8-5 Adding and Subtracting Rational Expressions

Quick Review

To add or subtract rational expressions with different denominators, write each expression with the LCD. A fraction that has a fraction in its numerator or denominator or in both is called a complex fraction. Sometimes you can simplify a complex fraction by multiplying the numerator and denominator by the LCD of all the rational expressions.

Example

Simplify the complex fraction. 1 x + 3 5 y + 4 fraction 1 over x , plus 3 , over 5 over y , plus 4 end fraction

1 x + 3 5 y + 4 = ( 1 x + 3 ) ⋅ x y ( 5 x + 4 ) ⋅ x y   = 1 x ⋅ x y + 3 ⋅ x y 5 x ⋅ x y + 4 ⋅ x y   = y + 3 x y 5 y + 4 x y table with 3 rows and 2 columns , row1 column 1 , fraction 1 over x , plus 3 , over 5 over y , plus 4 end fraction , column 2 equals . fraction open . 1 over x , plus 3 . close . dot x y , over open . 5 over x , plus 4 . close . dot x y end fraction , row2 column 1 , , column 2 equals . fraction 1 over x , dot x y plus 3 dot x y , over 5 over x , dot x y plus 4 dot x y end fraction , row3 column 1 , , column 2 equals . fraction y plus 3 x y , over 5 y plus 4 x y end fraction , end table

Exercises

Simplify the sum or difference. State any restrictions on the variable.

27. 3 x x 2 − 4 + 6 x + 2 fraction 3 x , over x squared , minus 4 end fraction . plus . fraction 6 , over x plus 2 end fraction
28. 1 x 2 − 1 − 2 x 2 + 3 x fraction 1 , over x squared , minus 1 end fraction . minus . fraction 2 , over x squared , plus 3 x end fraction

Simplify the complex fraction.

29. 2 − 2 x 3 − 1 x fraction 2 minus , 2 over x , over 3 minus , 1 over x end fraction
30. 1 x + y 4 fraction fraction 1 , over x plus y end fraction , over 4 end fraction

8-6 Solving Rational Equations

Quick Review

Solving a rational equation often requires multiplying each side by an algebraic expression. This may introduce extraneous solutions—solutions that solve the derived equation but not the original equation. Check all possible solutions in the original equation.

Example

Solve the equation. Check your solution.

1 2 x − 2 5 x = 1 2 10 x ( 1 2 x − 2 5 x ) = 10 x ( 1 2 ) 5 − 4 = 5 x x = 1 5 Check  1 2 ( 1 5 ) − 2 5 ( 1 5 ) = 5 2 − 2 = 1 2 ✓ table with 2 rows and 1 column , row1 column 1 , table with 4 rows and 2 columns , row1 column 1 , fraction 1 , over 2 x end fraction , minus , fraction 2 , over 5 x end fraction , column 2 equals , 1 half , row2 column 1 , 10 x . open . fraction 1 , over 2 x end fraction , minus , fraction 2 , over 5 x end fraction . close , column 2 equals 10 x . open , 1 half , close , row3 column 1 , 5 minus 4 , column 2 equals 5 x , row4 column 1 , x , column 2 equals , 1 fifth , end table , row2 column 1 , cap check . fraction 1 , over 2 . open , 1 fifth , close end fraction . minus . fraction 2 , over 5 . open , 1 fifth , close end fraction . equals , 5 halves , minus 2 equals , 1 half , check mark , end table

Exercises

Solve each equation. Check your solutions.

31. 1 x = 5 x − 4 1 over x , equals . fraction 5 , over x minus 4 end fraction
32. 2 x + 3 − 1 x = − 6 x ( x + 3 ) fraction 2 , over x plus 3 end fraction . minus , 1 over x , equals . fraction negative 6 , over x . open , x plus 3 , close end fraction
33. 1 2 + x 6 = 18 x 1 half , plus , x over 6 , equals , 18 over x
34. You travel 10 mi on your bicycle in the same amount of time it takes your friend to travel 8 mi on his bicycle. If your friend rides his bike 2 mi/h slower than you ride your bike, find the rate at which each of you is traveling.

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