and = 1( point (6,1)). yx−=14 − 11 3 2 yx−+=11 − 41 + 3 2 yx= − 3 3

2 Solution: The equation of the line is yx= − 3 3

44 FINDING THE EQUATION OF A LINE

When given two points on the line, first find the slope m, then use the slope and either one of the points to find the equation using the point-slope form y−= y11 mx( − x)

Example: Find the equation of the line through the points (2,3) and (1,1) .

31− 2 Slope m = = = 2 21− 1

−= − y y11 mx( x) Make the substitutions of

yx−=32( − 2) m= 2( slope is 2) and xy11= 2 and = 3

yx−=32 − 4 (point( 2,3)) . yx=21 −

Solution: The equation of the line is yx=21 −

45 FINDING THE EQUATION OF A LINE

Problems Answers

1. Put the following in slope-intercept form.

a. 84xy−= 4 yx=21 −

2 b. 233xy+= yx=−+1 3

2. Find the slope.

5 a. (3, 4) and (7,9) 4

4 b. (−2,1) and (3,− 3) − 5

3. Find the equation of the line through (5,3) yx=−+2 13 and parallel to yx=−+21.

1 13 4. Find the equation of the line through (−−1, 3) yx=−− 44 and perpendicular to yx=42 − .

46 FINDING THE EQUATION OF A LINE

Problems Answers

Find the equation of the line that satisfies the given conditions. Write the equation in both standard form and slope-intercept form.

Standard Form Slope-Intercept Form

1 15 5. slope = line passes through (3, 4) xy−=−25 yx= + 2 22

5 5 6. slope = − line passes through (0,0) 560xy+= yx= − 6 6

7. slope = 1 y-intercept –3 xy−=3 yx= − 3

8. horizontal line through (1, 4 ) y = 4

9. slope is undefined and passing through (−5,6) x = −5

3 −3 15 10. slope = − x-intercept –5 3xy+=− 2 15 yx= − 2 22

11. horizontal line through (5,− 3) y = −3

12. vertical line passing through (5,− 4) x = 5

13 13. line passing through (1, 2 ) and (5, 4) xy−=−23 yx= + 22

−5 17 14. line passing through (−3, 4) and (5,− 1) 5xy+= 8 17 yx= + 88

15. line passing through (4,− 3) and (4,− 7) x = 4

16. line parallel to 36xy+=and passing through 35xy+= yx=−+35 (1, 2 )

17. line passing through (5,− 3) and parallel to y = −3 y −=20

59 18. line perpendicular to 253xy+= and 52xy−=− 9 yx= + 22 passing through (1, 7 )

47 APPLICATION OF LINEAR EQUATIONS

Example: Determine Rate of Change

A credit union offers a checking account with a service charge for each check written. The relationship between the monthly charge for each check “y” and the number of checks written “x” is graphed below. At what rate does the monthly charge for the checking account change? Also, find the unit cost, another way to express the rate of change.

y

22

20

18

16 (75, 16) 14

12 (50, 14) for Charge Monthly

checking account ($)

10

8 x 10 20 30 40 50 60 70 80 90 100 Number of checks written during the month

Find the rate of change (slope of the line). The units will be dollars per number of checks.

From the graph, we see that two points on the line are (50, 14) and (75, 16). If we let

( xy11,) = ( 50,14) and( x2 , y 2) = ( 75,16) , we have

( yy−−)dollars( 16 14) dollars 2dollars Rate of = 21 = = The rate of change can be expressed change ( xx21−−)checks( 75 50) checks 25checks as $2 for every 25 checks.

Rate of change can also be expressed as $0.08 per check or 8¢ per check.

The monthly cost of the checking account increases $2 for every 25 checks written. To find the 2 cost of 1 check (unit cost), take the fraction which represents the rate of change, 25 , and divide both the numerator and denominator by 25.

2 dollars 2÷ 25 .08 dollars = = = $.08 / check. The unit cost = $.08 per check. 25 checks 25÷ 25 1 check

48

APPLICATION OF LINEAR EQUATIONS

Problems Answers

The graph models the number of members in an organization from 2001 to 2010.

(2001,200) 200 . 150 (2007,125) 100 . . 50 Number of Members

2001 2005 2010 Year

How many members did the 100 members organization have in 2009?

Find the rate of change. loss of 25 members every (slope of the line) two years

What is the rate of change per year? loss of 12.5 members per year

49 FUNCTIONS

Recall that relation is a of ordered pairs and that a is a special type of relation.

A function is a set of ordered pairs (a relation) in which to each first component, there corresponds exactly one second component.

The set of first components is called the of the function and the set of second components is called the range of the function.

Examples: Determine if the following are functions.

Domain Range x y This is a function, since for each −4 0 first component, there is exactly −8 1 one second component. −16 2

Domain Range x y This is not a function since for 2 1 the first component, 7, there 7 3 are two different second components, 3 and 5. 5

Vertical Line Tests: A graph in the plane represents a function if no vertical line intersects the graph at more than one point.

