Title: Default for 1.4 Arithmetic of Algebraic Fractions
Course: HELM Workbooks 2023-2024

Cancelling Common Factors

Consider the fraction \(\displaystyle{10\over 35}\). To simplify it we can factorise the numerator and the denominator and then cancel any common factors. Common factors are those factors which occur in both the numerator and the denominator. Thus

\(\qquad\displaystyle {10\over 35} = {\not{5}\times 2\over 7\times \not{5}} = {2\over 7}\)

Note that the common factor 5 has been cancelled. It is important to remember that only common factors can be cancelled. The fractions \(\displaystyle{10\over 35}\) and \(\displaystyle{2\over 7}\) have identical values - they are equivalent fractions - but \(\displaystyle{2\over 7}\) is in a simpler form than \(\displaystyle{10\over 35}\).

We apply the same process when simplifying algebraic fractions.

Example

Leave the expression in its most simplified form

1. \(\displaystyle{yx\over 2x}\)
2. \(\displaystyle{x\over xy}\)
3. \(\displaystyle{x\over x+y}\)

Solution

1. \(\dfrac{y}{2}\)
In the expression \(\dfrac{yx}{2x}\) \(x\) is a factor common to both numerator and denominator. This common factor can be cancelled to give
\(\dfrac{y\not{x}}{2\not{x}} = \dfrac{y}{2}\)

2. \(\dfrac{1}{y}\)
Note that \(\dfrac{x}{xy}\) can be written as \(\dfrac{1x}{xy}\). The common factor of x can be cancelled to give
\(\dfrac{1\not{x}}{\not{x}y} = \dfrac{1}{y}\)

3. \(\dfrac{x}{x+y}\)
In the expression \(\dfrac{x}{x+y}\) notice that an \(x\) appears in both numerator and denominator.
However \(x\) is not a common factor. Recall that factors of an expression are multiplied together whereas in the denominator \(x\) is added to \(y\). This expression cannot be simplified.

Exercise

Leave the expression in its most simplified form. If it can't get simplified rewrite the original expression.

1. \(\displaystyle{[value]}\)
2. \(\displaystyle{[value]}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. In the expression \(\displaystyle{[value]}\), the term \(\displaystyle{[value]}\) is common to both the numerator and the denominator and hence we have \(\displaystyle{[value]} = \displaystyle{[value]}\) 2. In the expression \(\displaystyle{[value]}\) notice that an \(a\) and \([value]\) appear in both numerator and denominator. However \(a\) and \([value]\) are not common factors. Recall that factors of an expression are multiplied together whereas in the denominator \(a\) is added to \([value]\). This expression cannot be simplified.

Exercise

Leave the expression in its most simplified form

\(\displaystyle{[value]} =\)
Feedback: Factorising and cancelling common factors gives: \(\displaystyle[value] = \dfrac{\not{[value]} \times [value] \times \not{x} \times x^[value]}{\not{[value]} \times [value] \times \not{x}} = \displaystyle[value]\)

Exercise

Leave the expression in its most simplified form

\(\displaystyle{[value]} =\)
Feedback: Factorising and cancelling common factors gives: \(\displaystyle[value] = \dfrac{\not{[value]} \times [value] \times \not{x}}{\not{[value]} \times [value] \times \not{x} \times x^[value]} = \displaystyle[value]\)

Example

Simplify \(\dfrac{3x+6}{6x+12}\)

Solution

First we factorise the numerator and the denominator to see if there are any common factors.

\(\qquad \dfrac{3x+6}{6x+12} = \dfrac{3(x+2)}{6(x+2)} = \dfrac{3}{6} = \dfrac{1}{2}\)

The factors \(x + 2\) and 3 have been cancelled.

Example

Show that the algebraic fraction \(\dfrac{3}{x+1}\) and \(\dfrac{3(x+4)}{x^2+5x+4}\) are equivalent.

