Review of Elementar Algebra Content 0
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Table of Contents Fractions...1 Integers...5 Order of Operations...9 Eponents...11 Polnomials...18 Factoring... Solving Linear Equations...1 Solving Linear Inequalities... Graphing Linear Equations...8 Finding the Equation of a Line... Application of Linear Equations...8 Functions...50 Sstems of Linear Equations...5 Application of a Sstem of Linear Equations...6 Solving Linear Inequalities in Two Variables...6 Solving Sstems of Linear Inequalities...67 Solving Absolute Value Equations...7 Solving Absolute Value Inequalities...76 Vocabular Used in Application Problems...79 Solving Application Problems...81 Comprehensive Review of Elementar Algebra...88 i Revised Ma 01
FRACTIONS Lowest Terms Eample: 1 Simplif: 1 = 15 5 = 5 Addition/Subtraction 1 Rule: To add and subtract fractions ou need a Common Denominator. Eample: + + = 7 7 7 5 = 7 Eample: 1 1 + = + 8 = + 1 1 11 = 1 For this eample, we alread have the common denominator. This eample, we need to find a common denominator, or a denominator which is DIVISIBLE b and. One such denominator is 1. Eample: 7 1 7 1 = 1 1 7 = 1 1 = 1 1 = Note: An number over itself is equal to 1. When ou multipl a number b 1, ou are not changing the number, just renaming it. 9 = 1 = 1 = 1 9 1
FRACTIONS Multiplication/Division Rule: Multiplication and division of fractions do not require a common denominator. Eample: 7 1 7 1 = 8 5 8 5 7 = 5 1 = 10 Cancel a common factor of Multipl across Eample: 5 6 = 6 5 1 6 = 5 = 15 = 5 To divide b 5, multipl b the 6 reciprocal which is 6 5
FRACTIONS Problems Answers Perform the following operations. 1. 7 5 + 1 1 1. 1 + 5 1 15. 1 5 1. 5 5 + 6 1 5 5. 5 1 7 6 6. 7 1 1 1 1 7. + 11 8. 5 11 9. 16 1 5 1 15 10. 16 1 7 76 1
FRACTIONS Problems (continued) Answers Perform the following operations. 11. 5 6 + 9 1. 1 8 1. 7 6 5 5 1. 15 15 15. 5 9 10 16. 8 16 17. 5 6 1 5 9 18. 9 5 7 0
INTEGERS Definition: The absolute value of a number is the distance between 0 and the number on the number line. The smbol for The absolute value of a is a. The absolute value of a number can never be negative. Eamples: 1) = ) = Addition of Signed Numbers: Like signs: Add their absolute values and use the common sign. Eamples: 1) + ( ) = ( + ) = 5 ) + 8= ( + 8) = 1 Unlike signs: Subtract their absolute values and use the sign of the number with the larger absolute value. Eamples: 1) + =+ ( ) = 1 ) + ( 6) = ( 6 ) = 5
INTEGERS Subtraction of Signed Numbers 1) Change the subtraction smbol to addition. ) Change the sign of the number being subtracted. ) Add the numbers using like signs or unlike signs rules for addition. Eamples: 1) = + ( ) = 7 ) 6 ( ) = 6+ ( ) = 8 ) ( ) = + = Multiplication and Division of Two Signed Numbers: Like signs: The product or quotient of two numbers with like signs is positive. Eamples: 1) ( )( ) = 1 ) 8 = ) 16 = = Unlike signs: The product or quotient of two numbers with unlike signs is negative. Eamples: 1) ( )( ) = 1 ) 15 = 5 Division b 0 is undefined. Eample: 0 is undefined. 6
INTEGERS Addition and Subtraction of Signed Numbers Problems Answers Perform the following operations. 1. 5+ ( ). 6+ ( ) 8. 7 1 5. 11 15 5. 6 + + ( 1) 5 6. 8 ( 5) 1 7. ( 11) 8. 5 1 6 9. ( ) ( ) 9 + 1 + 1 11 10.. 8.6 1 11. + ( 8) 10 1. 9 + 10 5 1. 8 ( 1) ( 9 ) 1. 8+ ( ) + 7+ ( 6) 15. + ( 1 + 1) ( 1 9) 10 5 7
INTEGERS Multiplication and Division of Signed Numbers Problems Answers Perform the following operations. 1. ( )( ) 1. 6. ( 10)( 1) 10. 5. 0 10 8 9 0 5 1 6. ( 9.8) ( 7) 1. 7. 0 8 5 8. ( 5.1 )(.0).10 9. 0 8 ( ) 10. ( 9 1)( ) ( 6) 6 8
ORDER OF OPERATIONS Please Ecuse M Dear Aunt Sall Grouping Smbols Parentheses ( ) Brackets [ ] Square Roots Absolute Value Eponents Multipl Divide "as ou see it, left to right" Add Subtract "as ou see it, left to right" Eamples: Note: When one pair of grouping smbols is inside another pair, perform the operations within the innermost pair of grouping smbols first. 6 ( 5 ) + 6 7 ( ) 5 1 + 6 ( ) + 6 7 ( ) 6 9 6 7 + [ ] 5 6 1 + 5 1 + + 18 1 + 6 0 + 7 Eamples: Note: For a problem with a fraction bar, perform the operations in the numerator and denominator separatel. 9 7+ 7 + 7 8 + 7 15 = = = = 9 5 5 ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 6 5 1 1 1 1 1 1+ 1 = = = = = 9 9 9 16 7 16 11 9
ORDER OF OPERATIONS Problems Answers Perform the following operations. 1. 5+ 5 18. 1 6 + 1 7. + 6 1. 6 8 + 1 5. 1 ( + ) 19 5 5 6. ( ) 9 6 5 17 7. ( ) 8. ( ) + 1 1 5 + 5 9. ( ) 19 10. 1 7 1 10
EXPONENTS Laws of Eponents Definition: n a = a a a... a n times a = the base n = the eponent Properties: 1. Product rule: m n m n a a = a + Eample: + a a = a a a a a = a = a 5 times times. Quotient rule: a a m n m n = a Eample: a a 7 5 1 1 1 1 1 aaaaaaa = 1 1 1 1 1 = a = a aaaaa 7 5 n. Power rule: ( ) m a nm = Eample: ( a ) = a = a 6 a Raising a product to a power: ( ) n n n ab = a b Raising a quotient to a power: n a a = b b n n. Zero power rule: 0 a = 1 Eample: ( ) 0 ( ) 0 0 0 = 1; = = 1 1 = 1 Eamples: ( )( ) 1 6 6 5 1+ + 5 7 ab a b c = a b c = a b c 8 16 = = 81 1 ( ) ( )( ) ( )( ) + 5 = + 5 = 6 + 15 = 6+ 15 + 1 + 1 + 11
EXPONENTS Problems Answers Perform the following operations and simplif. 1. 5 8 ( )( ) 10 8. ( 5ab 6 )( 7ab 8 ) 9 9 5a b. 10 6 z z. 7 10ab ab 5ab 5. ( ab c ) 1 6 ab c 6. 5 ( ) z 10 6 16 z 7. ( ) mn 9 mn 8. 0 ( n ) 9 9. ( 5 ) ( ) 0 1 7 10. z 8 z 6 1
EXPONENTS Negative Eponents Definitions: a m m a 1 = 1 a m 1 a = a m a 1 a b m n m b a m n m Note: The act of moving a factor from the numerator to the denominator (or from the denominator to the numerator) changes a negative eponent to a positive eponent. Eamples: 10 10 1 1 = = 1 10 10, 000 ab 1 1 ab a b a a = = = 1 1 b b Since b has a negative eponent, move it to the denominator to become b 5 5 z = = z 5 z Onl the factors with negative eponents are moved. 1