Examples: y This is not a graph of a

(3,2) function. For example, the x-value of 3 is y This is a graph of x assigned two different y a function. It (3,–2) values, 2 and −2. (for the x passes the vertical first component, 3, there line test. are two different second components, 2 and −2.)

Since we will often work with sets of ordered pairs of the form ( xy, ) , it is helpful to define a function using the variables x and y.

Given a relation in x and y, if to each value of x in the domain there corresponds exactly one value of y in the range, then y is said to be a function of x.

50 FUNCTIONS

Notation: To denote that y is a function of x, we write y= fx( ) . The expression “ fx( ) ” is read f of x. It does not mean f times x. Since y and fx( ) are equal, they can be used interchangeably. This means we can write yx= 2 , or we can write fx( ) = x2 .

Evaluate: To evaluate or calculate a function, replace the x in the function rule by the given x value from the domain and then compute according to the rule. For example:

Examples:

1. Given: fx( ) = 6 x + 5

Find:f ( 2) = 6( 2) += 5 17

f (0) = 60( ) += 5 5

f (−1) = 61( −+=−) 5 1

2. Given: gx( ) = 3 x2 −+ 5 x 8

2 Find:g ( 1) = 3( 1) − 5( 1) += 8 6

2 g (0) = 30( ) − 50( ) += 8 8

2 g (−1) = 3( − 1) − 5( −+= 1) 8 16

51 FUNCTIONS

Problems Answers

1. Determine whether or not each relation defines a function. If no, explain why not.

a. Domain Range a. Yes, for each first component, x y there corresponds exactly one 11 4 second component. 21 8

31 10

b. Domain Range b. No, for each first component, 3, there corresponds two different x y second components, 2 and 7. 3 2 5 6 3 7

c. {(1, 3) ,( 2,− 4) ,( 1, 0 )} c. No, for the first component, 1, there corresponds two different second components 3 and 0.

d. Yes, for each first component there corresponds exactly one second d. {(4,3) ,(− 2,3) ,( 1,3)} component. 2. Determine the domain and range of each of the following:

a. {(−−3,4) ,( 4,2) ,( 0,0) ,( 2,7)} a. Domain: {−−3, 2,0,4} Range: {0,2,4,7}

b. x y b. Domain :{−−−5,7,9} –5 2 Range: {2,4,6} –7 4 –9 6

c. x y c. Domain: {1,5,9} 1 7 Range: {7,13,21} 5 13 9 21

3. Given fxx( ) =+=−2 and gxx( ) 3, find the following: a. f (−2) a. 0 b. f (−4) b. –2 c. g (0) c. –3 d. g (2) d. –1

52 SYSTEMS OF LINEAR EQUATIONS

The following summary compares the graphing, elimination (addition), and substitution methods for solving linear systems of equations.

Method Example Notes Graphing 26xy+= You can see the solution is where the two xy−=28 lines intersect but if the solution does not y involve integers it’s impossible to tell exactly what the solution is.

5 2x+y=6 The solution for this system of linear 2 equations is (4,− 2) .

-5-4-3- 2 - 1 - 1 2 34 5 -2 (4,-2)

x-2y=8 -4 -

Substitution yx=31 − Gives exact solutions.  The solution for this system of equations 32xy−=− 4 is (2,5). Substitute 3xy− 1 for :

3xx− 23( −=− 1) 4 Solve for x: x = 2 Back-substitute: y =32( ) −= 1 5

Solution: (2,5)

Elimination 23xy+=− 8 Gives exact solution. (Addition)  Easy to use if a variable is on one 5xy+=− 4 34 side by itself. The solution for this system Multiply the top equation by 5 and of equations is −10, 4 . the bottom equation by −2. This will ( ) result in opposite of x. 10xy+=− 15 40  −10xy −= 8 68 Add the two equations: 7y = 28 y = 4

To find x, substitute y = 4 into either original equation. 2x +=− 34( ) 8

x = −10 Solution: (−10, 4)

53 SYSTEMS OF LINEAR EQUATIONS

Three Possible Solution Types

y No Solution (Parallel Lines) Inconsistent System Independent Equations 0 x # = different # (contradiction)

y Exactly One Solution (point of intersection) 0 x Consistent System . Independent Equations

Solution

Infinitely Many Solutions y (Lines Coincide-any point on the line is a solution) 0 x Consistent System Dependent Equations # = same # (identity)

54 SYSTEMS OF LINEAR EQUATIONS

Solving Linear Systems By Substitution Method

To solve a linear system of two equations in two variables by substitution:

1. Solve one of the equations for one of the variables. 2. Substitute the expression obtained in step 1 for that variable in the other equation, and solve the resulting equation in one variable. 3. Substitute the value for that variable into one of the original equations and solve for the other variable. a. if we get 0 = nonzero number or # = different # (contradiction), the system is inconsistent, the lines are parallel, the equations are independent, and there is no solution. b. if we get 00= or # = same # (identity), the system is consistent, the lines are the same (coincide), the equations are dependent, and there are infinitely many solutions.

