Algebraic Fractions

We need to find a common denominator between all three fractions before we can do the addition and subtraction.

As $30$ is the lowest common multiple of $2$, $3$ & $5$, we will choose $30x$ as the common denominator.

Hence we can multiply each term such that,

$dfrac{1}{2x}+dfrac{1}{3x}-dfrac{1}{5x} = bigg(dfrac{1}{2x}timesdfrac{15}{15}bigg)+bigg(dfrac{1}{3x}timesdfrac{10}{10}bigg)-bigg(dfrac{1}{5x}timesdfrac{6}{6}bigg)$

$begin{aligned} &= dfrac{15}{30x}+dfrac{10}{30x}-dfrac{6}{30x} &= dfrac{15+10-6}{30x} &=dfrac{19}{30x}end{aligned}$

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We need to find a common denominator between the fractions before we can do the addition, hence,

 $dfrac{8}{x}-dfrac{1}{x-3}=dfrac{8(x-3)}{x(x-3)}-dfrac{1(x)}{(x-3)(x)}$

This can be simplified to,

 $dfrac{8x-24-x}{x(x-3)}=dfrac{7x-24}{x(x-3)}$

There are no more common terms so this is fully simplified.

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First, we’ll look at the numerator, before we can factorise it, we must expand the brackets,

$2(8 – k) + 4(k – 1) = 16 – 2k + 4k – 4 = 2k + 12$

Then, the most we can do is take the $2$ out as a factor, leaving us with

$2k + 12 = 2(k + 6)$

Now, the denominator is a special type of quadratic expression referred to as the difference of two squares,

$k^2 – 36 = (k + 6)(k – 6)$

As there is a factor of $(k + 6)$ in both the numerator and denominator, these will cancel.

$dfrac{2cancel{(k + 6)}}{(cancel{k + 6)}(k – 6)} = dfrac{2}{k – 6}$

There are no more common factors, so we are done.

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Our first step when dividing any fractions should be to flip the second fraction over and turn the division into a multiplication.

$dfrac{7ab}{12} div dfrac{4a}{9b^2} = dfrac{7ab}{12} times dfrac{9b^2}{4a}$

Completing the multiplication,

$dfrac{7ab}{12} times dfrac{9b^2}{4a}=dfrac{63ab^3}{48a}$

There is a factor of $a$ that we can cancel and can also take out a factor of $3$ from $63$ and $48$,

$dfrac{3a times 21b^3}{3a times 16} = dfrac{21b^3}{16}$

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To do this subtraction, we need to find a common denominator.

Hence the left-hand fraction must be multiplied by $(z + 5)$ on top and bottom.

$dfrac{z + 2}{z – 1} = dfrac{(z + 2)(z + 5)}{(z – 1)(z + 5)}$

For the right-hand fraction we will multiply $(z – 1)$ on top and bottom.

$dfrac{z}{z + 5} = dfrac{z(z – 1)}{(z – 1)(z + 5)}$

Then, the subtraction is:

$dfrac{z + 2}{z – 1} – dfrac{z}{z + 5} = dfrac{(z + 2)(z + 5)}{(z – 1)(z + 5)} – dfrac{z(z – 1)}{(z – 1)(z + 5)}$  $= dfrac{(z + 2)(z + 5) – z(z – 1)}{(z – 1)(z + 5)}$

Expanding the numerator, we get:

$(z + 2)(z + 5) – z(z – 1) = z^2 + 7z + 10 – z^2 + z = 8z + 10$

Considering the denominator is $(z – 1)(z + 5)$, we can see that there are no common factors, meaning our final answer is:

$dfrac{2(4z + 5)}{(z – 1)(z + 5)}$

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