EXPONENTS Eample: 1 1 = Within the parenthesis, move the factors with negative eponents. = 6 Simplif within the parenthesis. ( ) = = 6 ( ) ( ) 1 Use the power rule for eponents. 1 = Move all factors with negative eponents. ( ) 16 1 = Simplif. 1
EXPONENTS Problems Answers Perform the following operations and simpl. 1. ( ) 9. 9. 0 1 8 1 1 = 81. 5 1 5 5 = 6 10 5. ( a 0 b )( ab c ) 7 9ab c 6. ( ebc ) ( ab 1 ) 18ac e 11 b 7. ( ) 6 9 8. 6ab c ( a) 0 1 bc 18a c b 15
EXPONENTS Problems (continued) Answers 9. 8 8 10. 1 8 1 11. 10 8 a a a 1. 6 ( a aa ) 8 1. 1. b 5 b 15. a 6 a 1 a 1 7 a 1 1 8 b a 16. p 5 p 6 p 5 17. 1 6 18. 8 8 5 1 19. 0. ab ab 1 5 8 1 ab 1. 5 5 16
EXPONENTS Problems (continued) Answers 8. b c bc 1 bc 6 6. 9 5. a 6 a 9 a 6 5. ( ) 5 6. ( ) a b 0 a 15 b 7. 5 ( ) 1 a a b 6 8. 6 ( a b ) ( a b) 1 9 7 a b ab 9. 5 ( ) a b 0. 5 ( ) a b a b 7 1. 1 ( ) 6 5a ab 5. ( ab ) ( a b ) 5. a ( a ) ( a ) b 16a a b 19 17 10 5b 19 a a b 7a 16 17
POLYNOMIALS n Definition: A sum of a finite number of terms of the form: a n where a n is a real number and n is a non-negative integer. (No negative eponents, no fractional eponents.) Eamples: 8 1 5 + + is a polnomial + is not a polnomial Tpes of Polnomials: Monomial: A polnomial with 1 term Eample: Binomials: A polnomial with terms Eample: + Trinomial: A polnomial with terms Eample: + Degree: Degree of a term the sum of the eponents on the variables. Eample: has degree + = 7 Degree of a polnomial the largest degree of an of the terms. In a polnomial with one variable it is the largest eponent. Eample: has degree + has degree since 5 + 8 has degree + 1= 18
POLYNOMIALS Addition of Polnomials Adding Polnomials: Combine like terms. Eample: ( ) ( ) + 5 + 8+ = + 5+ 8+ = 5 Subtraction of Polnomials Subtracting Polnomials: Distribute the negative sign and combine like terms. Eample: ( ) ( ) + 5 8+ = + 5 + 8 = + 1 7 Multiplication of Polnomials Multipling Polnomials: Use the distributive propert. Eample: ( + )( + ) = ( + ) + ( + ) 9 1 6 8 = + + + 7 6 8 = + + Eample: ( + )( ) = ( )( ) + ( )( ) + ( ) + ( ) = + 6 8 9 1 = + 6 1 Eample: ( )( ) + The multiplication of the sum and difference of two terms. ( ) ( ) ( ) ( ) = + + + = + 9 1 1 16 9 0 16 = + = 9 16 The answer is the difference of two squares. 19
POLYNOMIALS Eample: ( + ) The square of a binomial. ( + )( + ) = ( ) + ( ) + ( ) + ( ) Eample: 16 1 1 9 = + + + = 16 + 1 + 9 ( ) 16 9 = + + The answer is a perfect square trinomial. ( a b) The cube of a binomial. ( a b)( a b)( a b) = ( a b)( a b) ( a b) = aa + a( b) b( a) b( b) ( a b) ( ) = + a ab ab b a b ( ) ( ) ( ) ( ) = a ab + b a b = + a a b ab a b b a b = a a b a b + ab + ab b = a a b + ab b Use the distributive propert. Division of Polnomials Dividing a polnomial b a monomial. Eample: 5 5 9 + 9 = + 1 = 8 + 0 1 1 = 8 + = 8 + 5 1 0
POLYNOMIALS Problems Answers Find the degree of the following and determine what tpe of polnomial is given. 1. + 8 Trinomial, degree 5. 5 Binomial, degree 5. Monomial, degree 5. 8 Monomial, degree 0 Add: 5. ( + 5+ ) + ( 6 8) + 9 7 Subtract: 6. ( + 5+ ) ( 6 8) + + 1 Multipl: 7. ( + )( 5+ ) 8. ( + )( ) 9. ( + )( ) 10. ( 5 + 7) 11. ( 9 ) 1. ( ) 7+ 6 6 5 6 16 9 5 + 70+ 9 81 6+ 7 108 + 1 6 Divide: 1. 15 1 10 8 1 8 + 18 1 5 8 + 10 6 6 1
FACTORING Factoring Out the Greatest Common Factor 1. Identif the TERMS of the polnomial.. Factor each term to its prime factors.. Look for common factors in all terms.. Factor out the common factor. 5. Check b multipling. Eample: Factor 6 8 + Terms: 6, 8, and + Factor into primes + Look for common factors Factor out the common factor Answer: ( + ) Check: ( ) ( ) ( ) + Check b multipling + 6 8 Factoring Trinomials of the Form: + b + c To factor + 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 + = 5 1 6 = 6 Since the numbers and work for both, we will use them. So, + 5 + 6= + + + 6 ( ) ( ) ( )( ) = + + + = + + Substitute + in for 5. Factor b grouping. Answer: ( + )( + ) Check b multipling: ( )( ) + + = + + + 6 = + + 5 6
FACTORING Factoring Trinomials of the Form: a + b + c Eample: Factor 6 + 1+ 1. Factor out the GCF 6 + 1+ + 7+ ( ) GCF is. Identif the values of a, b, and c for the epression + 7+ a= ; b= 7; and c =. Find the product of ac ac= = 6. List all possible factor pairs that equal ac and identif the pair whose sum is b. 5. Substitute boed values from Step in for the b term. 1 6 1+6=7 + = 5 + 7 + + 1+ 6 + 1 & 6 and & are the onl factor pairs. 1 & 6 is the pair that adds up to 7 and multiplies up to 6. Substitute + 1+ 6 for + 7 6. Group the two pairs + 1+ 6+ First step in factor b grouping. 7. Factor the GCF out of each pair. ( 1) ( 1) 8. Factor out ( + 1) ( 1)( ) + + + The resulting common + 1 factor is ( ) + + These are the two factors of + 7+ 9. Recall the original GCF from Step 1. Answer: ( + + ) = ( + )( + ) 7 1 ( + )( + ) 1 There are factors of 6 + 1+ 10. Check b multipling. ( )( ) + 1 + = 6 + 1+
FACTORING Eample: Factor Note: If a cis negative, the factor pairs have opposite signs. 1. Factor out the GCF. There is no GCF.. Identif the values of a, b, and c. a= ; b= ; c=. Find the product of ac. ( ) ac= = 1. List all possible factor pairs that equal ac and identif the pair whose sum is b. ( ) ( ) + 1 1 + 1+ 1 = 11 Include all possible factor combinations, including 1 ( + 1) 1+ ( + 1) =+ 11 the + / combinations since ac is negative. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) + 6 + + 6 = + 6 + + 6 =+ + + + = 1 + + + =+ 1 5. Substitute the boed values from Step in for the b term. 6 = + Substitute 6 in for 6. Group the two pairs. = + 6 First step in factor b grouping. 7. Factor the GCF out of each pair ( ) ( ) = + + The resulting common +. factor is ( ) 8. Factor out ( + ) Answer: ( + )( ) ( + )( ) These are the factors. 9. Check: ( )( ) + =
