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Operations with algebraic fractions
We have already dealt with the interpretation of algebraic fractions. A fractional expression has a meaning if its denominator is not equal to zero. We should not forget about it later on.
Simplification of algebraic functions
Example 1
Let us simplify the fraction \latex{\frac{1,008}{1,512}}.
Solution
Simplification means dividing the numerator and the denominator of the fraction by the same number. To find the suitable number let us express both the numerator and the denominator as a product of prime numbers: \latex{1,008=2^4\times3^2\times7} and \latex{1,512=2^3\times3^3\times7} respectively, then we can simplify by the identical factors: \latex{\frac{2^4\times3^2\times7}{2^3\times3^3\times7}=\frac{2}{3}}.
The process is similar when simplifying algebraic fractions.
The process is similar when simplifying algebraic fractions.
Example 2
Let us simplify the fraction \latex{\frac{2x^2+6x}{x^2+6x+9}}.
Solution
So that we have a chance to simplify let us transform both the numerator and the denominator to a product; if we find identical factors, we can simplify by them:
\latex{\frac{2x^2+6x}{x^2+6x+9}=\frac{2x(x+3)}{(x+3)^2}=\frac{2x}{x+3}}.
The product form of the denominator also helps determining the domain of the fraction, since \latex{(x+3)^2\neq 0}, i.e. \latex{x+3\neq0}, which means that the original algebraic fraction has a meaning if \latex{x\neq-3.}
\latex{\frac{2x^2+6x}{x^2+6x+9}=\frac{2x(x+3)}{(x+3)^2}=\frac{2x}{x+3}}.
The product form of the denominator also helps determining the domain of the fraction, since \latex{(x+3)^2\neq 0}, i.e. \latex{x+3\neq0}, which means that the original algebraic fraction has a meaning if \latex{x\neq-3.}
Example 3
Let us simplify the following fractions.
- \latex{\frac{3a^2-12a}{a^2-16}}
- \latex{\frac{b^2-10b+25}{ab-5a+2b-10}}
Solution (a)
\latex{\frac{3a^2-12a}{a^2-16}=\frac{3a(a-4)}{(a-4)(a+4)}=\frac{3a}{a+4}}.
From the product form of the denominator it can be seen that \latex{a\neq4} and \latex{a\neq-4.}Solution (b)
\latex{\frac{b^2-10b+25}{ab-5a+2b-10}=\frac{(b-5)^2}{a(b-5)+2(b-5)}=\frac{(b-5)^2}{(b-5)(a+2)}=\frac{b-5}{a+2}}.
The product form of the denominator shows that \latex{b\neq5} and \latex{a\neq-2}.
Multiplication of algebraic fractions
Example 4
Let us find the product \latex{\frac{12}{25}\times\frac{10}{9}}.
Solution
\latex{\frac{12}{25}\times\frac{10}{9}=\frac{120}{225}}.
The result can be simplified, but it could be done even before the operation, by simplifying by 3 and 5;
\latex{\frac{12}{25}\times\frac{10}{9}=\frac{3\times4}{5\times5}\times\frac{2\times5}{3\times3}=\frac{4}{5}\times\frac{2}{3}=\frac{8}{15}}.
Example 5
Let us find the product \latex{\frac{3x-1}{2x+3}\times\frac{x-4}{x+2}}.
Solution
The fractions have a meaning if \latex{x\neq-\frac{3}{2}} and \latex{x\neq-2}. The numerators and the denominators are all first-order expressions and we cannot factor out anything, so it is not possible to simplify.
When doing the respective multiplications:
\latex{\frac{(3x-1)(x-4)}{(2x+3)(x+2)}=\frac{3x^2-12x-x+4}{2x^2+4x+3x+6}=\frac{3x^2-13x+4}{2x^2+7x+6}}.
When doing the respective multiplications:
\latex{\frac{(3x-1)(x-4)}{(2x+3)(x+2)}=\frac{3x^2-12x-x+4}{2x^2+4x+3x+6}=\frac{3x^2-13x+4}{2x^2+7x+6}}.
Example 6
Let us find the product \latex{\frac{3x+12}{2x-2}\times\frac{4x-4}{6x-12}}.
Solution
The fractions have a meaning if \latex{x\neq1} and \latex{x\neq2}. All the factors are first-order expressions again, but we can try factoring out:
\latex{\frac{3(x+4)}{2(x-1)}\times\frac{4(x-1)}{6(x-2)}}.
