7.7 Rational Expressions - Solving Rational Equations

1 Expressions - Solving Rational EquationsObjective: Solve Rational Equations by identifying and multiplying bythe least common Solving Equations that are made up of Rational Expressions we will solvethem using the same strategy we used to solve linear Equations with we solved problems like the next example, we cleared thefraction by multi-plying by the least common denominator (LCD)Example 56=34 Multiply each term by LCD,122(12)3x 5(12)6=3(12)4 Reduce fractions2(4)x 5(2) = 3(3)Multiply8x 10= 9 Solve+10+10 Add 10 to both sides8x=19 Divide both sides by888x=198 Our SolutionWe will use the same process to solve Rational Equations , theonly difference is ourLCD will be more involved.

2 We will also have to be aware of domain issues. If our1 LCD equals zero, the solution is undefined. We will always check our solutions inthe LCD as we may have to remove a solution from our solution + 5x+ 2+ 3x=x2x+ 2 Multiply each term by LCD,(x+ 2)(5x+ 5)(x+ 2)x+ 2+ 3x(x+ 2) =x2(x+ 2)x+ 2 Reduce fractions5x+ 5 + 3x(x+ 2) =x2 Distribute5x+ 5 + 3x2+ 6x=x2 Combine like terms3x2+11x+ 5 =x2 Make equation equal zero x2 x2 Subtractx2from both sides2x2+11x+ 5 = 0 Factor(2x+ 1)(x+ 5) = 0 Set each factor equal to zero2x+ 1 = 0orx+ 5 = 0 Solve each equation 1 1 5 52x= 1orx= 522x= 12or 5 Check solutions,LCD can tbe zero 12+ 2 =32 5 + 2 = 3 Neither make LCD zero,both are solutionsx= 12or 5 Our SolutionThe LCD can be several factors in these problems.

3 As the LCD gets more com-plex, it is important to remember the process we are using to solve is still + 2+1x+ 1=5(x+ 1)(x+ 2)Multiply terms by LCD,(x+ 1)(x+ 2)x(x+ 1)(x+ 2)x+ 2+1(x+ 1)(x+ 2)x+ 1=5(x+ 1)(x+ 2)(x+ 1)(x+ 2)Reduce fractions2x(x+ 1) + 1(x+ 2) = 5 Distributex2+x+x+ 2 = 5 Combine like termsx2+ 2x+ 2 = 5 Make equatino equal zero 5 5 Subtract6from both sidesx2+ 2x 3 = 0 Factor(x+ 3)(x 1) = 0 Set each factor equal to zerox+ 3 = 0orx 1 = 0 Solve each equation 3 3+ 1 + 1x= 3orx= 1 Check solutions,LCD can tbe zero( 3 + 1)( 3 + 2) = ( 2)( 1) = 2 Check 3in(x+ 1)(x+ 2),it works(1 + 1)(1 + 2) = (2)(3) = 6 Check1in(x+ 1)(x+ 2),it worksx= 3or1 Our SolutionIn the previous example the denominators were factored for us.

4 More often wewill need to factor before finding the LCDE xample 1 1x 2=11x2 3x+ 2 Factor denominator(x 1)(x 2)LCD= (x 1)(x 2)Identify LCDx(x 1)(x 2)x 1 1(x 1)(x 2)x 2=11(x 1)(x 2)(x 1)(x 2)Multiply each term by LCD,reducex(x 2) 1(x 1) =11 Distributex2 2x x+ 1 =11 Combine like termsx2 3x+ 1 =11 Make equation equal zero 11 11 Subtract 11 from both sidesx2 3x 10= 0 Factor(x 5)(x+ 2) = 0 Set each factor equal to zerox 5 = 0orx+ 2 = 0 Solve each equation+ 5 + 5 2 2x= 5orx= 2 Check answers,LCD can tbe0(5 1)(5 2) = (4)(3) =12 Check5in(x 1)(x 2),it works( 2 1)( 2 2) = ( 3)( 4) =12 Check 2in(x 1)(x 2),it worksx= 5or 2 Our Solution3 World View Note:Maria Agnesi was the first women to publish a math text-book in 1748, it took her over 10 years to write!