FACTORING Factoring Perfect Square Trinomials Eample: Factor + 18+ 81 STEP 1. Determine if the trinomial is a Perfect Square. Is the first term a perfect square? NOTES Yes = ( ) Is the third term a perfect square? +81 Yes ( ) Is the second term twice the product of the square roots of the first and third terms? 9 9 = 9 = 81 + 18 Yes ( 9) = 18. To factor a Perfect Square Trinomial: Find the square root of the first term = Identif the sign of the second term + Find the square root of the third term 9 81 = 9 Write the factored form + 18+ 81 = + 9 + 9 = + 9 Answer: ( 9) + Check our work ( )( ) ( )( ) ( ) + 9 + 9 = + 18+ 81 Eample: Factor 5 + 0 + STEP 1. Is the trinomial a Perfect Square Trinomial? Is the first term a perfect square? Is the third term a perfect square? Is the second term twice the product of the square roots of the first and third terms?. Factor the trinomial Square root of the first term 5 = 5 Sign of the second term + 0 + Square root of the third term = Factored form Answer: NOTES 5 Yes ( ) 5 = 5 Yes ( ) = 0 Yes ( ) 5 = 0 5 + 0+ = ( 5+ )( 5+ ) = ( 5+ ) ( 5+ ) Check our work ( )( ) 5+ 5+ = 5 + 0+ 5
FACTORING Factoring the Difference of Two Squares Eample: Factor 6 STEP NOTES 1. Are there an common factors? No. Determine if the binomial is the Difference of Two Squares Is the first term a perfect square? Yes = Is the second term a perfect square? 6 Yes 6 6 = 6 Is this the difference of the two terms? 6 Yes, means the difference of the two terms.. To factor the Difference of Two Squares: Find the square root of the first term. Find the square root of the second term. 6 6 = 6 = Write the factored form. Answer: 6 = ( + 6)( 6) ( + 6)( 6) Check our work. ( )( ) + 6 6 = 6 The factors are the product of the sum (+) and difference ( ) of the terms square roots. Eample: Factor 6a 11b STEP NOTES 1. Are there an common factors? No. Is the binomial a Difference of Two Squares? Is the first term a perfect square? 6a Yes 6a 6a= 6a Is the second term a perfect square? 11b Yes 11b 11b= 11b Is this the difference of the two terms? 6a 11b Yes, means the difference of the two terms.. To factor the Difference of Two Squares: Find the square root of the first term. 6a 6a = 6a Find the square root of the second term. 11b 11b = 11b Write the factored form. ( )( ) 6 11 = 6 + 11 6 11 a b a b a b Answer: ( 6a+ 11b)( 6a 11b ) Check our work. ( )( ) 6a+ 11b 6a 11b = 6a 11b The factors are the product of the sum (+) and difference ( ) of the terms square roots. 6
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 coefficient is positive.. How man terms does the polnomial have? If it has two terms, look for the following problem tpe: a. The difference of two squares If it has three terms, look for the following problem tpes: 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, tr to factor b grouping.. Can an factors be factored further? If so, factor them completel.. Does the factorization check? Check b multipling. 7
FACTORING STRATEGY Alwas check for Greatest Common Factor First Then continue: Terms Terms Terms Difference of Two Squares ( ) ( ) + ( )( ) Perfect Square Trinomials + 0+ 5 ( ) + ( )( ) + ( + 5)( + 5) ( + 5) 5 5 Form: + b + c 6+ 5 List factors of 5 51 5 1 Choose the pair which adds to 6 5+ 1= 6 ( 5)( 1) 8 Form: a + b+ c 5 7 6 5 6 = 0 List factors of 0 1 0 1 0 15 15 10 10 56 5 6 Choose the pair which adds to 7 + 10= 7 Substitute this pair in for the middle term. 5 + 10 6 Factor b Grouping. ( 5 + ) ( 5 + ) ( 5+ )( ) Factor b Grouping 7 + 1 6 1 ( + ) ( + ) 7 6 ( + )( 7 6)
FACTORING Problems Answers Factor completel. 1. + 6+ 8 ( + )( + ). u + 15u+ 56 ( u+ 8)( u+ 7). + 8 0 ( + 10)( ). 6 0 ( 10)( + ) 5. m 15m+ 5 ( m 9)( m 6) 6. 5 1 + ( + 7)( ) 7. a 16ab 8b + + ( a+ 1b)( a+ b) 8. 18 ( 6)( + ) 9. 1 ( 7+ )( 1) 10. 19 0 + ( 5)( ) 9
FACTORING Problems (continued) Answers 11. 16 96 ( 1)( + ) 1. 5 5 108 5 ( )( + 1) 5 1. 6 + + 1 ( + + ) 6 1. + 18( + )( 1) 6 18 5 15. 1 ( 1)( + 1) 16. 16 81 ( + 9)( + )( ) 17. 7 8 ( + )( ) 18. 6 5 + + ( + 5)( + ) 19. 18+ 81 ( 9) 0. 9 + + 16 ( + ) 1. 0 + 60+ 5 5 ( + ) 0
SOLVING LINEAR EQUATIONS To Solve a Linear Equation 1. Remove grouping smbols b using the distributive propert.. Combine like terms to simplif each side.. Clear fractions b multipling both sides of the equation b the Least Common Denominator.. Move the variable terms to one side and the constants to the other side. Do this b adding or subtracting terms. 5. Solve for the variable b multipling b the inverse or dividing b the coefficient of. 6. Check b substituting the result into the original equation. + 8 = + Eample Solve ( ) ( ) Solution: 1 8= + 0 = + = = = 8 Use the distributive propert. Combine terms Move variable to one side, numbers to the other Divide to solve for the variable. Check: ( ) ( ( ) + ) = ( ) + ( 1) ( 16 + 8) = + 6 ( 8) = 8 8 8 8 8 8 = 8 Eample: Solve 1 = Solution: Check: 1 ( ) ( ) = ( ) 1 1 1 8 6= 9 8 15 = 8 8 = 15 8 15 1 = 8 = 1 8 5 = = 1 5 15 1 1 Clear fractions b multipling b the LCD
SOLVING LINEAR EQUATIONS Special Cases: 1. When the variable terms drop out and the result is a true statement, (i.e., = or 0= 0), there are an infinite number of solutions. The equation is called an identit. Eample: Solve ( ) ( ) + 9 = + 1 1 + 9 = + 1 1 + 9= 9+ 1 + 9= 9+ 0= 0 Infinite number of solutions Solution: All real numbers. When the variable terms drop out and the result is a false statement (i.e., 10 = or 8 = 0), there is no solution. The equation is called a contradiction. Eample: Solve ( ) 5= + 5= 6+ 5= 5= Solution: No solution
SOLVING LINEAR EQUATIONS Problems Answers Solve. 1. ( 5) ( 5) + = 17 =. ( 1 )( 5) ( 1 )( ) = + = 19. + = + 6 + No solution. + = + 6 Infinite number of solutions: All real numbers 5. 5 ( ) + + = + = 0