Example:

Solve the system of linear equations using the substitution method.

35xy+= 3  xy=84 − This equation is solved for x.

38( − 4yy) += 5 3 Substitute 84− y in for x in the first equation. 24− 12yy += 5 3   24−= 7y 3   Solve for y. −=−7y 21 y = 3 

x =−=−8 43( ) 4 Substitute 3 in for y to find the value of x.

The solution is (− 4,3) .

Check in both original equations: 353xy+= xy=84 − 3(−+ 4) 53( ) = 3 −=4 8 − 43( )

−+=12 15 3 −=4 8 − 12 33= −=−44

55

SYSTEMS OF LINEAR EQUATIONS Example:

Solve the system of linear equations using the substitution method.

4xy+= 12 4 Since y has a coefficient of 1, it will be easy to solve this  −5xy += 11 second equation for y.

4xy+= 12 4   yx=5 + 11

4xx+ 12( 5 += 11) 4 Substitute 5x + 11 in for y

4xx++= 60 132 4  64x = − 128  Solve for x.  x = −2 

y =5( −+ 2) 11 Substitute – 2 in for x to find the value of y.

y =1

The solution is (−2,1) .

Check in both original equations:

4xy+= 12 4 −5xy += 11 4(−+ 2) 12( 1) = 4 −5( − 2) += 1 11

−+8 12 = 4 10+= 1 11 = 44 11= 11

56

SYSTEMS OF LINEAR EQUATIONS

Solving Linear Systems By Elimination (Addition) Method

To solve linear systems of two equations in two variables by elimination:

1. Write the system so that each equation is in standard form.

ax+= by c

2. Multiply one equation (or both equations if necessary), by a number to obtain (opposite) coefficients of one of the variables.

3. Add the resulting equations and solve the new equation in one variable.

4. Substitute the value for that variable into one of the original equations and solve for the other variable.

a. If we get 0 = nonzero number or # = different # (contradiction), the system is inconsistent; there are no solutions, the equations are independent, and the lines are parallel.

b. If we get 0 = 0 or # = same # (identity), the system is consistent; there are infinitely many solutions, the equations are dependent and the lines are the same (coincide). Example:

Solve the system of linear equations using the elimination (addition) method.

xy+=9 The coefficients of y are opposites (additive  inverses). Add these equations. 23xy−=−  36x = Solve for x. x = 2

xy+=9 Substitute 2 in for x and solve for y.

29+=y y = 7 The solution is (2,7).

Check in both original equations: xy+=9 23xy−=− 279+= 22( ) −=− 7 3 99= 47−=− 3 −=−33

57

SYSTEMS OF LINEAR EQUATIONS Example:

Solve this system of linear equations using the elimination (addition) method.

3xy+= 2 11 Multiplying the first equation by 4 yields 8y.This will result in  28xy−=− 2 opposite coefficients (additive inverses) of y.

12xy+= 8 44 Add these equations.   28xy−=− 2 14x = 42 x = 3 Solve for x.

3( 3) += 2y 11 Substitute 3 in for x and solve for y. 9+= 2y 11

22y =

y =1

The solution is (3,1).

Check in both original equations:

3xy+= 2 11 28xy−=− 2 3( 3) += 2( 1) 11 23( ) −=− 81( ) 2

9+= 2 11 68−=− 2 11= 11 −=−22

58 SYSTEMS OF LINEAR EQUATIONS

Problem

Solve this System of Linear Equations using all three methods: Graphing Method Substitution Method Elimination (Addition) Method

xy+=−1  −= xy1

Graphing Method:

y

x 0

Substitution Method:

Elimination (Addition) Method:

Answer is on the following page.

59 SYSTEMS OF LINEAR EQUATIONS

y Answer:

2

–2 0 2 x

–2

Solution:

60 SYSTEMS OF LINEAR EQUATIONS

Problems Answers Solve these systems of linear equations using either the substitution or elimination (addition) method.

1. xy+=11 xy=6 and = 5

xy−=1 (6,5)

2. 29xy+= xy=4 and = 1

3xy−= 11 (4,1)

3. 25xy+= xy=2 and = 1

52xy−= 8 (2,1)

4. 34xy+= 7 xy=1 and = 1

87xy−= (1,1)

11 xy= and = 5. 23xy+= 2 23

10xy−= 6 3 11 , 23

6. 2xy+= 7 23 xy=1 and = 3

xy−=−4 11 (1, 3)

7. xy+=3 xy=1 and = 2

238xy+= (1, 2 )

8. 55xy+= xy=1 and = 0

23xy+= 2 (1, 0 )

9. 448xy−= Inconsistent system with no

xy−=1 solution.

10. 24xy−= Dependent system with

6xy−= 3 12 infinitely many solutions.

61 APPLICATION OF A SYSTEM OF LINEAR EQUATIONS

The following steps are helpful when solving problems involving two unknown quantities.