Solution

The denominator \(x^2+5x+4\) can be factorised as \((x+4)(x+1)\) so that

\(\qquad \dfrac{3(x+4)}{x^2+5x+4} = \dfrac{3(x+4)}{(x+4)(x+1)}\)

Note that \((x + 4)\) is a factor common to both the numerator and the denominator and can be cancelled to leave \(\dfrac{3}{x+1}\). Thus \(\dfrac{3}{x+1}\) and \(\dfrac{3(x+4)}{x^2+5x+4}\) are equivalent fractions

Example

Simplify \(\dfrac{6(4-8x)(x-2)}{1-2x}\)

Solution

The factor \(4-8x\) can be factorised to \(4(1-2x)\). Thus

\(\qquad \dfrac{6(4-8x)(x-2)}{1-2x} = \dfrac{(6)(4)(1-2x)(x-2)}{1-2x} = 24(x-2)\)

Exercise

Leave the expression in its most simplified form

\(\displaystyle{[value]} =\)
Feedback: Factorise the numerator and denominator, and cancel any common factors: \(\displaystyle[value] = \dfrac{[value] \times [value]}{[value](x+[value])} = \displaystyle[value]\)

Exercise

Leave the expression in its most simplified form

\(\displaystyle{[value]} =\)
Feedback: First factorise the numerator and denominator and then cancel any common factors: \(\qquad \displaystyle{[value]} = \displaystyle{\dfrac{(x+[value])(x+[value])}{(2x+[value])(x+[value])}} = [value]\)

Exercise

Leave the expression in its most simplified form

1. \(\displaystyle{[value]}\)
2. \(\displaystyle{[value]}\)
3. \(\displaystyle{[value]}\)
4. \(\displaystyle{[value]}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. In the expression \(\displaystyle{[value]}\) the numerator is a factor of the denominator and hence we can divide the top and bottom of the fraction by the numerator to give: \(\displaystyle{[value]} = \displaystyle{[value]}\) 2. Again the numerator of the expression \(\displaystyle{[value]}\) is a factor of the denominator and so we can divide the top and bottom of the fraction by the numerator to give: \(\displaystyle{[value]} = \displaystyle{[value]}\) 3. The expression \(\displaystyle{[value]}\) has no factors common to both the numerator and the denominator and hence it does not simplify further. 4. Finally the numerator of the expression \(\displaystyle{[value]}\) again is a factor of the denominator and so we can divide the top and bottom of the fraction by the numerator to give: \(\displaystyle{[value]} = \displaystyle{[value]}\)

Exercise

Leave the expression in its most simplified form

1. \(\displaystyle{[value]}\)
2. \(\displaystyle{[value]}\)
3. \(\displaystyle{[value]}\)
4. \(\displaystyle{[value]}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. We can note that the denominator of \(\displaystyle{[value]}\) is a product of the numerator and \([value]\) and hence dividing the top and bottom of the fraction by the numerator we have: \(\displaystyle{[value]} = \displaystyle{[value]}\) 2. Now we have that [value] is a factor of both the numerator and the denominator of \(\displaystyle{[value]}\) and therefore dividing the top and bottom of the fraction by the [value] we have: \(\displaystyle{[value]} = \displaystyle{[value]}\) 3. The expression \(\displaystyle{[value]}\) has no factors common to both the numerator and the denominator and hence it does not simplify further. 4. Similarly the expression \(\displaystyle{[value]}\) has no factors common to both the numerator and the denominator and hence it does not simplify further.

Exercise

Leave the expression in its most simplified form

1. \(\dfrac{[value]z}{z}\)
2. \(\dfrac{[value]z}{[value]z}\)
3. \(\dfrac{[value]}{[value]z^2}\)
4. \(\dfrac{[value]z}{[value]z^2}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. In the expression \(\dfrac{[value]z}{z}\), the term \(z\) is common to both the numerator and the denominator and hence we have: \(\dfrac{[value]z}{z} = \displaystyle{[value]}.\) 2. In the expression \(\dfrac{[value]z}{[value]z}\), the terms \(z\) and \([value]\) are both common to the numerator and the denominator and hence we have: \(\dfrac{[value]z}{[value]z} = \displaystyle{[value]}.\) 3. The expression \(\dfrac{[value]}{[value]z^2}\) has the common factor \([value]\) in both the numerator and the denominator and hence we have: \(\dfrac{[value]}{[value]z^2} = \displaystyle{[value]}.\) 4. In the expression \(\dfrac{[value]z}{[value]z^2}\), \(z\) and \([value]\) are both common to the numerator and the denominator and hence we have: \(\dfrac{[value]z}{[value]z^2} = \displaystyle{[value]}.\)