SOLVING LINEAR INEQUALITIES Solving Linear Inequalities Linear inequalities are solved almost eactl like linear equations. There is onl one eception: if it is necessar to divide or multipl b a NEGATIVE number, the inequalit sign must be reversed. The solution can be written as a statement of inequalit or in interval notation. It can also be shown as a graph. In interval notation and graphing: Use a bracket, [ ], or closed circle, if the endpoint is included in the solution Use parenthesis, ( ), or open circle, if the endpoint is not included in the solution Eample: 1 < Interval Notation: (,1) Eample: ( ) or 0 1 0 1 5 Distribute + 6 5 11 Move variable term to one side, constants to the other 11 Divide b and reverse the inequalit because of division 11 Simplif Solution: 11 11 Interval Notation:, [ 0 1 0 1 11 5 or 11 5
SOLVING LINEAR INEQUALITIES Problems Answers Solve. 1. a ( a ) 7> + 5 10 a < or (, 10). ( ) ( ) 5 + 6 + 15 or [, ). 5 ( ) + + < + 0 > or ( 0, ). 10 + 8 6 + 10 1 or (,1] 5. ( ) ( ) 5+ 6 + 1 or [, ) 6. ( ) ( ) 5 + < 6 + 11 < or (, 11) 5
Compound Inequalities SOLVING LINEAR INEQUALITIES Compound inequalities are solved the same wa as simple inequalities. An operation (addition, subtraction, multiplication, division) must be performed on all three pieces of the inequalit. Never remove the variable from the middle piece of the inequalit. Alwas remember that when dividing or multipling b a NEGATIVE number, the inequalit sign must be reversed. Eample: Solve 8 < + + + 6 < 6 6 6 < < Add to all three parts of the inequalit Divide all three parts of the inequalit b Simplif In interval notation, this is (, ], because is between and, including the endpoint, but not including the endpoint. Solution: < Interval Notation: (,] ( - - -1 0 1 or - - -1 0 1 Eample: Solve 5 < < 1 5 1 > > 5 > > 5 < < 5, When dividing an inequalit b a negative number, reverse the inequalit sign. Keep the variable in the middle piece of the inequalit. Arrange the inequalit so that the lesser value is on the left and the greater value on the right, and change the inequalit smbols to preserve the relationship. Solution: 5 5 < < Interval Notation:, ( ) -5-0 or -5-0 6
SOLVING LINEAR INEQUALITIES Problems Answers Solve 1. < + 1< 1 1 < < or ( 1,1). 5 + 5 5 5 0 or [ 5 ],0. < + 8 < or (, ]. < 5 < 7 1 1 < < or ( 1,1 ) 5. 1 < 6 5 1 < or ( 5, 1] 6. < < 0 < < or ( ), 7. 0< 1 < 7 1 < < or ( 1, ) 8. 1 7 or 7 [, ] 9. 7< < 11 < < or 11 (, ) 10. < < 6 < < 0 or ( ),0 7
GRAPHING LINEAR EQUATIONS Standard Form: a + b = c A line is made up of an infinite number of points in the form (,). The coordinates for each of these points will satisf the equation for the line (make it true). Two special points on the line are its and intercepts. These are the points on the line where the line crosses the ais and the ais. The intercept is in the form (,0), and the intercept is in the form (0,). Find the intercept of the line b substituting 0 for, and solving for, (,0). Find the intercept of the line b substituting 0 for, and solving for, (0,). Eample: Graph the line using the intercept method = 1 Find the intercept: Substitute 0 for and solve for : (,6 ) = 1 ( ) 0 = 1 0= 1 = 1 1 = = The intercept is (,0). Find the intercept: Substitute 0 for and solve for : = 1 ( ) 0 = 1 0 = 1 = 1 1 = = The intercept is (0,). 0 To graph the line, graph these two points, and connect them. As a check, find one more point on the line. Let = and find = 6. The point (,6) is on the line and collinear with (,0 ) and ( 0, ). 8
GRAPHING LINEAR EQUATIONS Slope Intercept Form: = m + b If the equation of the line is in slope-intercept form, we use the slope m, and the intercept, (0,b), to graph the line. The slope of a line passing through points (, ) and (, ) vertical change rise change in 1 1 is 1 m= = = = if 1 horizontal change run change in 1 Eample: Graph the line: = +. The slope calculation is called a Rate of Change. The rate of change describes how one quantit (the numerator) changes with respect to another (the denominator). Identif the slope as a fraction: In the equation = m + b, m is the slope. In the equation = +, is the slope, so m =. 1 Identif the intercept as a point: In the equation, m b, ( 0, b) = + is the intercept. In the equation, = +, (0,) is the intercept Graph the intercept: (0,) is the first point to be graphed 8 Graph the slope: rise of m = = 1 run of 1 is positive; move up (rise) from the intercept, (0,), to the point (0,5). 6 (0,) (1,5) 0 6 8 1 is positive; then move 1 to the right (run) from (0,5) to (1,5). (1,5) is the second point of the line. Graph the line through the two points: Graph the line through the points (0,) and (1,5). 9
GRAPHING LINEAR EQUATIONS Special Cases: The graph of the linear equation the( 0, k ) Eample: = = k, where k is a real number, is the horizontal line going through The graph of the linear equation = k, where k is a real number, is the vertical line going through the point ( k,0). Eample: = 0
GRAPHING LINEAR EQUATIONS Problems Graph the following equations: 1. 5 = 5. + = 5. = +. = 5. 6 = 1 6. = 1
GRAPHING LINEAR EQUATIONS Answers 1. 5 = 5. + = 5. = +. = 5. 6 = 1 6. = 5
FINDING THE EQUATION OF A LINE The standard form of the equation of a line is written as a + b = c where a and b are not both 0. Definition of slope: Let L be a line passing through the points (, ) and (, ) of L is given b the formula: m = 1 1 1 1. Then the slope The slope-intercept form of the equation of a line is written as = m + b where m is the slope and the point ( 0,b) is the -intercept. Point-Slope Form: The equation of the straight line passing through ( 1, 1) is given b = m( ). 1 1 and having slope m Parallel Propert: Parallel lines have the same slope. Perpendicular Propert: When two lines are perpendicular, their slopes are negative (opposite) reciprocals of one another. The product of their slopes is 1.