1. Analyze the problem by reading it carefully to understand the given facts. Often a diagram or table will help you visualize the facts of the problem. 2. Define variables to represent the two unknown quantities. 3. Translate the words of the problem to form two equations involving each of the two variables. 4. Solve the system of equations using graphing, substitution, or elimination (addition) method. 5. State the conclusion. 6. Check the results in the words of the problem.

Example: Determine the cost of a quart of pineapple and a container of frozen raspberries for the punch Diane is making. Three (3) quarts of pineapple juice and 4 containers of raspberries will cost $10. Five (5) quarts of pineapple juice and 2 containers of raspberries will cost $12.

Let: p = cost of a quart of pineapple juice Define the variables r = cost of a container of raspberries

3pr+= 4 10 Solve the system for “p” using  5pr+= 2 12 elimination (addition) method

 3pr+= 4 10  −10pr −=− 4 24 −=−7p 14 p = 2

3( 2) += 4r 10 Substitute 2 in for p and solve for r r =1

Answer: A quart of pineapples costs $2 and a container of frozen raspberries costs $1.

Check:

Using $2 as the cost for a quart of pineapple Check the results in the words juice, 3 quarts costs $6. Using $1 as the cost of the problem. of a container of raspberries, 4 containers cost $4. $6 for pineapple juice and $4 for raspberries is $10 total cost. Do the same technique for checking the $12 cost.

62 APPLICATION OF A SYSTEM OF LINEAR EQUATIONS

Problem Answer

Use the steps for solving a system of linear equations.

People have begun purchasing tickets to a production A child’s ticket costs $5 of a musical at a regional theatre. A purchase of 9 and an adult ticket costs $9. adult tickets and 7 tickets for the children costs $116. Another purchase of 5 adult tickets and 8 tickets for the children costs $85. Find the cost of an adult ticket and a child’s ticket.

63 SOLVING LINEAR INEQUALITIES IN TWO VARIABLES

Graphing Linear Inequalities In Two Variables

1. Replace the inequality symbol with an equal symbol = and graph the boundary line of the region. If the original inequality allows the possibility of equality (the symbol is either ≤≥or ), draw the boundary line as a solid line. If equality is not allowed (< or >), draw the boundary line as a dashed line.

2. Pick a test point that is on one side of the boundary line. (Use the origin if possible; it is easier). Replace x and y in the inequality with the coordinates of that point. If a true statement results, shade the side (half-plane) that contains that point. If a false statement results, shade the other side (other half-plane).

Example:

Solve 42xy+≥

Step 1: 42xy+= ≥ sign, draw the solid boundary line yx=−+42 y

2 x -2 0 2 -2

Step 2: pick (0,0) Pick test point. 40( ) +≥ 0 2 Shade the side that contains (0,0). 4≥ 2is true y

2. 0 2 x

(0,0)

64 SOLVING LINEAR INEQUALITIES IN TWO VARIABLES

Problems 1 1. Solve yx≤−1 2 y

x

2. Solve 236yx+<

y

x

3. Solve y ≥ 5 y

x

65 SOLVING LINEAR INEQUALITIES IN TWO VARIABLES

Answers y 1.

4 2 4 x

y 2.

5

5 x

y 3.

2 2 x

66 SOLVING SYSTEMS OF LINEAR INEQUALITIES

1. Graph each inequality on the same rectangular .

2. Use shading to highlight the intersection of the graphs (the region where the graphs overlap). The points in this region are the solutions of the system.

3. As an informal check, pick a point from the region where the graphs intersect and verity that its coordinates satisfy each inequality of the original system.

xy+≥4 Example: Solve this system of inequalities.  43xy−> 9

Look at the graph of each inequality separately. xy+≥4 y ≥ sign, draw solid boundary line. Use (0, 0) as a test point. 04≥ is false. 4 Shade the half-plane that does not include (0, 0). x 4

43xy−> 9 y > sign, draw dashed boundary line. Use (0, 0) as a test point. 0 > 9 is false. Shade the half-plane that does not include (0, 0).

3 x −3

67 SOLVING SYSTEMS OF LINEAR INEQUALITIES

xy+≥4  43xy−> 9

y Graph each inequality on the same coordinate system.

Notice the region where the graphs overlap.

The points in this region are the solutions of the system. 0

Solution of the System:

y

4

0 3 x

68 SOLVING SYSTEMS OF LINEAR INEQUALITIES

Informal Check: y

The point (5,2) is in the region of overlap. Substitute (5,2) in to both inequalities. 4 xy+≥4 43xy−> 9 52+ 4 45( ) − 32( ) 9 • (5,2)

≥ 7 4 is true 20− 6 9 0 3 x 14> 9 is true

(5,2) is one of the solutions.

69 SOLVING SYSTEMS OF LINEAR INEQUALITIES

Problems

1. Solve this system of linear inequalities.

 4 yx≥− +2  3  5 yx>+1  6 y

x

2. Solve this system of linear inequalities.

35xy+ ≥−  −35xy − ≥− y

x

Answers on the next page

70

SOLVING SYSTEMS OF LINEAR INEQUALITIES

Answers

y 1.