This product form shows that we can simplify by \latex{2} , by \latex{3} and by \latex{(x-1)}:\latex{\frac{x+4}{1}\times\frac{1}{x-2}=\frac{x+4}{x-2}}.
Division of algebraic fractions
Example 7
Let us do division \latex{\frac{14}{15}\div\frac{42}{25}}.
Solution
Dividing by a fraction means multiplying by its reciprocal, where after simplifying:
\latex{\frac{14}{15}\times\frac{25}{42}=\frac{2\times7}{3\times5}\times\frac{5^2}{2\times3\times7}=\frac{5}{9}}.
The process is similar in the case of algebraic fractions too.
Example 8
Let us do the following divisions.
a) \latex{\frac{x^2-3x}{x^2+4x+4}\div\frac{2x-6}{x^2-4}}
b) \latex{\frac{5x^2+10x+5}{12x-6}\div\frac{x^3+3x^2+3x+1}{12x^2-12x+3}}
Solution (a)
Watch out, not only the denominators but the numerator of the divisor cannot be equal to 0 either. If we factorise then we can easily give the domain too:
\latex{x^2+4x+4=(x+2)^2}, therefore \latex{x\neq-2}; \latex{2x-6=2(x-3)}, therefore \latex{x\neq3}; \latex{x^2-4=(x-2)(x+2)}, which means \latex{x\neq2;\, x\neq-2}.
To summarise: the fractions mentioned in the example have a meaning if \latex{x\neq-2;\, x\neq2;\, x\neq3}.
We only have to factorise the numerator of the dividend now and then we can simplify:
\latex{\frac{x(x-3)}{(x+2)^2}\times\frac{(x-2)(x+2)}{2(x-3)}=\frac{x(x-2)}{2(x+2)}}.
Solution (b)
Let us first concentrate on the domain of the fractions.
\latex{12x-6\neq0}, which means \latex{x\neq\frac{1}{2}; \; 12x^2-12x+3\neq0}.
Factorisation helps:
\latex{3(4x^2-4x+1)=3(2x-1)^2\neq0}, therefore \latex{x\neq\frac{1}{2}}.
Let us examine the numerator of the divisor:
\latex{x^3+3x^2+3x+1=(x+1)^3\neq0}, which means \latex{x\neq-1}.
So the problem has a meaning if \latex{x\neq\frac{1}{2}} and \latex{x\neq-1}.
To do the operation and to simplify we also have to factorise the numerator of the dividend:
\latex{5x^2+10x+5=5(x^2+2x+1)=5(x+1)^2}.
The original task after factorising the numerators and the denominators:
\latex{\frac{5(x+1)^2}{6(2x-1)^2}\div\frac{(x+1)^3}{3(2x-1)^2}=\frac{5(x+1)^2}{6(2x-1)}\times\frac{3(2x-1)^2}{(x+1)^3}=\frac{5}{2}\times\frac{2x-1}{x+1}=\frac{10x-5}{2x+2}}.
Addition of algebraic factions
Example 9
Let us do the following operations: \latex{\frac{1}{4}+\frac{7}{6}-\frac{1}{3}}.
Solution
The common denominator of the fractions is the lowest common multiple of the denominators: \latex{12}. Doing the operation: \latex{\frac{3}{12}+\frac{14}{12}-\frac{4}{12}=\frac{13}{12}}.
The process is similar when dealing with algebraic fractions.
The process is similar when dealing with algebraic fractions.
Example 10
Let us do the following operations.
- \latex{\frac{1}{x}-\frac{2}{3x}+\frac{3}{2x}}
- \latex{\frac{x-3}{x+3}-\frac{x-1}{x-3}+\frac{12x}{x^2-9}}
Solution (a)
The denominators cannot be equal to zero, i.e. \latex{x\neq0}. The common denominator is the “lowest common multiple” of the denominators (\latex{ 6x }); after doing the respective expansions:
\latex{\frac{1}{x}-\frac{2}{3x}+\frac{3}{2x}=\frac{6}{6x}-\frac{4}{6x}+\frac{9}{6x}=\frac{11}{6x}}.
Solution (b)
The denominators cannot be equal to zero, so
\latex{x\neq-3} and \latex{x\neq3}.