FINDING THE EQUATION OF A LINE To find the equation of a line, use: Slope-Intercept form: m b m Slope-Intercept Form = +, slope m, intercept ( 0,b ) or Point-Slope form: = ( ), slope m, point (, ) 1 1 When given the slope and intercept: use the slope-intercept form = m + b 1 1 Eample: Find the equation of the line with slope and intercept ( 0, ) Use the slope-intercept form, = m + b, and fill in the values: Solution: The equation of the line is: = Make the substitutions of m = slope is and b= intercept is 0, ( ( )) Point-Slope Form When given the slope and a point on the line: use the point-slope form = m( ) Eample: Find the equation of the line with slope, through the point ( 6,1 ) Use the point-slope form, = m ( ) 1 1, and fill in the values. 1 1 1= ( 6) 6 1 = 1 1= 1+ 1= + 1 = Solution: The equation of the line is Make the substitutions of m = slope is and = 6 and = 1 point 6,1 ( ( )) 1 1. =
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 = m points to find the equation using the point-slope form ( ) 1 1 Eample: Find the equation of the line through the points (, ) and ( 1,1 ). 1 Slope m = = = 1 1 ( ) ( ) = m 1 1 = = = 1 Make the substitutions of m= slope is and = and = ( ) ( point (, )). 1 1 Solution: The equation of the line is = 1 5
FINDING THE EQUATION OF A LINE Problems Answers 1. Put the following in slope-intercept form. a. 8 = = 1 b. + = = + 1. Find the slope. a. (, ) and ( 7,9 ) 5 b. (,1) and (, ) 5. Find the equation of the line through ( ) and parallel to = + 1. 5, = + 1. Find the equation of the line through ( 1, ) and perpendicular to =. 1 1 = 6
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. 5. slope = 1 line passes through ( ) Standard Form, = 5 Slope-Intercept Form 1 5 = + 6. slope = 5 line passes through ( ) 6 0,0 5+ 6 = 0 5 = 6 7. slope = 1 -intercept = = 8. horizontal line through ( 1, ) = 9. slope is undefined and passing through ( 5,6) = 5 10. slope = -intercept 5 + = 15 15 = 11. horizontal line through ( 5, ) = 1. vertical line passing through ( 5, ) = 5 1. line passing through ( 1, ) and ( ) 1. line passing through (, ) and ( 5, 1) 5, = 5+ 8 = 17 1 = + 5 17 = + 8 8 15. line passing through (, ) and (, 7) = 16. line parallel to + = 6and passing through + = 5 = + 5 ( 1, ) 17. line passing through ( 5, ) and parallel to = = 0 18. line perpendicular to + 5 = and 5 = 9 passing through ( 1, 7 ) 5 9 = + 7
APPLICATION OF LINEAR EQUATIONS Eample: Determine Rate of Change A credit union offers a checking account with a service charge for each check written. The relationship between the monthl charge for each check and the number of checks written is graphed below. At what rate does the monthl charge for the checking account change? Also, find the unit cost, another wa to epress the rate of change. 0 Monthl Charge for checking account ($) 18 16 1 1 (50, 1) (75, 16) 10 8 10 0 0 0 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, 1) and (75, 16). If we let, = 50,1 and, = 75,16, we have ( ) ( ) ( ) ( ) 1 1 Rate of change ( 1) ( ) 1 ( ) ( ) dollars 16 1 dollars dollars = = = checks 75 50 checks 5checks The rate of change can be epressed as $ for ever 5 checks. Rate of change can also be epressed as $0.08 per check or 8 per check. The monthl cost of the checking account increases $ for ever 5 checks written. To find the cost of 1 check (unit cost), take the fraction which represents the rate of change, 5, and divide both the numerator and denominator b 5. dollars 5.08 dollars = = = $.08 / check. 5 checks 5 5 1 check The unit cost = $.08 per check. 8
APPLICATION OF LINEAR EQUATIONS Problems Answers The graph models the number of members in an organization from 001 to 010. Number of Members 00 150 100 50.(001,00).. (007,15) 001 005 Year 010 How man members did the organization have in 009? 100 members Find the rate of change. (slope of the line) loss of 5 members ever two ears What is the rate of change per ear? loss of 1.5 members per ear 9
FUNCTIONS Recall that relation is a set of ordered pairs and that a function is a special tpe of relation. A function is a set of ordered pairs (a relation) in which to each first component, there corresponds eactl one second component. The set of first components is called the domain of the function and the set of second components is called the range of the function. Eamples: Determine if the following are functions. Domain Range 0 8 1 16 This is a function, since for each first component, there is eactl one second component. Domain Range 1 7 5 This is not a function since for the first component, 7, there are two different second components, and 5. Vertical Line Tests: A graph in the plane represents a function if no vertical line intersects the graph at more than one point. Eamples: This is a graph of a function. It passes the vertical line test. Since we will often work with sets of ordered pairs of the form (,, ) it is helpful to define a function using the variables and. Given a relation in and, if to each value of in the domain there corresponds eactl one value of in the range, then is said to be a function of. (,) (, ) This is not a graph of a function. For eample, the -value of is assigned two different values, and. (for the first component,, there are two different second components, and.) 50
FUNCTIONS Notation: To denote that is a function of, we write = f ( ). The epression f ( ) f of. It does not mean f times. Since and f ( ) are equal, the can be used interchangeabl. This means we can write =, or we can write f ( ) =. is read Evaluate: To evaluate or calculate a function, replace the in the function rule b the given value from the domain and then compute according to the rule. For eample: Eamples: 1. ( ) Given: f = 6+ 5 ( ) ( ) Find: f = 6 + 5 = 17 f f ( ) ( ) 0= 60+ 5= 5 ( ) ( ) 1 = 6 1 + 5= 1. ( ) g = + Given: 5 8 ( ) ( ) ( ) Find: g 1 = 1 5 1 + 8 = 6 ( ) ( ) ( ) g 0 = 0 50+ 8= 8 ( ) ( ) ( ) g 1 = 1 5 1 + 8 = 16 51
FUNCTIONS Problems Answers 1. Determine whether or not each relation defines a function. If no, eplain wh not. a. Domain Range 11 1 8 1 10 a. Yes, for each first component, there corresponds eactl one second component. b. Domain Range b. No, for each first component,, there corresponds two different second components, and 7. 5 6 7 c. {( 1, ),(, ),( 1, 0) } d. {(, ),(, ),( 1,) }. Determine the domain and range of each of the following: c. No, for the first component, 1, there corresponds two different second components and 0. d. Yes, for each first component there corresponds eactl one second component. a. {(, ),(, ),( 0,0 ),(,7) } a. Domain: {,,0, } Range: { 0,,,7 } b. b. 5 7 9 6 c. c. 1 7 5 1 9 1 Domain:{ 5, 7, 9} Range: {,,6 } Domain: { 1,5,9 } Range: { 7,1,1 }. Given ( ) ( ), f = + and g = find the following: a. f ( ) a. 0 b. f ( ) b. c. g ( 0) c. d. g ( ) d. 1 5