Exercise

Leave the expression in its most simplified form

1. \(\displaystyle{[value]}\)
2. \(\displaystyle{[value]}\)
3. \(\displaystyle{[value]}\)
4. \(\displaystyle{[value]}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. We have that \((x+1)\) is a common factor between the numerator and the denominator of \(\displaystyle{[value]}\) and so we have: \(\displaystyle{[value]} = \displaystyle{[value]}\) There are no more common factors between the numerator and denominator and so this is the final answer 2. If we first factorise the numerator and the denominator of \(\displaystyle{[value]}\) we see that \((x+1)\) is again a common factor and so can be cancelled to give: \(\displaystyle{[value]} = \displaystyle{[value]}\) 3. Here we can see that \([value]\) divides every term in the expression \(\displaystyle{[value]}\) and so dividing the numerator and the denominator by \([value]\) gives us: \(\displaystyle{[value]} = \displaystyle{[value]}\) Just as before this cannot be simplified further 4. By factorising the numerator of \(\displaystyle{[value]}\) we can see that the term \((x-[value])\) is a common factor between the numerator and the denominator and hence we have: \(\displaystyle{[value]} = \displaystyle{[value]}\)

Exercise

Leave the expression in its most simplified form

1. \(\dfrac{[value]x}{x}\)
2. \(\dfrac{[value]x}{x^2}\)
3. \(\dfrac{[value]s}{s^[value]}\)
4. \(\dfrac{[value]z^{[value]}}{[value]z^2}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. In the expression \(\dfrac{[value]x}{x}\), the term \(x\) is common to both the numerator and the denominator and hence we have: \(\dfrac{[value]x}{x} = \displaystyle{[value]}.\) 2. In the expression \(\dfrac{[value]x}{x^2}\), the term \(x\) is common to both the numerator and the denominator and hence we have: \(\dfrac{[value]x}{x^2} = \dfrac{[value] \times x}{x \times x} = \displaystyle{[value]}.\) 3. The expression \(\dfrac{[value]s}{s^[value]}\) has the common factor \(s\) in both the numerator and the denominator and hence we have: \(\dfrac{[value]s}{s^[value]} = \dfrac{[value] \times s}{s \times s^[value]} = \displaystyle{[value]}.\) 4. In the expression \(\dfrac{[value]z^{[value]}}{[value]z^2}\), \(z\) and \([value]\) are both common to the numerator and the denominator and hence we have: \(\dfrac{[value]z^{[value]}}{[value]z^2} = \dfrac{[value] \times 3 \times z^2 \times z^{[value]}}{[value] \times z^2} = \displaystyle{[value]}.\)

Exercise

Leave the expression in its most simplified form. If it can't get simplified rewrite the original expression.

1. \(\dfrac{[value]x+[value]}{[value]x+[value]}\)
2. \(\dfrac{[value]x+[value]}{[value]x}\)
3. \(\dfrac{[value]x+[value]}{[value]}\)
4. \(\dfrac{[value]x+[value]}{[value]x+1}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. First notice that \([value]\) can be factored out of every term in the numerator and the denominator of \(\dfrac{[value]x+[value]}{[value]x+[value]}\) and so we have: \[\dfrac{[value]x+[value]}{[value]x+[value]} = \dfrac{[value] \times x+[value] \times 3}{[value] \times 5 \times x+[value]} = \displaystyle{[value]}\] There are no more common factors between the numerator and denominator and so this is the final answer. 2. Again we have that \([value]\) can be factored out of every term in the numerator and the denominator of \(\dfrac{[value]x+[value]}{[value]x}\) and so we have: \[\dfrac{[value]x+[value]}{[value]x} = \dfrac{[value] \times x+[value] \times 3}{[value] \times 5 \times x} = \displaystyle{[value]}\] There are no more common factors between the numerator and denominator and so this is the final answer. 3. Similarly \([value]\) can be factored out of every term in the numerator and the denominator of \(\dfrac{[value]x+[value]}{[value]}\): \[\dfrac{[value]x+[value]}{[value]} = \dfrac{[value] \times x+[value] \times 3}{[value] \times 5} = \displaystyle{[value]}\] Just as before this cannot be simplified further. 4. There are no common factors between the numerator and denominator of \(\dfrac{[value]x+[value]}{[value]x+1}\) and therefore it is already fully simplified.

