Page 1 of 14 Review of Rational Epressions and Equations A rational epression is an epression containing fractions where the numerator and/or denominator may contain algebraic terms 1 Simplify 6 14 Identification/Analysis Identify this as an epression Notice it has only one term, so simplify means be sure the numerator and denominator do not have any common factors In other words, we need to make sure the greatest common factor (GCF) is a 1 Solution 6 and 14 7 Factor numerator and denominator The GCF is 1 7 7 7 7 7 Identify GCF Factor the GCF out of both the numerator and denominator Reduce Epression simplified Check/Verification Since and 7 have a GCF of 1, we know the fraction is simplified
Page of 14 Simplify 5 7 6 + 0 Identification/Analysis Identify this as an epression Simplify means combine the two terms into a single term by adding fractions Recall to add fractions, either we have common denominators or we must write equivalent fractions that have a common denominator In this case the denominators are not the same, therefore factor each denominator into a product of prime numbers and determine the least common denominator (LCD) Solution 6 and 0 5 The LCD is 5 or 60 Factor denominators Find LCD 5 ( 10) 5 50 6 ( 10) 6 60 and 7 ( 7 ) 1 0 ( ) 0 60 Now 5 7 6 50 1 can be written as + 0 60 60 Write equivalent fractions with LCD 50 1 71 Add fractions together + 60 60 60 71 60 Epression simplified Check/Verification Check to see if the fraction can be simplified, that is, if there is a common factor in both the numerator and denominator Since the GCF of 71 and 60 is 1, the fraction is simplified so we are finished
Page of 14 5 1 Simplify + 1 8 Identification/Analysis This is an epression with two terms Simplify means combine the two terms into a single term by adding fractions In order to add fractions, you must have a common denominator Therefore my first step is to find the LCD Solution 1 and 8 Factor denominators The LCD is or 4 Find LCD 5 ( ) 5 10 1 ( 1 ) and 1 1 4 8 8 4 Write equivalent fractions with LCD ( ) 5 1 10 10+ + + 1 8 4 4 4 10 + 4 ( ) Since they have common denominators, add the numerators Epression simplified Check/Verification The numerator is simplified because the two terms are not like terms The fraction is simplified because the GCF of the numerator and denominator is 1
Page 4 of 14 + 4+ 7 4 Simplify + 10 Identification/Analysis This is an epression with two terms Simplify means combine the two terms into a single term by subtracting the fractions To subtract fractions, you must have a common denominator Therefore the first step is to find the LCD Notice the denominators are different than the previous eamples is a twoterm polynomial, called a binomial and + 10 is a three-term polynomial called a trinomial To factor these denominators, we must remember how to factor polynomials Solution Factor each denominator GCF of the two terms is 1 It is not a difference of squares Prime (will not factor further) + 10 GCF of the three terms is 1 1 10 and 5 are factors of 10 Also, + 5 Quadratic of form + b + c Find factors + 10 ( + 5)( ) that multiply to 10 and add to The most number of factors in either epression is one Find the LCD of and ( + 5)( ) The most number of + 5 factors in either epression is one Determine the most number of each factor Therefore the LCD is ( + 5)( ) in either denominator + ( + 5)( + ) + 8+ 15 Write equivalent fractions with the LCD ( + 5)( ) ( + 5)( ) 4+ 7 4+ 7 This fraction already has the LCD, so no + 10 ( + 5)( ) equivalent fraction is needed + 4+ 7 + 8+ 15 4+ 7 Write both fractions with a common + 10 + 5 + 5 denominator ( )( ) ( )( ) + 8+ 15 ( 4+ 7) ( + 5)( ) + 4+ 8 ( + 5)( ) Subtract Be careful with the minus sign! Combine like terms
Page 5 of 14 Check/Verification Finally, we need to ensure the fraction + 4+ 8 + 5 ( )( ) is simplified That means make sure the GCF of the numerator and denominator is 1 Therefore, try to factor the numerator Since there is no pair of numbers that both multiply to 8 and add to 4, the numerator does not factor Therefore the GCF is 1, and we are finished Review of Rational Equations: Equations are fundamentally different than epressions both in appearance and in purpose First, equations include an equal sign Second, an equation is a statement about the equality of two epressions Typically your job is to determine the value or values for the variable which make the equation true By definition, rational equations include fractions, often with variables in the denominator It is important to remember that division by zero is not defined Therefore one of the first things to do is consider what values would make the denominator zero, and thus make the fraction undefined The values that make the denominator to equal zero are not in the domain The general strategy for solving a rational equation is to find the LCM of all the denominators and multiply the entire equation by this value This will eliminate all the denominators and leave you with an equation that will be easier to solve Frequently the resulting equation is either linear or quadratic, though other possibilities eist The following eamples will illustrate this general strategy