SYSTEMS OF LINEAR EQUATIONS The following summar compares the graphing, elimination (addition), and substitution methods for solving linear sstems of equations. Method Eample Notes Graphing + = 6 You can see the solution is where the two = 8 lines intersect but if the solution does not involve integers it s impossible to tell eactl what the solution is. 5 +=6 The solution for this sstem of linear,. equations is ( ) -5----1-1 5 - (,-) -=8 - - Substitution = 1 = Substitute 1 for : ( 1) = Solve for : = Back-substitute: = ( ) 1= 5 Solution: (,5) Gives eact solutions. The solution for this sstem of equations is (,5 ). Elimination (Addition) + = 8 5+ = Multipl the top equation b 5 and the bottom equation b. This will result in opposite coefficients of. 10+ 15 = 0 10 8 = 68 Add the two equations: 7 = 8 = To find, substitute = into either original equation. + ( ) = 8 = 10 10, Solution: ( ) Gives eact solution. Eas to use if a variable is on one side b itself. The solution for this sstem 10,. of equations is ( ) 5
Three Possible Solution Tpes SYSTEMS OF LINEAR EQUATIONS 0 No Solution (Parallel Lines) Inconsistent Sstem Independent Equations # = different # (contradiction). 0 Eactl One Solution (point of intersection) Consistent Sstem Independent Equations Solution 0 Infinitel Man Solutions (Lines Coincide-an point on the line is a solution) Consistent Sstem Dependent Equations # = same # (identit) 5
SYSTEMS OF LINEAR EQUATIONS Solving Linear Sstems B Substitution Method To solve a linear sstem of two equations in two variables b substitution: 1. Solve one of the equations for one of the variables.. Substitute the epression obtained in step 1 for that variable in the other equation, and solve the resulting equation in one variable.. 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 sstem is inconsistent, the lines are parallel, the equations are independent, and there is no solution. b. if we get 0= 0 or # = same # (identit), the sstem is consistent, the lines are the same (coincide), the equations are dependent, and there are infinitel man solutions. Eample: Solve the sstem of linear equations using the substitution method. + 5 = = 8 This equation is solved for. ( ) 8 + 5 = 1+ 5 = 7 = 7 = 1 = Substitute 8 in for in the first equation. Solve for. = 8 ( ) = Substitute in for to find the value of. The solution is (,). Check in both original equations: + 5 = = 8 ( ) + 5 ( ) = = 8 1 + 15 = = 8 1 = = ( ) 55
Eample: SYSTEMS OF LINEAR EQUATIONS Solve the sstem of linear equations using the substitution method. + 1 = 5 + = 11 Since has a coefficient of 1, it will be eas to solve this second equation for. + 1 = = 5+ 11 ( ) + 1 5 + 11 = Substitute 5 + 11 in for + 60+ 1 = 6 = 18 = Solve for. ( ) = 5 + 11 = 1 Substitute in for to find the value of. The solution is (,1). Check in both original equations: + 1 = ( ) ( ) + 1 1 = 8 + 1 = = 5+ = 11 ( ) 5 + 1 = 11 10 + 1 = 11 11 = 11 56
SYSTEMS OF LINEAR EQUATIONS Solving Linear Sstems B Elimination (Addition) Method To solve linear sstems of two equations in two variables b elimination: 1. Write the sstem so that each equation is in standard form. a + b = c. Multipl one equation (or both equations if necessar), b a number to obtain additive inverse (opposite) coefficients of one of the variables.. Add the resulting equations and solve the new equation in one variable.. 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 sstem is inconsistent; there are no solutions, the equations are independent, and the lines are parallel. Eample: b. If we get 0 = 0 or # = same # (identit), the sstem is consistent; there are infinitel man solutions, the equations are dependent and the lines are the same (coincide). Solve the sstem of linear equations using the elimination (addition) method. + = 9 = = 6 = The coefficients of are opposites (additive inverses). Add these equations. Solve for. + = 9 + = 9 = 7 The solution is (,7). Substitute in for and solve for. Check in both original equations: + = 9 + 7= 9 9= 9 = ( ) 7= 7= = 57
Eample: SYSTEMS OF LINEAR EQUATIONS Solve this sstem of linear equations using the elimination (addition) method. + = 11 8 = Multipling the first equation b ields 8.This will result in opposite coefficients (additive inverses) of. 1+ 8 = 8 = 1 = = Add these equations. Solve for. ( ) + = 11 9 + = 11 = = 1 Substitute in for and solve for. The solution is (,1). Check in both original equations: + = 11 ( ) ( ) + 1 = 11 9 + = 11 11 = 11 8 = ( ) ( ) 81 = 6 8= = 58
SYSTEMS OF LINEAR EQUATIONS Problem Solve this Sstem of Linear Equations using all three methods: Graphing Method Substitution Method Elimination (Addition) Method + = 1 = 1 Graphing Method: 0 Substitution Method: Elimination (Addition) Method: Answer is on the following page. 59
SYSTEMS OF LINEAR EQUATIONS Answer: 0 Solution: 60
SYSTEMS OF LINEAR EQUATIONS Problems Solve these sstems of linear equations using either the substitution or elimination (addition) method. Answers 1. + = 11 = 6 and = 5 1 6,5 = ( ). + = 9 = and = 1 11,1 = ( ). + = 5 = and = 1 5 8,1 = ( ). + = 7 = 1 and = 1 8 7 1,1 = ( ) 5. + = 10 6 = 1 1 = and = 1 1, 6. + 7 = = 1 and = 11 1, = ( ) 7. + = = 1 and = 8 1, + = ( ) 8. 5+ = 5 = 1 and = 0 1, 0 + = ( ) 9. = 8 = 1 10. = 6 = 1 Inconsistent sstem with no solution. Dependent sstem with infinitel man solutions. 61