5 This textbookcovered everythingfrom arithmetic thorugh differential Equations and was over1,000 pages!If we are subtracting a fraction in the problem, it may be easier to avoid a futuresign error by first distributing the negative through the 2x 3 x+ 2x+ 2=58 Distribute negative through numeratorx 2x 3+ x 2x+ 2=58 Identify LCD,8(x 3)(x+ 2),multiply each term(x 2)8(x 3)(x+ 2)x 3+( x 2)8(x 3)(x+ 2)x+ 2=5 8(x 3)(x+ 2)8 Reduce8(x 2)(x+ 2) + 8( x 2)(x 3) = 5(x 3)(x+ 2)FOIL8(x2 4) + 8( x2+x+ 6) = 5(x2 x 6)Distribute8x2 32 8x2+ 8x+48= 5x2 5x 30 Combine like terms8x+16= 5x2 5x 30 Make equation equal zero 8x 16 8x 16 Subtract8xand 160 = 5x2 13x 46 Factor0 = (5x 23)(x+ 2)

6 Set each factor equal to zero5x 23= 0orx+ 2 = 0 Solve each equation+23+23 2 25x=23 orx= 255x=235or 2 Check solutions,LCD can tbe08(235 3)(235+ 2)= 8(85)(335)=211225 Check235in8(x 3)(x+ 2),it works8( 2 3)( 2 + 2) = 8( 5)(0) = 0 Check 2in8(x 3)(x+ 2),can tbe0!x=235 Our SolutionIn the previous example, one of the solutions we found made the LCD this happens we ignore this result and only use the results that make therational Expressions and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) Practice - Solving Rational EquationsSolve the following Equations for the given variable:1)3x 12 1x= 03)x+20x 4=5xx 4 25)x+6x 3=2xx 37)2x3x 4=4x+ 56x 1 33x 49)3m2m 5 73m+ 1=3211)4 x1 x=123 x13)7y 3 12=y 2y 415)1x+ 2 12 x=3x+ 8x2 417)x+ 1x 1 x 1x+ 1=5619)32x+ 1+2x+ 11 2x= 1 8x24x2 121)x 2x+ 3 1x 2=1x2+x 623)3x+ 2+x 1x+ 5=5x+206x+2425)xx 1 2x+ 1=4x2x2 127)2xx+ 1 3x+ 5= 8x2x2+ 6x+ 529)x 5x 9+x+ 3x 3= 4x2x2 12x+2731)x 3x 6+x+ 5x+ 3= 2x2x2 3x 1833)4x+ 1x+ 3+5x 3x 1=8x2x2+ 2x 32)x+ 1 =4x+ 14)x2+ 6x 1+x 2x 1= 2x6)x 4x 1=123 x+ 18)6x+ 52x2 2x 21 x2=3xx2 110)4x2x 6 45x 15=1212)73 x+12=34 x14)23 x 68 x= 116)x+ 23x 1 1x=3x 33x2 x18)x 1x 3+x+ 2x+ 3=3420)3x 55x 5+5x 17x 7 x 41 x= 222)x 1x 2+x+ 42x+ 1=12x2 3x 224)xx+ 3 4x 2= 5x2x2+x 626)

7 2xx+ 2+2x 4=3xx2 2x 828)xx+ 1 3x+ 3= 2x2x2+ 4x+ 330)x 3x+ 6+x 2x 3=x2x2+ 3x 1832)x+ 3x 2+x 2x+ 1=9x2x2 x 234)3x 1x+ 6 2x 3x 3= 3x2x2+ 3x 18 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) - Solving Rational Equations1) 12,232) 3,13)34) 1,45)26)137) 18) 139) 510) 71511) 5,012)5,1013)163,514) 2, 1315) 816) 217) 15,518) 95,119)3220) 1021) 0, 522) 2,5323)4,724) 125)2326)1227)31028) 129) 2330) 131)13432)133) 1034)74 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( )6