Page 6 of 14 1 1 Solve the equation + 8 Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction Clear the fractions by multiplying both sides of the equation by the LCD Solution There are no variables in the denominator, so the domain Find the domain includes all real numbers is prime and 8 The LCD is or 8 Find the LCD Factor all the denominators Solve 1 Multiply both sides of the equation by the 8 + 8 8 LCD to eliminate (reduce) the fractions The resulting equation is linear 1 8 + 8 8( ) Solve for 8 4 + 4 0 Check/Verification 1 0 Substitute 0 into the original equation + 8 and simplify the results 1 5 + The statement is true so 0 is a solution
Page 7 of 14 Solve 5 1 1 + 8 Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction, so clear the fractions by multiplying both sides of the equation by the LCD Since the denominator contains a variable, determine the domain restrictions Solution 1 0 if and only if 0 Find the domain restrictions 8 0 if and only if 0 Set denominators with variables equal to 0 if and only if 0 zero Therefore 0 The domain does not include zero Zero cannot be a solution to the equation 1 Factor each denominator 8 LCD is 4 Find the LCD Build the LCD using the 5 1 4 + 4 1 8 5 1 4 + 4 4 1 8 5 + 1 4 ( )( ) ( )( ) ( )( ) 10 + 7 6 6 factors above Multiply both sides of the equation by LCD to eliminate (reduce) the fractions Distribute the LCD to both terms on the left side Simplify the fractions This is a linear equation Solve for The solution is in the domain Check/Verification Substitute into the original equation and simplify the results to verify it is a solution 6
Page 8 of 14 Solve 5 1 1 + 8 1 Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction, so clear the fractions by multiplying both sides of the equation by the LCD Since the denominator contains a variable, determine the domain restrictions Solution 1 0 if and only if 0 Find the domain 8 0 if and only if 0 Set denominators with variables equal to has no variable factor, so the denominator cannot equal 0 zero Therefore 0 The domain does not include zero Zero Solve for cannot be a solution to the equation State the restrictions for the domain 1 Factor each denominator 8 1 8 LCD is 4 5 1 1 4 + 4 1 8 5 1 1 4 + 4 4 1 8 5 + 1 8 1 ( )( ) ( )( ) ( )( ) 10 + 8 8 10 0 4+ 1 0 4 + 1 0 or 0 1 or 4 ( )( ) Find the LCD Build the LCD using the factors above Multiply both sides of the equation by LCD to eliminate (reduce) the fractions Distribute the LCD to both terms on the left side Simplify the fractions This is a nd degree (quadratic) equation Write the equation in standard form If possible, factor the trinomial Set each factor equal to zero Solve for The solutions are in the domain
Page 9 of 14 Check/Verification Substitute 1 and 4 into the original equation and simplify the results to verify both are solutions + 4 + 7 4 Solve 1 + 10 Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction, so clear the fractions by multiplying both sides of the equation by the LCD Since the denominator contains a variable, determine the domain restrictions Solution + 10 0 Find the domain 0 Since there is a polynomial in the ( + 5)( ) 0 and denominator factor each denominator + 5 0 or 0 Set each factor equal to zero 5 or Solve for Therefore or 5 The domain does not include either State the restrictions to the domain value They cannot be a solution to the equation The most number of factors in either denominator is one, so the LCD must have one The most number of + 5 factors in either denominator is one, so the LCD must have one + 5 + 5 The LCD is ( )( ) + 4+ 7 ( + 5)( ) ( + 5)( ) 1 + 10 + 4+ 7 + 5 + 5 5 + + 10 + 5 + 4+ 7 + 5 ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) + 8 + 15 4 7 + 10 + 4 + 8 + 10 18 Find the LCD Build the LCD using the factors above Multiply both sides of the equation by LCD to eliminate (reduce) the fractions Distribute the LCD to both terms on the left side Simplify the fractions and distribute Collect like terms The equation is linear Solve for The solution is in the domain Check/Verification Substitute 18 into the original equation and simplify the results to verify it is a solution