APPLICATION OF A SYSTEM OF LINEAR EQUATIONS The following steps are helpful when solving problems involving two unknown quantities. 1. Analze the problem b reading it carefull to understand the given facts. Often a diagram or table will help ou visualize the facts of the problem.. Define variables to represent the two unknown quantities.. Translate the words of the problem to form two equations involving each of the two variables.. Solve the sstem of equations using graphing, substitution, or elimination (addition) method. 5. State the conclusion. 6. Check the results in the words of the problem. Eample: Determine the cost of a quart of pineapple and a container of frozen raspberries for the punch Diane is making. Three () quarts of pineapple juice and containers of raspberries will cost $10. Five (5) quarts of pineapple juice and containers of raspberries will cost $1. Let: p = cost of a quart of pineapple juice Define the variables r = cost of a container of raspberries p+ r = 10 5 p+ r = 1 Solve the sstem for p using elimination (addition) method p+ r = 10 10 p r = 7 p = 1 p = ( ) + r = 10 r = 1 Substitute in for p and solve for r Answer: A quart of pineapples costs $ and a container of frozen raspberries costs $1. Check: Using $ as the cost for a quart of pineapple juice, quarts costs $6. Using $1 as the cost of a container of raspberries, containers cost $. $6 for pineapple juice and $ for raspberries is $10 total cost. Do the same technique for checking the $1 cost. Check the results in the words of the problem. 6
APPLICATION OF A SYSTEM OF LINEAR EQUATIONS Problem Answer Use the steps for solving a sstem of linear equations. People have begun purchasing tickets to a production of a musical at a regional theatre. A purchase of 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. A child s ticket costs $5 and an adult ticket costs $9. 6
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES Graphing Linear Inequalities In Two Variables 1. Replace the inequalit smbol with an equal smbol = and graph the boundar line of the region. If the original inequalit allows the possibilit of equalit (the smbol is either or ), draw the boundar line as a solid line. If equalit is not allowed (< or >), draw the boundar line as a dashed line.. Pick a test point that is on one side of the boundar line. (Use the origin if possible; it is easier). Replace and in the inequalit 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). Eample: Solve + Step 1: + = = + sign, draw the solid boundar line - 0 - Step : pick (0,0) ( ) 0+ 0 is true Pick test point. Shade the side that contains (0,0).. 0 ( 0,0) 6
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES Problems 1 1. Solve 1. Solve + < 6. Solve 5 65
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES Answers 1.. 5 5. 66
SOLVING SYSTEMS OF LINEAR INEQUALITIES 1. Graph each inequalit on the same rectangular coordinate sstem.. 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 sstem.. As an informal check, pick a point from the region where the graphs intersect and verit that its coordinates satisf each inequalit of the original sstem. Eample: Solve this sstem of inequalities. + > 9 Look at the graph of each inequalit separatel. + sign, draw solid boundar line. Use (0, 0) as a test point. 0 is false. Shade the half-plane that does not include (0, 0). > 9 > sign, draw dashed boundar line. Use (0, 0) as a test point. 0 > 9 is false. Shade the half-plane that does not include (0, 0). 67
SOLVING SYSTEMS OF LINEAR INEQUALITIES + > 9 0 Graph each inequalit on the same coordinate sstem. Notice the region where the graphs overlap. The points in this region are the solutions of the sstem. Solution of the Sstem: 0 68
SOLVING SYSTEMS OF LINEAR INEQUALITIES Informal Check: The point (5,) is in the region of overlap. Substitute (5,) in to both inequalities. + 5+ 7 is true > 9 5 ( ) ( ) 9 0 6 9 1 > 9 is true 0 (5,) (5,) is one of the solutions. 69
SOLVING SYSTEMS OF LINEAR INEQUALITIES Problems 1. Solve this sstem of linear inequalities. + 5 > + 1 6. Solve this sstem of linear inequalities. + 5 5 Answers on the net page 70
SOLVING SYSTEMS OF LINEAR INEQUALITIES Answers 1. 5 -. 7 71
SOLVING ABSOLUTE VALUE EQUATIONS Absolute Value: The absolute value of a number is its distance from 0 on the number line For an positive number k and an algebraic epression X: To solve X = k, solve the equivalent compound equation X = k or X = k Eample: = 7 Solution: =7 or =-7 7 0 7 Eample: = = or = Solution: =5 =-1 1 0 5 Check: = 5 = 5 = = = Check: = 1 = 1 = = = 7
SOLVING ABSOLUTE VALUE EQUATIONS Eample: Solve the following: 6 = Solve the equation b rewriting it as two separate equations. 6 6 = or = When X and X = k = k, then X = k Solve each equation for. 6 = 18 6 = 18 Multipl both sides b. Solutions: = 19 = 6 Add 6 to both sides. = 8 or = -16 Divide both sides b. Check: Check = 8 Check = 16 ( ) 8 6 = ( ) 16 6 = 19 6 = 6 6 = 18 = 18 = = = = = The two solutions check. 7
SOLVING ABSOLUTE VALUE EQUATIONS Eample: Solve the following: 5 7 = ( + 1) Solve the equation b rewriting it as two separate equations. ( ) ( ) 5 7= + 1 or 5 7= + 1 When X X = k, then = k and X = k Solve each equation for. Solutions: [ ] 5 7= + 5 7= + 7 = 5 7= = 11 9 7 = 9 = = 11 or 1 = Use the Distributive Propert. Check: Check = 11 Check 1 = ( ) = ( + ) 5 11 7 11 1 ( ) 55 7 = 1 8 = 8 8 = 8 1 1 5 7 = + 1 5 7 = 16 16 = 16 16 = The two solutions check. 7
SOLVING ABSOLUTE VALUE EQUATIONS Problems Answers 1. = 5 7 = or = 1. 10 = 0 No solution. You can t solve an absolute value equation when the absolute value is equal to a negative quantit.. 10 + + = 9 7 = or =. 1 5 = = 10 5. 5+ = + 5 7 = 11 or = 6. 1 ( 1) + = + = or = 5 75
SOLVING ABSOLUTE VALUE INEQUALITIES To solve inequalities with absolute value signs, there are two cases: For an positive number k and an algebraic epression X Case 1: To solve X < k, solve the equivalent compound inequalit k < X < k. To solve X k, solve the equivalent compound inequalit k X k. Eample: Solve: < 7 Solution: -7<<7 Interval Notation: (-7,7) -7 0 7 or -7 0 7 For an positive number k and an algebraic epression X Case : To solve X > k, solve the equivalent compound inequalit X < k or X > k. To solve X k, solve the equivalent compound inequalit X k or X k. Eample: Solve: 7 Solution: 7 or 7 Interval Notation:, 7] [7, -7 0 7 or -7 0 7 76
SOLVING ABSOLUTE VALUE INEQUALITIES Eample: Solve: Case 1 Equivalent compound inequalit 1 5 Addition Propert of Equalit Solution: 1 5 Interval Notation: [ 1,5] -1 0 5 or -1 0 5 Eample: Solve: > Case < or > Equivalent inequalities < 1 > 5 Interval Notation: (, 1) ( 5, ) Solution: < 1 or > 5 - -1 0 5 or - -1 0 5 77
SOLVING ABSOLUTE VALUE INEQUALITIES Problems Answers 1. 5 < < < 8,8 ( ). 7 > 10 < or > 17, 17, ( ) ( ). 5 1 19 19,. < 9 < < 6, 6 ( ) 5. 5 6 7 or (, 7], ) 6. 8 0 or,, [ ) 78
