Adding and subtracting rational expressions - Higher
Adding and subtracting algebraic fractions is a similar process to adding and subtracting normal fractions.
Fractions can only be added or subtracted when there is a common denominatorA common denominator exists when the denominator (the number at the bottom of a fraction) of two or more fractions is the same. Common denominators help to compare or add/subtract two or more fractions. Common denominators are made by using equivalent fractions, eg a common denominator for 1/4 and 1/3 would be twelfths. and algebraic fractions are the same in this method.
Example
Write \(\frac{2}{y} + \frac{1}{y}\) as a single fraction.
In this example, the denominatorThe bottom part of a fraction. For 鈪, the denominator is 8, which represents 'eighths'. of the two fractions are the same, so the numeratorThe top part of a fraction. For 鈪 , the numerator is 5. can simply be added.
This gives \(\frac{2}{y} + \frac{1}{y} = \frac{2 + 1}{y} = \frac{3}{y}\).
This fraction cannot be simplified so \(\frac{3}{y}\) is the final answer.
Question
Write \(\frac{5}{3t} - \frac{2}{7t}\) as a single fraction.
The denominators of each fraction are different, \(3t\) and \(7t\), so a common denominator must be created. This is found by working out the lowest common multiple of \(3t\) and \(7t\) which is \(21t\).
Remember, if it is difficult to work out the lowest common multiple of two expressions, a common denominator can always be found by simply multiplying the denominators together. Remember this may mean the fractions need simplifying at the end.
\(\frac{5}{3t} - \frac{2}{7t}\)
To create a common denominator of \(21t\), the first fraction's numerator and denominator must be multiplied by 7 and the second fraction's numerator and denominator must be multiplied by 3:
\(\frac{35}{21t} - \frac{6}{21t}\)
Now the denominators are the same, the numerators can be subtracted:
\(\frac{35 - 6}{21t} = \frac{29}{21t}\)
This fraction cannot be simplified any further, so \(\frac{29}{21t}\) is the final answer.
Question
Simplify \(\frac{2}{y + 4} + \frac{3}{y - 2}\).
These fractions do not have a common denominator. There are also no common factors between the denominators, so the only way to create a common denominator is to multiply the two expressions together.
\((y + 4) \times (y - 2)\) can be written as \((y + 4)(y - 2)\).
\(\frac{2}{y + 4}\) (multiply numerator and denominator by \(y - 2\)) \(+ \frac{3}{y - 2}\) (multiply numerator and denominator by \(y + 4\)) \(= \frac{2(y - 2)}{(y + 4)(y - 2)} + \frac{3(y + 4)}{(y - 2)(y + 4)}\)
Expanding brackets means everything inside the bracket has to be multiplied by the term outside the bracket.
\(2(y - 2) = 2 \times y + 2 \times -2 = 2y + -4 = 2y - 4\)
\(3(y + 4) = 3 \times y + 3 \times 4 = 3y + 12\)
This gives \(\frac{2y - 4}{(y + 4)(y - 2)} + \frac{3y + 12}{(y - 2)(y + 4)}\).
Now the denominators are the same, the numerators can be simplified:
\(= \frac{2y - 4 + 3y + 12}{(y - 2)(y + 4)}\)
Collect the like terms: \(2y - 4 + 3y + 12\). \(2y + 3y = 5y\). \(-4 + 12 = 8\).
\(= \frac{5y + 8}{(y - 2)(y + 4)}\)
This fraction cannot be simplified, so this is the final answer.
More guides on this topic
- Algebraic expressions - AQA
- Algebraic formulae - AQA
- Solving linear equations - AQA
- Solving simultaneous equations - AQA
- Solving quadratic equations - AQA
- Inequalities - AQA
- Sequences - AQA
- Straight line graphs - AQA
- Other graphs - AQA
- Transformation of curves - Higher- AQA
- Using and interpreting graphs - AQA