# Adding algebraic fractions:  Level 2

Back to Section 1

Problem 8.

Do the problem yourself first!

 a) 1x + 2 3x = 3 + 2  3x = 5 3x b) 3 2x − 4 2x2 = 3x − 4   2x2
 c) 1     (x + 1)2 + 2    (x + 1) = 1 + 2(x + 1)   (x + 1)2 = 1 + 2x + 2   (x + 1)2 = 2x + 3 (x + 1)2
 d) 6     x(x − 1) + 2     x(x − 2) = 6(x − 2) + 2(x − 1)   x(x − 1)(x − 2) = 6x − 12 + 2x − 2  x(x − 1)(x − 2) = _8x − 14_   x(x − 1)(x − 2)
 Example 7.   Add 4     x2 − 25 − 3        x2 − 6x + 5

Note that we tend to say "add" in algebra, even though the operation is subtraction.

Solution.   To construct the LCM of the denominators, we must first factor the denominators.  In fact, whenever we add fractions in algebra, the denominators must be composed of factors.

 4     x 2 − 25 − 3        x2 − 6x + 5 = _4_       (x + 5)(x − 5) − _3_       (x − 5)(x − 1) = 4(x − 1) − 3(x + 5) (x + 5)(x − 5)(x − 1) = _4x − 4 − 3x − 15_(x + 5)(x − 5)(x − 1) = __x − 19__      (x + 5)(x − 5)(x − 1)

Problem 9.   Add.   First factor each denominator.

 a) x2 + _5_  2x + 2 = x2 + _5_   2(x + 1) = x(x + 1) + 5   2(x + 1) = x2 + x + 5 2(x + 1)
 b) 1   x2 − x + 2x = _1_   x(x − 1) + 2x = 1 + 2(x − 1)  x(x − 1) = 1 + 2x − 2  x(x − 1) = 2x − 1 x(x − 1)
 c) 2   x + 3 + 12   x2 − 9 = 2   x + 3 + __12__    (x + 3)(x − 3) = 2(x − 3) + 12(x + 3)(x − 3) = 2x − 6 + 12   (x + 3)(x − 3) = ___2x + 6___ (x + 3)(x − 3) = __ 2(x + 3) __ (x + 3)(x − 3) = _2_   x − 3
 d) ___6___  x2 + 5x + 6 + ___2___ x2 − x − 6 = ___6___    (x + 2)(x + 3 ) + ___ 2___   (x + 2)(x − 3) = 6(x − 3) + 2(x + 3) (x + 2)(x + 3)(x − 3) = _ 6x − 18 + 2x + 6 _(x + 2)(x + 3)(x − 3) = _____8x − 12_____(x + 2)(x + 3)(x − 3) = ___ 4(2x − 3) ___  (x + 2)(x + 3)(x − 3)

Factor the numerator to see if anything might cancel.

 e) ___3___  x2 − 7x + 10 − __2__ x2 − 25 = ___3___    (x − 2)(x − 5) − ___ 2___   (x + 5)(x − 5) = 3(x + 5) − 2(x − 2) (x − 2)(x − 5)(x + 5) = _ 3x + 15 − 2x + 4 _(x − 2)(x − 5)(x + 5) = ___ __x + 19__ ___(x − 2)(x − 5)(x + 5)
 f) ___7___   3x2 − 5x + 2 − ___4___  3x2 + x − 2 = ___7___    (3x − 2)(x − 1) − ___ 4___    (3x − 2)(x + 1) = 7(x + 1) − 4(x − 1)  (3x − 2)(x − 1)(x + 1) = _ 7x + 7 − 4x + 4_  (3x − 2)(x − 1)(x + 1) = ___ __3x + 11__ ___(3x − 2)(x − 1)(x + 1)

Example 8.   Symmetrically, if the numerator of a fraction is a sum, then we can rewrite that fraction as a sum of fractions:

 a + b + c      d = ad + bd + cd .

Problem 10.   Rewrite each of the following as a sum of fractions, and simplify.

 a) 1 + 2 + 3      6 = 16 + 26 + 36 = 16 + 13 + 12
 b) 2n2 − 4n + 1      n2 = 2n2 n2 − 4nn2 + 1 n2 = 2 − 4n + 1 n2
 c) x³ + 4x2 + 2       2x5 = x³ 2x5 + 4x22x5 + 2 2x5 = 1 2x2 + 2 x3 + 1 x5
 d) x − 1x + 1 = x   x + 1 − 1   x + 1
 Example 9.   Simplify
 Solution. = = c· ab   b + a (Definition of division) = cab  b + a or abc  a + b

Problem 11.   Simplify.

 a) = = 16 · 1 10 = 1 60
 b) = =  z· xy  y + x = zxy  y + x
 = x − (x + h)  (x + h)x · 1h
 = x − x − h  (x + h)x · 1h
 = −h    (x + h)x · 1h
 = − 1     (x + h)x , on canceling the h's.
 d) = = (x + 1)(x − 1)         x2 · x  x − 1 = x + 1   x

Or, on recognizing the numerator as the Difference of Two Squares:

 = =  1 + 1x = x + 1   x
 e) = = (a + b)(a − b)        ba · ba  a + b = a − b

Next Lesson:  Equations with fractions

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