To find the common denominator we have to factorise the third denominator:
\latex{x^2-9=(x+3)(x-3)}, this product can be used as the common denominator:
\latex{\frac{x-3}{x+3}-\frac{x-1}{x-3}+\frac{12x}{x^2-9}=\frac{(x-3)^2}{(x+3)(x-3)}-\frac{(x-1)(x+3)}{(x+3)(x-3)}+\\[10pt]}\latex{+\frac{12x}{(x+3)(x-3)}=\frac{x^2-6x+9-x^2-3x+x+3+12x}{(x+3)(x-3)}=\\[10pt]}\latex{=\frac{4x+12}{(x+3)(x-3)}=\frac{4(x+3)}{(x+3)(x-3)}=\frac{4}{x-3}}.
Realise that after the combining we had a chance to simplify, we only had to factorise the numerator.
Example 11
Simplify the following fraction: \latex{\frac{x^3-8}{x^3-4x^2+4x}}.
Solution
Factorise the denominator:
\latex{x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2}.
The fraction has a meaning if \latex{x\neq0} and \latex{x\neq2}.
The numerator can also be factorised, since:
The numerator can also be factorised, since:
\latex{x^3-8=x^3-2^3=(x-2)(x^2+2x+4)}.
Using the product form of the numerator and the denominator we can simplify the fraction:
\latex{\frac{x^3-8}{x^3-4x^2+4x}=\frac{(x-2)(x^2+2x+4)}{x(x-2)^2}=\frac{x^2+2x+4}{x(x-2)}}.
Exercises
{{exercise_number}}. Simplify the following fractions.
- \latex{\frac{17x^3y^2}{34xy^4}}
- \latex{\frac{24(2x-3)^2}{8(2x-3)(2x+3)}}
- \latex{\frac{9x^2-6x+1}{9x^2-1}}
- \latex{\frac{2x^2-x-15}{4x^2+20x+25}}
- \latex{\frac{x^2+8x+15}{x^2-2x-15}}
- \latex{\frac{6ab+3+2b+9a}{2ab-2b+3a-3}}
{{exercise_number}}. Perform the following multiplications and divisions.
- \latex{\frac{12xy^5}{7ab}\times\frac{21a^2b^3}{32x^2y^3}}
- \latex{\frac{8a^4x^3}{15y^6}\div\frac{16a^2x^4}{25y^7}}
- \latex{\frac{6x^2-4x}{24+6x^2}\times\frac{x+4}{9x^2-12x+4}}
- \latex{\frac{4b^2+12b+9}{9b^2-12b+4}\div\frac{6b^2+9b}{18b^2-12b}}
- \latex{\frac{25-c^2}{2c^2-10c}\times\frac{20c+4c^2}{25+10c+c^2}}
- \latex{\frac{3a^3+18a^2+27a}{5a+15}\div\frac{a^3-9a}{10a-30}}
- \latex{\frac{3a^2-3ab}{a^3-3a^2b+3ab^2-b^3}\times\frac{a^2-b^2}{5a^2+5ab}}
- \latex{\frac{x^4-1}{2x^3+2x}\div\frac{x^2+2x+1}{4x}}
{{exercise_number}}. Do the combining of the following fractional expressions.
- \latex{\frac{3}{2x}+\frac{1}{x^2}}
- \latex{\frac{1}{3a^2}-\frac{2}{5a}-\frac{1}{a}}
- \latex{\frac{x-1}{3x+1}+\frac{x+2}{6x+2}}
- \latex{\frac{b}{b^2+4b+4}-\frac{b-3}{b+2}}
- \latex{\frac{a-1}{4a^2-9}+\frac{a}{4a^2-12a+9}}
- \latex{\frac{2a-1}{a^2-49}-\frac{a+1}{a^2-14a+49}-\frac{2}{-a-7}}
- \latex{\frac{1}{9y^2-1}+\frac{1}{9y^2+6y+1}-\frac{2}{9y^2-6y+1}}
- \latex{\frac{2}{x^3-3x^2+3x-1}-\frac{1}{x^2-1}+\frac{1}{x^2-2x+1}}
Quiz
Sophie’s age is \latex{ 16 } more than the sum of Terry’s and Gladys’ age. The square of Sophie’s age is \latex{ 1,632 } more than the square of the sum of Terry’s and Gladys’ age. What is the sum of their ages?