Exercise

Leave the expression in its most simplified form

1. \(\dfrac{[value]}{[value]}\)
2. \(\dfrac{[value]}{[value]}\)
3. \(\dfrac{[value]}{[value]}\)
4. \(\dfrac{[value]}{[value]}\)

Solution
Feedback: To simplify these expressions we first factorise their numerators and denominators before cancelling any common factors. Using this process we have: 1. \(\dfrac{[value]}{[value]} = \dfrac{(x-4)(x-[value])}{(x-4)(x-[value])} = \displaystyle{[value]}\) 2. \(\dfrac{[value]}{[value]} = \dfrac{(x-[value])(x+[value])}{(x-[value])(x-8)} = \displaystyle{[value]}\) 3. \(\dfrac{[value]}{[value]} = \dfrac{(2x-3)(x-[value])}{(2x-3)(x-[value])} = \displaystyle{[value]}\) 4. \(\dfrac{[value]}{[value]} = \dfrac{5(z-[value])}{(z-6)(z-[value])} = \displaystyle{[value]}\)

Exercise

Simplify

1. \(\displaystyle [value] \times [value]\)
2. \(\displaystyle [value] \times [value]\)
3. \(\displaystyle [value] \times [value]\)
4. \(\displaystyle [value] \times [value]\)

Solution
Feedback: 1. \(\displaystyle [value] \times [value] = [value] = [value]\) 2. \(\displaystyle [value] \times [value] = [value] = [value]\) 3. \(\displaystyle [value] \times [value] = [value] = [value]\) 4. \(\displaystyle [value] \times [value] = [value] = [value]\)

Multiplication and division of algebraic fractions

To multiply together two fractions (numerical or algebraic) we multiply their numerators together and then multiply their denominators together. That is

Key Point

Multiplication

\[ \dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{ac}{bd} \]

Any factors common to both numerator and denominator can be cancelled. This cancellation can be performed before or after the multiplication.

To divide one fraction by another (numerical or algebraic) we invert the second fraction and then multiply.

Key Point

Division of fractions

\[ \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c} = \dfrac{ad}{bc} \quad b\neq 0, \, c\neq0, \, d\neq 0 \]

Example

Simplify:

1. \(\dfrac{2a}{c} \times \dfrac{4}{c}\)
2. \(\dfrac{2a}{c} \times \dfrac{c}{4}\)
3. \(\dfrac{2a}{c} \div \dfrac{4}{c}\)

Solution

1. \(\dfrac{2a}{c} \times \dfrac{4}{c} = \dfrac{8a}{c^2}\)
2. \(\dfrac{2a}{c} \times \dfrac{c}{4} = \dfrac{2ac}{4c} = \dfrac{2a}{4}= \dfrac{a}{2}\)
3. Division is performed by inverting the second fraction and then multiplying.
\(\dfrac{2a}{c} \div \dfrac{4}{c} = \dfrac{2a}{c} \times \dfrac{c}{4}= \dfrac{a}{2}\)

Example

Simplify:

1. \(\dfrac{1}{5x} \times 3x\)
2. \(\dfrac{1}{x} \times x\)

Solution

1. Note that \(3x = \dfrac{3x}{1}.\)
Then \(\dfrac{1}{5x} \times 3x = \dfrac{1}{5x} \times \dfrac{3x}{1} = \dfrac{3x}{5x}= \dfrac{3}{5}.\)
2. \(x\) can be written as \( \dfrac{x}{1}.\)
Then \(\dfrac{1}{x} \times x = \dfrac{1}{x} \times \dfrac{x}{1}= \dfrac{x}{x} = 1.\)

Exercise

Leave the expressions.

1. \(\displaystyle{[value]}\)
2. \(\displaystyle{[value]}\)

When simplifying remember only common factors can be cancelled.