5

2 3 x

-2

y 2. 7

x 4

71 SOLVING ABSOLUTE VALUE EQUATIONS

Absolute Value: The absolute value of a number is its distance from 0 on the number line

For any positive number k and any X: To solve Xk= , solve the equivalent compound equation Xk=or X = − k

Example: x = 7

Solution: x=7 or x=-7

–7 0 7

Example: x −=23

xx−=2 3 or −=− 2 3

Solution: x=5 x=-1

–1 0 5

Check: x = 5 Check: x = −1 x −=23 x −=23 52−= 3 −−12 = 3 33= −=33 33= 33=

72 SOLVING ABSOLUTE VALUE EQUATIONS

4x − 64 Example: Solve the following: = 32 4

Solve the equation by 4xx−− 64 4 64 = = = = − When Xk, then Xk rewriting it as two separate 32 or 32 44 and Xk= − equations.

Solve each equation for x. 4x −= 64 128 4x −=− 64 128 Multiply both sides by 4.

Solutions: 4x = 192 4x = − 64 Add 64 to both sides.

x = 48 or x = -16 Divide both sides by 4.

Check: Check x = 48 Check x = −16

4( 48) − 64 4(−− 16) 64 = 32 = 32 4 4

192− 64 −−64 64 = 32 = 32 4 4

128 −128 = 32 = 32 4 4

32= 32 −=32 32

32= 32 32= 32 The two solutions check.

73 SOLVING ABSOLUTE VALUE EQUATIONS

Example: Solve the following: 574xx−=( + 1)

Solve the equation by 574xx−=( + 1or57) x −=− 4( x + 1) When Xk= , then rewriting it as two separate  Xk= and Xk= − equations.

Solve each equation for x. 5744xx−= + 57 x −=−[ 44 x +] Use the Distributive Property. x−=74 5xx−=−− 7 4 4

xx=11 9−=− 7 4

93x = 1 Solutions: xx= 11 or = 3

1 Check: Check x =11 Check x = 3

11   5( 11) −= 7 4( 11 + 1) 5−= 74  + 1  33  

55−= 7 4( 12) 54 −=74

33

48= 48 −16 16 = 33

48= 48 16 16 The two solutions = 33 check.

74 SOLVING ABSOLUTE VALUE EQUATIONS

Problems Answers

1. 3x −= 25 7 xx= or = − 1 3

2. 10xx−=− 40 No solution. You can’t solve an absolute value equation when the absolute value is equal to a negative quantity.

2 9 27 3. x ++=3 4 10 xx= or = − 3 22

1 x =10 4. 3x −−=− 54 4 2

5. 5xx+= 3 3 + 25 7 xx=11 or = − 2

6. 213xx+=( + 1) 4 xx=−=−2 or 5

75 SOLVING ABSOLUTE VALUE INEQUALITIES

To solve inequalities with absolute value signs, there are two cases:

For any positive number k and any algebraic expression X Case 1: To solve X< k, solve the equivalent compound inequality −

Example: Solve: x < 7

Solution: -7

-7 0 7 or

-7 0 7

For any positive number k and any algebraic expression X Case 2: To solve X> k, solve the equivalent compound inequality X<− k or Xk > . To solve X≥ k, solve the equivalent compound inequality X≤− k or Xk ≥ .

Example: Solve: x ≥ 7

﴿∞,Solution: x≤− 7 or x ≥ 7 Interval Notation: ﴾–∞,–7] ∪ [7

-7 0 7

or

-7 7 0

76 SOLVING ABSOLUTE VALUE INEQUALITIES

Example: Solve: x −≤23 Case 1 −≤3x − 23 ≤ Equivalent compound inequality −≤15x ≤ Addition Property of Equality

Solution: −≤1x5 ≤ Interval Notation: [–1,5]

5 -1 0 or

5 -1 0

Example: Solve: x −>23 Case 2 xx−2 <− 3 or − 2 > 3 Equivalent inequalities xx<−15 >

Solution: xx<−1 or > 5 Interval Notation: (∞−, 1) ∪( 5, ∞)

-2 -1 0 5

or

-2 -1 0 5

77 SOLVING ABSOLUTE VALUE INEQUALITIES

Problems Answers 28<

xx<−3 or > 17 2. x −>7 10 (−∞, − 3) ∪( 17, ∞)

−≤ 19 3. 5 3x 14 −≤3 x ≤ 3 19 −3, 3

−<36x < 4. 2x −< 39 (−3, 6)

3− x xx≤−27 or ≥ 33 5. ≥ 6 5 (−∞, − 27] ∪ 33, ∞) 

−−≥ 4 6. 3x 840 xx≤≥or 4 3 4  −∞, ∪[ 4, ∞) 3

78 VOCABULARY USED IN APPLICATION PROBLEMS

Addition add English Algebra sum The sum of a number and 4. x + 4 total Seven more than a number. x + 7 plus Six increased by a number. 6 + x in all A number added to 8. 8 + x more than A number plus 4. x + 4 together increased by all together combined

Subtraction subtracted from English Algebra difference The difference of a number and 3. x − 3 take away The difference of 3 and a number. 3− x less than Five less than a number. x − 5 minus A number decreased by 3. x − 3 remain A number subtracted from 8. 8 − x decreased by Eight subtracted from a number. x −8 have left Two minus a number. 2 − x are left more fewer Be careful with subtraction. The order is important. Three less than a number is x−3.