VOCABULARY USED IN APPLICATION PROBLEMS Addition add English Algebra sum The sum of a number and. + total Seven more than a number. + 7 plus Si increased b a number. 6 + in all A number added to 8. 8 + more than A number plus. + together increased b all together combined Subtraction subtracted from English Algebra difference The difference of a number and. take awa The difference of and a number. less than Five less than a number. 5 minus A number decreased b. remain A number subtracted from 8. 8 decreased b Eight subtracted from a number. 8 have left Two minus a number. are left more fewer Be careful with subtraction. The order is important. Three less than a number is. 79
VOCABULARY USED IN APPLICATION PROBLEMS Multiplication English Algebra product of The product of and a number. multiplied b Three-fourths of a number. times Four times a number. of A number multiplied b 6. 6 Double a number. Twice a number. Division English Algebra divided b quotient of separated into equal parts shared equall The quotient of a number and. The quotient of and a number. A number divided b 6. Si divided b a number. or or 6 or 6 or 6 6 Be careful with division. The order is important. A number divided b 6 is 6 80
SOLVING APPLICATION PROBLEMS Strateg for solving word problems: 1. Analze the problem: Read the problem carefull.. Visualize the facts of the problem (if needed): Use diagrams and/or tables.. Define the variable/s: Identif the unknown quantit (or quantities) and label them, i. e., let = something.. Write an equation: Use the defined variable/s. 5. Solve the equation: Make sure ou have answered the question that was asked. 6. Check our answer(s): Use the original words of the problem. Eamples: 1. The sum of three times a number and 11 is 1. Find the number. Let = the number + 11 = 1 Define the variable. Write an equation.. + 11 11 = 1 11 = = 8 Solve the equation. Solution: The number is 8. Check: ( ) 8 = + 11 = 1. Together, a lot and a house cost $0,000. The house costs seven times more than the lot. How much does the lot cost? The house? Let = the cost of the lot 7 = the cost of the house + 7= 0,000 8 = 0,000 = 5,000 7 = 7 5000 = 5,000 Solution: The lot costs $5,000. The house costs $5,000. Define the variable. Write an equation Solve the equation. Check: 5,000 + 5,000 = 0,000 7 5000 = 5,000 81
SOLVING APPLICATION PROBLEMS Problems: 1. Five plus three more than a number is nineteen. What is the number?. When 18 is subtracted from si times a certain number, the result is 96. What is the number?. If ou double a number and then add 85, ou get three-fourths of the original number. What is the original number?. 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 pa? 6. A student pas $78 for a calculator and a keboard. If the calculator costs $6 less than the keboard, how much did each cost? 8
SOLVING APPLICATION PROBLEMS Answers ( ) 1. 5 + + = 19 The number is 11. = 11. 6 18 = 96 = 19 The number is 19.. + 85 = = 68 The number is 68.. ( ) + + = 180 = 0 The lengths are 0 m, 0 m, and 10 m. 5. + + 17 = 107 = 5 Melissa paid $5. 6. ( ) + 6 = 78 = 171 The keboard costs $171, and the calculator costs $107. 8
SOLVING APPLICATION PROBLEMS Problems Solve the following word problems using one variable or two variables in a sstem. Integer Problems 1. The sum of three consecutive integers is 1. Find the integers.. The sum of three consecutive even integers is 8. Find the integers.. The sum of three consecutive odd integers is 111. Find the integers. Perimeter Problems. The length of a rectangle is seven more than the width. The perimeter is inches. Find the length and width. 5. The length of a rectangle is 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 inches more than the shortest side. The longest side is two inches more than twice the length of the shortest side. The perimeter is 6 inches. Find the length of all three sides. 8
SOLVING APPLICATION PROBLEMS Miture Problems 7. For Valentine s Da, Cand, a cand store owner, wants a miture of cand hearts and foilwrapped chocolates. If she has 10 pounds of cand hearts, which sell for $ per pound, how man pounds of the chocolates, which sell for $6 per pound, should be mied to get a miture selling at $5 per pound? 8. The same cand store owner has pounds of chocolate creams which sell for $1 per pound. How man pounds of chocolate caramel nut clusters, which sell for $9 per pound, should he mi to get a miture selling at $10 per pound? 9. Sharon wants to make 100 pounds of holida mi for her baskets. She purchases cashews at $8.75 per pound and walnuts at $.75 per pound. She feels she can afford a miture which costs $6.5 per pound. How much of each tpe of nut should she purchase to make the mi? Distance Problems 10. Two cars leave Chicago at 11 a.m. headed in opposite directions. At p.m., the two cars are 75 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 950 miles apart. If the first plane is fling 100 mph slower than the second plane, find their rates. 1. 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 :00 and arrives at its destination two hours later. When the arrive at their destination, the are 00 miles apart. How fast is each of the cars traveling? 1. A bicclist can ride miles with the wind in hours. Against the wind, the return trip takes him hours. Find the speed of the wind. (Hint: Solve using sstems of equations.) 85
SOLVING APPLICATION PROBLEMS Investment Problems 1. A couple wants to invest $1,000 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 $95? 15. An investor has put $,500 in a credit union account earning % annual interest. How much should he invest in an account which pas 10% annual interest to receive total annual interest of $1,000 from the two accounts. 16. Three accounts generate a total annual interest of $1,9.50. The investor deposited an equal amount of mone 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 Crstal Lake are $9 for adults, $8 for seniors, and $5 for children. A famil purchased twice as man children s tickets as adults and the same number of senior tickets as adults. The total cost of the tickets was $5. How man of each tpe of ticket was purchased? 18. Marie has $.0 worth of quarters, dimes, and nickels. She has times as man nickels as quarters and fewer dimes than quarters. How man of each tpe of coin does she have? 19. Deb went shopping at the school bookstore and purchased $5 of computer items. The CD s cost $ each, the DVD s cost $ each, and the flash drives were $15 each. The total cost was $5. He purchased one more DVD than CD s and half as man flash drives as CD s. How man of each did he purchase? 86