E2.3 - Calculating with algebraic fractions (Extended)

E2.3 - Calculating with algebraic fractions (Extended)

Why can you cancel across a product but not across a sum? Algebraic fractions follow the same arithmetic as numerical fractions, provided every denominator and divisor is valid. This Extended E2.3 lesson develops sums, products and quotients, then uses factorisation to simplify rational expressions.

Fraction sums

Half a quantity plus a third of it is five sixths of it: x2+x3=3x+2x6=5x6\frac x2+\frac x3=\frac{3x+2x}6=\frac{5x}6. The denominator names the equal-sized pieces, so you need a common denominator before adding their numerators.

With different numerical denominators, use their lowest common multiple. For example,

x4+x−23=3x+4(x−2)12=7x−812.\frac{x}{4}+\frac{x-2}{3}=\frac{3x+4(x-2)}{12}=\frac{7x-8}{12}.

The entire numerator x−2x-2 is multiplied by 4; its brackets protect both terms.

Adding and subtracting fractions

For B≠0B\ne0 and D≠0D\ne0,

AB±CD=AD±BCBD.\frac AB\pm\frac CD=\frac{AD\pm BC}{BD}.

A smaller common denominator may be available if BB and DD share factors. Multiply top and bottom of each fraction by the same factor, so its value is preserved.

Not on the supplied formula lists: learn the method.

Linear denominators work in the same way. Record their excluded values first. A common denominator for 3 and x−1x-1 is 3(x−1)3(x-1).

Worked example 1: a numerical and a linear denominator

Question. Write 13−2x−1\frac13-\frac2{x-1} as a single fraction.

Plan. Use denominator 3(x−1)3(x-1), with x≠1x\ne1, then subtract the complete second numerator.

13−2x−1=x−13(x−1)−63(x−1)=(x−1)−63(x−1)=x−73(x−1)\begin{aligned} \frac13-\frac2{x-1} &=\frac{x-1}{3(x-1)}-\frac6{3(x-1)}\\ &=\frac{(x-1)-6}{3(x-1)}\\ &=\frac{x-7}{3(x-1)} \end{aligned}

Answer. x−73(x−1)\frac{x-7}{3(x-1)}, for x≠1x\ne1.

Check. At x=4x=4, the original is 13−23=−13\frac13-\frac23=-\frac13. The single fraction is 4−79=−13\frac{4-7}{9}=-\frac13.

Replace only the first denominator, 3, by x+3x+3. Both denominators are now linear expressions, so the common denominator changes and there is an additional excluded value.

Worked example 2: two linear denominators

Question. Write 1x+3−2x−1\frac1{x+3}-\frac2{x-1} as a single fraction.

Plan. Use denominator (x+3)(x−1)(x+3)(x-1), with x≠−3,1x\ne-3,1, and subtract the complete second product.

1x+3−2x−1=(x−1)−2(x+3)(x+3)(x−1)=x−1−2x−6(x+3)(x−1)=−x−7(x+3)(x−1)\begin{aligned} \frac1{x+3}-\frac2{x-1} &=\frac{(x-1)-2(x+3)}{(x+3)(x-1)}\\ &=\frac{x-1-2x-6}{(x+3)(x-1)}\\ &=\frac{-x-7}{(x+3)(x-1)} \end{aligned}

Answer. −x−7(x+3)(x−1)\frac{-x-7}{(x+3)(x-1)}, for x≠−3,1x\ne-3,1.

Check. At x=4x=4, the original is 17−23=3−1421=−1121\frac17-\frac23=\frac{3-14}{21}=-\frac{11}{21}. The result gives −4−77×3=−1121\frac{-4-7}{7\times3}=-\frac{11}{21}.

For two linear denominators, the same decision gives

1x−2−x+1x+3=(x+3)−(x+1)(x−2)(x−2)(x+3),\frac1{x-2}-\frac{x+1}{x+3} =\frac{(x+3)-(x+1)(x-2)}{(x-2)(x+3)},

where x≠2,−3x\ne2,-3. Keep the subtraction bracket until after expanding: it applies to every term in the second numerator.

Watch out

AB+CD≠A+CB+D\frac AB+\frac CD\ne\frac{A+C}{B+D}. Denominators identify piece sizes; adding them does not make equivalent fractions. Also, −(x−4)=−x+4-(x-4)=-x+4, not −x−4-x-4.

Find a common denominator before combining these two fractions.

Products and quotients

Multiplication needs no common denominator. Multiply the numerators and denominators, cancelling common factors before or after the products. Division instead multiplies by the reciprocal of the divisor, exactly as with numerical fractions.

Multiplication and division

AB×CD=ACBD,B,D≠0AB÷CD=AB×DC,B,C,D≠0.\begin{aligned} \frac AB\times\frac CD&=\frac{AC}{BD},\qquad B,D\ne0\\ \frac AB\div\frac CD&=\frac AB\times\frac DC, \qquad B,C,D\ne0. \end{aligned}

For division, the divisor must be defined and non-zero. Keep the first fraction unchanged and reverse only the second fraction.

Not on the supplied formula lists: learn these methods.

You may cancel across a product because all numerator factors together form one numerator. An added term does not have that role.

Worked example 3: multiplying and cancelling

Question. Simplify 3a8×125a\frac{3a}{8}\times\frac{12}{5a}.

Plan. Record a≠0a\ne0, combine the products, and cancel the common factors.

3a8×125a=36a40a=910divide top and bottom by 4a\begin{aligned} \frac{3a}{8}\times\frac{12}{5a} &=\frac{36a}{40a}\\ &=\frac9{10} &&\text{divide top and bottom by }4a \end{aligned}

Answer. 910\frac9{10}, for a≠0a\ne0.

Check. At a=2a=2, 68×1210=7280=910\frac68\times\frac{12}{10}=\frac{72}{80}=\frac9{10}. The result's independence from aa follows because its two appearances cancel as non-zero factors.

Change only multiplication to division. The same two fractions now give a result containing a2a^2, since reversing the divisor puts its aa in the numerator.

Continue this lesson.

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