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Addition and subtraction of algebraic fractions

Addition and subtraction of algebraic fractions

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Recall that to get the fractions to the same denominator we looked to multiply or divide the entire fraction (both top and bottom) by the same number. This ensures that we do not change the value of the fraction as this is equivalent to multiplying or dividing by 1.

Example

Simplify the following:

1) \frac{a}{4} + \frac{a}{3}

2) \frac{7y}{15}-\frac{y}{5}

3) \frac{3}{2x}+\frac{1}{x}

1) To solve this we just need to change the two fractions so that they will have the same denominator. 4 and 3 have a common factor of 12, so we can convert both of these fractions so that we have \frac{3a}{12}+\frac{4a}{12} and then we can easily put the two together so that we get the answer of \frac{7a}{12}.

2) Again, we need to convert both of the fractions that are in our sum to having the same denominator by converting the second into also having a denominator of 15. So we get \frac{7y}{15}-\frac{3y}{15} which allows us to do the calculation so that we get \frac{4y}{15}.

3) Here the unknowns are on the bottom of the fraction but this does not make any difference, we just need to get the same denominator still. By multiplying the top and bottom of the second fraction by two we get \frac{3}{2x}+\frac{2}{2x} which gives us the result \frac{5}{2x}.

This technique of getting the same denominator for all fractions involved in the calculation still works for when we have more complex fractions. In order to add or subtract fractions, we must make the denominators the same.

So even if we have two fractions with completely different denominators, we can simply multiply the two together and use this as the denominator.

Example

Simplify the following:

1) \frac{1}{x}+\frac{1}{y}

2) \frac{3}{x-1}-\frac{2}{x}

1) Despite the two denominators being completely different we can still multiply them by each other so that the denominator is the same. When doing this we need to multiply the numerators by the correct values also (these will be different for the two fractions). So the denominator must be x \times y=xy which will give the two fractions as \frac{y}{xy}+\frac{x}{xy}. The numerators are the opposite fractions denominator. Then adding the two we get \frac{y+x}{xy}.

2) For this calculation we get the denominator as x(x-1) and so we have to solve \frac{3x}{x(x-1)}-\frac{2(x-1)}{x(x-1)} giving an answer of \frac{3x-2x+2}{x(x-1)}=\frac{x+2}{x(x-1)}.

The rule that we must follow is therefore:

Multiply the top and bottom of the 1st fraction by the denominator of the second and vice versa for the second fraction.

\frac{a}{b} \pm\frac{c}{d}=(\frac{ad}{bd}) \pm(\frac{bc}{bd})

By making use of this method we can find a common denominator of any two fractions and can then add or subtract one from the other as needed. For example, if we had a bracket on each of the numerators with a more complex function, we can simply multiply the two together to find the correct common denominator. Then we would have to find what is to be the numerator accordingly.

If we have

\frac{1}{x+3}+\frac{2}{y+1}

Then the correct denominator would be (x+3)(y+1) and the numerators would be the original multiplied by the other fractions denominator. So we would have:

\frac{1(y+1)}{(x+3)(y+1)}+\frac{2(x+3)}{(y+1)(x+3)}

This has the same effect of multiplying both of the fractions by 1, because we have multiplied both the denominator and numerator of each by the same number.

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