79 VOCABULARY USED IN APPLICATION PROBLEMS

Multiplication

English Algebra product of The product of 3 and a number. 3x multiplied by Three-fourths of a number. 3 x 4 times Four times a number. 4x of A number multiplied by 6. 6x

Double a number. 2x

Twice a number. 2x

Division

English Algebra

x divided by The quotient of a number and 3. x ÷ 3 or 3 quotient of 3 The quotient of 3 and a number. 3÷ x or x separated into equal x A number divided by 6. x ÷ 6 or parts 6 6 shared equally Six divided by a number. 6÷ x or x

Be careful with division. The order is important. A number divided x by 6 is 6

80 SOLVING APPLICATION PROBLEMS

Strategy for solving word problems:

1. Analyze the problem: Read the problem carefully. 2. Visualize the facts of the problem (if needed): Use diagrams and/or tables. 3. Define the variable/s: Identify the unknown quantity (or quantities) and label them, i. e., let x = something. 4. Write an equation: Use the defined variable/s. 5. Solve the equation: Make sure you have answered the question that was asked. 6. Check your answer(s): Use the original words of the problem.

Examples:

1. The sum of three times a number and 11 is –13. Find the number.

Let x = the number Define the variable.

3x +=− 11 13 Write an equation..

3x +−=−− 11 11 13 11 3x = − 24 Solve the equation. x = −8

Solution: The number is – 8.

Check: 3(−=− 8) 24

−+=−24 11 13

2. Together, a lot and a house cost $40,000. The house costs seven times more than the lot. How much does the lot cost? The house?

Let x = the cost of the lot Define the variable. 7x = the cost of the house

Write an equation xx+=7 40,000

8x = 40,000 x = 5,000 Solve the equation. 7x = 7•5000 = 35,000

Solution : The lot costs $5,000. The house costs $35,000.

Check: 5,000 + 35,000 = 40,000 7•5000 = 35,000

81 SOLVING APPLICATION PROBLEMS

Problems:

1. Five plus three more than a number is nineteen. What is the number?

2. When 18 is subtracted from six times a certain number, the result is 96. What is the number?

3. If you double a number and then add 85, you get three-fourths of the original number. What is the original number?

4. A 180-m rope is cut into three pieces. The second piece is twice as long as the first. The third piece is three times as long as the second. How long is each piece of rope?

5. Donna and Melissa purchased rollerblades for a total of $107. Donna paid $17 more for her rollerblades than Melissa did. What did Melissa pay?

6. A student pays $278 for a calculator and a keyboard. If the calculator costs $64 less than the keyboard, how much did each cost?

82 SOLVING APPLICATION PROBLEMS

Answers

1. 5++=( x 3) 19 The number is 11.

x =11

2. 6x −= 18 96 The number is 19. x =19

3 3. 2xx+= 85 The number is –68. 4 x = −68

4. xx++2 3( 2 x) = 180 The lengths are 20 m, 40 m, and 120 m. x = 20

5. xx++17 = 107 Melissa paid $45. x = 45

6. xx+−( 64) = 278 The keyboard costs $171, and the calculator costs $107. x =171

83 SOLVING APPLICATION PROBLEMS

Problems

Solve the following word problems using one variable or two variables in a system.

Integer Problems

1. The sum of three consecutive integers is 144. Find the integers.

2. The sum of three consecutive even integers is 84. Find the integers.

3. The sum of three consecutive odd integers is 111. Find the integers.

Perimeter Problems

4. The length of a rectangle is seven more than the width. The perimeter is 34 inches. Find the length and width.

5. The length of a rectangle is 3 inches less than twice the width. The perimeter is 18 inches. Find the length and width.

6. The length of one side of a triangle is 4 inches more than the shortest side. The longest side is two inches more than twice the length of the shortest side. The perimeter is 26 inches. Find the length of all three sides.

84 SOLVING APPLICATION PROBLEMS Mixture Problems

7. For Valentine’s Day, Candy, a candy store owner, wants a mixture of candy hearts and foil- wrapped chocolates. If she has 10 pounds of candy hearts, which sell for $2 per pound, how many pounds of the chocolates, which sell for $6 per pound, should be mixed to get a mixture selling at $5 per pound?

8. The same candy store owner has 24 pounds of chocolate creams which sell for $12 per pound. How many pounds of chocolate caramel nut clusters, which sell for $9 per pound, should he mix to get a mixture selling at $10 per pound?