Page 10 of 14 5 Solve + 1 1 + + 5 + 6 + 1 + Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction, so clear the fractions by multiplying both sides of the equation by the LCD Since the denominator contains a variable, determine the domain restrictions Solution + 5+ 6 ( + )( + ) Find the domain Since there are polynomials in the + and + are prime denominator,factor each denominator There are only two different factors Set each factor equal to zero + 0 or + 0 Solve for or State the restrictions for the domain Thus or The domain does not include either value They cannot be a solution to the equation In factored form, the three denominators are ( + ), ( + ), Therefore the LCD is + + and ( + )( + ) ( )( ) + 1 1 1 + 5+ 6 + + + 1 1 + + + + + + 5+ 6 + 1 ( + )( + ) + + 1 + + 1 + 1 ( + )( + ) + ( + )( + ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) + 1+ + + + 0 ( + )( 1) 0 + 1 or 1 ( ) 0 or ( ) 0 Find the LCD Build the LCD using the factors above Multiply both sides of the equation by LCD to eliminate (reduce) the fractions Distribute the LCD to both terms on the left side Simplify the fractions and distribute Collect like terms This is a nd degree (quadratic) equation Write the equation in standard form If possible, factor the trinomial Set each factor equal to zero Solve for The solution is not in the domain while 1 is in the domain
Page 11 of 14 Check/Verification is not a solution to the equation, since is not in the domain When is substituted into the original equation the denominator is zero The solution 1 is in the domain Substitute 1 into the original equation and simplify the results to verify it is a solution 6 Solve + 5 + + 5 4 + 5 + 6 + Identification/Analysis This is an equation (contains an equal sign) Determine the values for that make the statement true This implies you need to isolate There is a fraction, so clear the fractions by multiplying both sides of the equation by the LCD Since the denominator contains a variable, determine the domain restrictions Solution + 5 ( 1)( + ) Find the domain Since there are polynomials in the + 5+ 6 ( + )( + ) denominator, factor each denominator + ( 1)( + ) There are three different factors Set each different factor equal to zero 1 0 or + 0 or + 0 1 Solve for or or 1 Thus, or The domain does not include State the restrictions for the domain these values They cannot be a solution to the equation 1 + + Find the LCD The LCD is ( )( )( ) ( 1)( + )( + ) + + 5 4 + 5 + 5+ 6 ( 1)( + )( + ) + Build the LCD using the factors above Multiply both sides of the equation by LCD to eliminate (reduce) the fractions
Page 1 of 14 Solution 1 + + 5 4 Distribute the LCD to both terms on + + ( 1)( + )( + ) + 5 + 5+ 6 the left side ( 1)( + )( + ) + + + 5 + 1 4 + Simplify the fractions and distribute continued ( )( )( ) ( )( ) ( )( ) ( )( ) + 5 + 4 + 10 + 8 4 + + 14 + 6 0 Find factors that multiply to 6 and add to 14 There is no pair of numbers that does both, so the quadratic does not factor The quadratic formula states the solutions for the equation b± b 4ac a + b + c 0 in the above equation a a 1, b 14, and c 6 thus ( 14) ( )( ) 1 () 14 ± 4 1 6 14 ± 196 4 14 ± 17 14 ± 4 7 ± 4 Collect like terms This is a nd degree (quadratic) equation Write the equation in standard form If possible, factor the trinomial Use the quadratic formula The solutions are in the domain Check/Verification Substitute 7+ 4 and 7 4 into the original equation and simplify the results to verify that each is a solution
Page 1 of 14 Function Notation: Function notation replaces the dependent variable y with f ( ) where is the independent variable Eample: Suppose we have the equation y + 7 and want to know what the value of y is when we replace with Using function notation, the equation would be stated as f 7 f 1 Given g( ) Of course the answer is 1, so we would respond f ( ) 1 + 5, find g ( 4) ( ) + and the question would be stated as: Find ( ) Identification/Analysis In this case we have an epression to evaluate That means substitute 4 in for and then simplify Solution/evaluate ( 4) + 11 Put 4 in for and simplify g ( 4) 11 4 5 1 Find the domain of g( ) + 5 Identification/Analysis The domain is the set of all possible inputs for the independent variable (in this case ) Since dividing by zero is not possible, we need to set the denominator equal to zero and determine which values of make the denominator zero Solution 5 0 5 Set denominator equal to zero and solve g is all real numbers ecept 5 State the domain The domain of ( ) In set-builder notation the domain is written { 5} In interval notation the domain is written (,5) ( 5, ) U
Page 14 of 14 Given g( ) + 5, find an so that g ( ) 4 Identification/Analysis Now we are asked to find the value(s) for so that the output or answer is 4 Solve the equation + 4 This is an equation with a fraction and a variable in the denominator 5 Solution 5 0 5 Thus 5 The domain does not include 5 so 5 cannot be a solution to the equation Find the domain Set denominator equal to zero Solve for State the restrictions Since the only denominator is 5, that is the LCD Find the LCD + Multiply both sides of the equation by LCD ( 5) ( 5) 4 5 to eliminate (reduce) the fractions + 4 0 Simplify the fraction and distribute This is a linear equation Isolate The solution is in the domain Check/Verification + 1 g + 6 6 4 1 5 1 Thus g 4 so the solution is Substitute into the function Evaluate g