Solution
Feedback: 1. \(\displaystyle{[value]} = \dfrac{1}{y} \times \dfrac{x}{1}= \displaystyle{[value]}\) 2. \(\displaystyle{[value]} = \dfrac{y}{x} \times \dfrac{x}{1}= \dfrac{yx}{x} = \displaystyle{[value]}\)

Example

Simplify: \( \dfrac{\dfrac{2x}{y}}{\dfrac{3x}{2y}}\)

Solution

We can write the fraction as \(\dfrac{2x}{y} \div \dfrac{3x}{2y}.\)

Inverting the second fraction and multiplying we find:

\[\dfrac{2x}{y} \times \dfrac{2y}{3x} = \dfrac{4xy}{3xy} = \dfrac{4}{3}\]

Example

Simplify: \[ \dfrac{4x+2}{x^2+4x+3} \times \dfrac{x+3}{7x+5} \]

Solution

Factorising the numerator and denominator we find

\[\dfrac{4x+2}{x^2+4x+3} \times \dfrac{x+3}{7x+5} = \dfrac{2(2x+1)}{(x+1)(x+3)} \times \dfrac{x+3}{7x+5}\]

\[ = \dfrac{2(2x+1)(x+3)}{(x+1)(x+3)(7x+5)}\]

\[= \dfrac{2(2x+1)}{(x+1)(7x+5)}\]

Exercise

Leave the expression in its most simplified form

\(\displaystyle{[value] \div [value]} =\)
Feedback: To divide we invert the second fraction and multiply: \(\displaystyle{[value] \div [value]} = \displaystyle{[value] \times [value]} = [value]\)

Exercise

Simplify

1. \(\displaystyle [value] \div [value]\)
2. \(\displaystyle [value] \div [value]\)
3. \(\displaystyle [value] \div [value]\)
4. \(\displaystyle [value] \div [value]\)

Solution
Feedback: 1. \(\displaystyle [value] \div [value] = [value] \times [value] = [value] = [value]\) 2. \(\displaystyle [value] \div [value] = [value] \times \dfrac[value][value] = [value] = [value]\) 3. \(\displaystyle [value] \div [value] = [value] \times [value] = [value] = [value]\) 4. \(\displaystyle [value] \div [value] = [value] \times [value] = [value] = [value]\)

Exercise

Simplify

1. \([value] \times \dfrac[value][value] =\)
2. \( \displaystyle [value] \times [value]([value]) =\)
3. \( \displaystyle [value] \times ([value]) =\)
Feedback: 1. \(\displaystyle [value] \times \dfrac[value][value] = [value]\) 2. \( \displaystyle [value] \times [value]([value]) = [value]\) 3. \( \displaystyle [value] \times ([value]) = [value]\)

Exercise

Leave the expression in its most simplified form

\( \dfrac{\displaystyle [value]}[value] =\)
Feedback: We can write the fraction as: \(\dfrac{\displaystyle [value]}[value] = \displaystyle [value] \div \dfrac[value][value]\) Inverting the second fraction and multiplying we find: \(\displaystyle [value] \times \dfrac{1}[value] = [value]\)

Exercise

Leave the expression in its most simplified form

\( \displaystyle{ [value] \div [value]} =\)
Feedback: Inverting the second fraction and multiplying we find: \(\displaystyle{ [value] \div [value]} = \displaystyle [value] \times [value] = [value]\)

Leave the expression in its most simplified form

\( \displaystyle{ [value] \div [value]} =\)
Feedback: Inverting the second fraction and multiplying we find: \(\displaystyle{ [value] \div [value]} = \displaystyle [value] \times [value] = [value]\)

Addition and subtraction of algebraic fractions

To add two algebraic fractions the lowest common denominator must be found first. This is the simplest algebraic expression that has the given denominators as its factors. All fractions must be written with this lowest common denominator. Their sum is found by adding the numerators and dividing the result by the lowest common denominator.

To subtract two fractions the process is similar. The fractions are written with the lowest common denominator. The difference is found by subtracting the numerators and dividing the result by the lowest common denominator.

Example

State the simplest expression which has \( x+1\) and \( x+4\) as its factors.

Solution

The simplest expression is \((x+1)(x+4).\) Note that both \(x+1\) and \(x+4\) are factors.

Example

State the simplest expression which has \( x-1\) and \( (x-1)^2\) as its factors.

Solution

The simplest expression is \( (x-1)^2.\) Clearly \( (x-1)^2 \) must be a factor of this expression. Also, because we can write \( (x-1)^2=(x-1)(x-1)\) it follows that \( x-1)\) is a factor too.

Example

Express as a single fraction \( \dfrac{3}{x+1}+\dfrac{2}{x+4} \)

Solution

The simplest expression which has both denominators as its factors is \( (x+1)(x+4).\)

This is the lowest common denominator. Both fractions must be written using this denominator.