9. Sharon wants to make 100 pounds of holiday mix for her baskets. She purchases cashews at $8.75 per pound and walnuts at $3.75 per pound. She feels she can afford a mixture which costs $6.35 per pound. How much of each type of nut should she purchase to make the mix?

Distance Problems

10. Two cars leave Chicago at 11 a.m. headed in opposite directions. At 2 p.m., the two cars are 375 miles apart. If one car is traveling 5 mph faster than the other, what are their speeds?

11. A plane leaves Chicago headed due west at 10 a.m. At 11 a.m., another plane leaves Chicago headed due east. At 1 p.m., the two planes are 2950 miles apart. If the first plane is flying 100 mph slower than the second plane, find their rates.

12. A car leaves Milwaukee at noon headed north. Five hours later it arrives at its destination. A second car traveling south at a rate 10 mph slower leaves Milwaukee at 3:00 and arrives at its destination two hours later. When they arrive at their destination, they are 400 miles apart. How fast is each of the cars traveling?

13. A bicyclist can ride 24 miles with the wind in 2 hours. Against the wind, the return trip takes him 3 hours. Find the speed of the wind. (Hint: Solve using systems of equations.)

85 SOLVING APPLICATION PROBLEMS

Investment Problems

14. A couple wants to invest $12,000 in two retirement accounts, one earning 6% and the other 9%. How much should be invested in each account for them to earn an annual interest of $945?

15. An investor has put $4,500 in a credit union account earning 4% annual interest. How much should he invest in an account which pays 10% annual interest to receive total annual interest of $1,000 from the two accounts.

16. Three accounts generate a total annual interest of $1,249.50. The investor deposited an equal amount of money in each account. The accounts paid an annual rate of return of 7%, 8%, and 10.5%. How much was invested in each account?

Number-Value Problems

17. The admission prices for a movie theater in Crystal Lake are $9 for adults, $8 for seniors, and $5 for children. A family purchased twice as many children’s tickets as adults and the same number of senior tickets as adults. The total cost of the tickets was $54. How many of each type of ticket was purchased?

18. Marie has $2.20 worth of quarters, dimes, and nickels. She has 3 times as many nickels as quarters and 3 fewer dimes than quarters. How many of each type of coin does she have?

19. Deb went shopping at the school bookstore and purchased $53 of computer items. The CD’s cost $2 each, the DVD’s cost $3 each, and the flash drives were $15 each. The total cost was $53. He purchased one more DVD than CD’s and half as many flash drives as CD’s. How many of each did he purchase?

86 SOLVING APPLICATION PROBLEMS

Answers

1. The 3 consecutive integers are 47, 48, and 49.

2. The 3 consecutive even integers are 26, 28, and 30.

3. The 3 consecutive odd integers are 35, 37, and 39.

4. The length is 12 inches, and the width is 5 inches.

5. The length is 5 inches, and the width is 4 inches.

6. The lengths are 5inches, 9 inches, and 12 inches.

7. The store owner should add 30 pounds of the chocolates worth $6 per pound.

8. The store owner should mix 48 pounds of chocolate caramel nut clusters worth $9 per pound.

9. 52 pounds of cashews and 48 pounds of walnuts.

10. One car is traveling 60 mph; the other is traveling 65 mph.

11. The first plane is traveling 550 mph. The second plane is traveling 650 mph.

12. One car is traveling 60 mph; the other is traveling 50 mph.

13. The wind speed is 2 miles per hour.

14. $4,500 invested at 6% and $7,500 at 9%.

15. $8,200 at 10%.

16. $4,900 in each account.

17. 2 adult tickets, 4 children’s, and 2 senior tickets.

18. 5 quarters, 2 dimes, and 15 nickels.

19. 4 CD’s, 5 DVD’s, and 2 flash drives.

87 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Problems: 1. Evaluate the following:

a. −+32 45 22 −

20 b. (45− )

−−−37( ) c. 232 −

d. 1221− −−+( 82)

2. If x = 2 and y = -4, evaluate:

a. x− xy

xy22− b. 3xy+

3. Solve the following equations for x:

a. 3( xx− 5) += 22

x − 5 b. −=57 3

21 c. xx+=1 + 53

−5 d. x =15 8

88 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

4. Solve the following for the indicated variable:

a. Solve: P=22 l + w for w . 1 b. Solve: A= bh for h. 2 c. Solve: 32x+= y 5 for y . ab c d. Solve: += for a. 234

5. Solve the following inequalities. Leave your answers in interval form.

a. 32−( x + 4) <− 2( xx + 3) +

b. −5x +≥ 46

6. Simplify the following:

2 a. (−3xy22)

−4 b. (2y)

2 c. ( xx34)

d. (ab−−34 c)( ab 4 c 2)

ab40 e. a−3

3 4t−−34 tt 5 f. − 3tt26 

89 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

7. Perform the indicated operations:

a. (432342cc22+−+) ( cc ++)