Note that \( \dfrac{3}{x+1} \) is equivalent to \(\dfrac{3(x+4)}{(x+1)(x+4)}\) and \(\dfrac{2}{x+4}\) is equivalent to \(\dfrac{2(x+1)}{(x+1)(x+4)}. \)

Thus writing both fractions with the same denominator we have \[ \dfrac{3}{x+1}+\dfrac{2}{x+4} = \dfrac{3(x+4)}{(x+1)(x+4)} + \dfrac{2(x+1)}{(x+1)(x+4)}\]

The sum is found by adding the numerators and dividing the result by the lowest common denominator \[ \dfrac{3(x+4)}{(x+1)(x+4)} + \dfrac{2(x+1)}{(x+1)(x+4)} = \dfrac{3(x+4)+2(x+1)}{(x+1)(x+4)} = \dfrac{5x+14}{(x+1)(x+4)}\]

Key Point

Addition of two algebraic fractions

Step 1: Find the lowest common denominator

Step 2: Express each fraction with this denominator

Step 3: Add the numerators and divide the result by the lowest common denominator

Example

Express \( \dfrac{1}{x-1}+\dfrac{5}{(x-1)^2} \) as a single fraction.

Solution

The simplest expression which has both denominators as its factors is \( (x-1)^2.\) We write both fractions with this denominator.

\[ \dfrac{1}{x-1}+\dfrac{5}{(x-1)^2} = \dfrac{x-1}{(x-1)^2} + \dfrac{5}{(x-1)^2} = \dfrac{x-1+5}{(x-1)^2} = \dfrac{x+4}{(x-1)^2}\]

Example

Express \( \dfrac{5x}{7} - \dfrac{3x-4}{2} \) as a single fraction.

Solution

In this example both denominators are simply numbers. The lowest common denominator is \(14\), and both fractions are re-written with this denominator. Thus

\[ \dfrac{5x}{7} - \dfrac{3x-4}{2} = \dfrac{10x}{14} - \dfrac{7(3x-4)}{14} = \dfrac{10x-7(3x-4)}{14} = \dfrac{28-11x}{14}\]

Exercise

Simplify

1. \(\displaystyle [value] = \)
2. \(\displaystyle [value] = \)
3. \(\displaystyle [value] = \)
4. \(\displaystyle [value] = \)
Feedback: 1. Both denominators are simply numbers. The lowest common denominator is [value], and both fractions are re-written with this denominator: \(\displaystyle [value] = \dfrac[value][value] + \dfrac[value][value] = [value]\) 2. Both denominators are simply numbers. The lowest common denominator is [value], and both fractions are re-written with this denominator: \(\displaystyle [value] = \dfrac[value][value] - \dfrac[value][value] = [value]\) 3. The lowest common denominator is the product of the denominators, [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] + \dfrac[value][value] = [value]\) 4. The lowest common denominator is the product of the denominators, [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] - \dfrac[value][value] = [value]\)

Exercise

Simplify

1. \(\displaystyle [value] = \)
2. \(\displaystyle [value] = \)
3. \(\displaystyle [value] = \)
Feedback: 1. The lowest common denominator is the product of the denominators, [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] + \dfrac[value][value] = [value].\) 2. The lowest common denominator is the product of the denominators, [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] - \dfrac[value][value] = [value].\) 3. The simplest expression having both denominators as its factors is [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] + \dfrac[value][value] = [value].\)

Exercise

Find

\(\displaystyle [value] = \)
Feedback: The simplest expression having both denominators as its factors is [value]. Thus writing both fractions with the same denominator we have: \(\displaystyle [value] = \dfrac[value][value] + \dfrac[value][value] = [value].\)

Exercise

Simplify

1. \(\displaystyle [value] = \)
2. \(\displaystyle [value] = \)
Feedback: 1. Both denominators are simply numbers. The lowest common denominator is [value], and both fractions are re-written with this denominator: \(\displaystyle [value] = \dfrac[value][value] - \dfrac[value][value] + \dfrac[value][value]= [value]\) 2. Both denominators are simply numbers. The lowest common denominator is [value], and both fractions are re-written with this denominator: \(\displaystyle [value] = \dfrac[value][value] - \left( \dfrac[value][value] + \dfrac[value][value] \right) = \dfrac[value][value] - \dfrac[value][value] - \dfrac[value][value] = [value]\)