2 b. 32xx( + 3)

c. (233t+− s)( ts)

d. (10xx2 + 11 +÷ 3) ( 5 x + 3)

e. (58xy−) −−( 25 xy + )

f. (−+3x y)( x22 − 8 xy + 16 y )

g. (2xx−÷ 32) 16

h. ( xyxy+)( −+) xxy( +)

8. Graph the following:

a. 4xy−= 3 12

b. x = 4

c. 38xy−= 0

d. y = −2

9. Find the x- and y-intercept of the following:

a. 32xy−= 5

b. 23xy−= 0

c. x = 4

d. y = −2

90 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

10. Find the equation of the following lines:

a. A line passing through (-2,4) and (6, 10)

2 b. A line with a slope of passing through (0, 5) 3 1 c. A line through (1, 2) with a slope of 2 d. A horizontal line through (-3, 5)

e. A vertical line through (-3, 5)

f. A line with an undefined slope through (3, -2)

11. Find the slope of the following:

a. A line through (-3, 7) and (2, -5)

b. The line 38xy−= 4

c. The line y = 7

d. The line x = -4

12. Solve the following systems:

xy+=−42 a.  yx=−−5

32xy−= b.  28xy+=

35160xy−−=  c.  x 51  −=y 26 3

xy=−+36 d.  2xy+= 6 12

xy−=4 e.  −=−2yx 42

91 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

13. Solve the following inequalities. Leave your answer in interval notation.

a. 3x −< 24 and −<33x

b. −≤22x −< 56

c. −2x +> 48 or x −≥12

d. 3x −< 24

e. 2x +≥ 74

f. −23x −> 1 6

g. 2x − 432 −<

14. Factor.

a. 14xyz− 16 x22 y z

b. −−rr2 14 − 45

c. 6sss543−− 26 20

d. 91z2 −

15. The length of a rectangle is one foot longer than twice the width. The perimeter is 20 inches. Find the length and width.

16. The longest side of a triangle is two centimeters longer than twice the shortest side. The medium length side is three centimeters longer than the shortest side. The perimeter is 25 centimeters. Find the lengths of the three sides.

17. One train leaves Chicago at 7 p.m. At 9 p.m., another train leaves Chicago heading in the opposite direction, traveling 10 mph faster. At 11 p.m., the two trains are 530 miles apart. How fast was each train moving?

18. Scotty inherited $20,000 which he invested in two different accounts. The first account paid out an interest rate of 6%. The second account paid out an interest rate of 3%. After 1 year, Scotty had earned $750 interest. How much did he invest in each account

92 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Answers

1. a. 0

b. 1

c. 4

d. –2

2. a. 10

b. –6

3. a. 13

b. 41

10 c. 9

d. –24

P Pl− 2 4. a. wl= − or w = 2 2

2A b. h = b

35 c. yx=−+ 22

12 34cb− d. a= cb − or a = 23 6

5. a. (5, ∞)

2 b.  −∞, − 5

93 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Answers (continued)

6. a. 9xy44

1 b. 16y4

c. x14

d. a22 bc

e. a7

64 f. 27

7. a. 77cc2 +

b. 12xxx32++ 36 27

c. 673t22+− st s

d. 2x + 1

e. 7x – 13y

f. −+3x32 25 x y − 56 xy 23 + 16 y

12 g. − 8 x

h. 2x22+− xy y

94 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Answers (continued)

8. a. 4x – 3y = 12

y

x 3

−4

8. b. x = 4

y

5

x 5

95 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Answers (continued)

8. c. 3x -8y = 0

y

4

x 2 7

8. d. y = −2

y

2

x 2

96 COMPREHENSIVE REVIEW y = -2 OF ELEMENTARY ALGEBRA

Answers (continued)

5 9. a. x-intercept = ,0 3

5 y-intercept = 0, − 2

b. x-intercept = (0, 0) y-intercept = (0, 0)

c. x-intercept = (4, 0) No y-intercept

d. No x-intercept y-intercept (0, -2)

3 11 10. a. yx= + 42

2 b. yx= + 5 3

13 c. yx= + 22

d. y = 5

e. x = -3

f. x = 3

12 11. a. m = − 5

3 b. m = 8

c. m = 0

d. Slope undefined

97 COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA

Answers (continued)

12. a. ( -6, 1 )

b. ( 2, 4 )

c. No solution

d. Infinite solutions

e. No solution

13. a. (-1,2)

3 11 b. ,  22 c. (−∞, − 2) ∪[ 3, ∞) 2 d. − ,2 3 11  3  e. −∞−∪∞,, 22  

f. no solution

19 g. − , 22

14. a. 2xyz( 78− xy)

b. −+(rr95)( +)

c. 232ss3 ( +−)( s 5)

d. (3zz+− 13)( 1)

15. The length is 7 feet and the width is 3 feet.

16. Short side is 5 cm, medium side is 8 cm, and the longest side is 12 cm.

17. The train that left at 7:00 p.m., travelled at 85 mph. The train that left at 9:00 p.m., travelled at 95 mph.

18. $15,000 at 3% and $5,000 at 6%.

98