Question #3b032

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Question #3b032

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Answer 1 · 2017-02-03T20:54:28

see explanation.

Explanation:

If you check again it appears that the question is.

$\frac{5 x}{6} - \frac{3}{10} and not \frac{5}{x} - \frac{3}{10}$ $(5x)/6-3/10" and not " 5/x-3/10$

This accounts for the LCD of 30

Before we can add/subtract fractions they must have a $common denominator$ $color(blue)"common denominator"$ That is they have to be the same number on the denominator of both fractions.

We have to find a number then that is common to both 6 and 10, usually called the $lowest common multiple$ $color(blue)"lowest common multiple"$

One way to find it, is to divide the larger of the 2 numbers by the smaller and if it divides into it exactly , with a remainder of 0 then the larger number is the lowest common multiple. Repeat this process with multiples of the larger number until this happens.

$10 \div 6 = 1 remainder 4 so not 10 , now try 20$ $10÷6=1" remainder 4 so not 10 , now try 20"$

$20 \div 6 = 3 remainder 2 so not 20, now try 30$ $20÷6=3" remainder 2 so not 20, now try 30"$

$30 \div 6 = 5 remainder 0$ $30÷6=5" remainder 0"$ thus 30 is the value.

The 2 fractions can now be expressed with a LCD of 30

$\Rightarrow \frac{5 x}{6} \times \frac{5}{5} = \frac{25 x}{30} and \frac{3}{10} \times \frac{3}{3} = \frac{9}{30}$ $rArr(5x)/6xx5/5=(25x)/30" and " 3/10xx3/3=9/30$

Now that the fractions have the same denominator we can add/subtract the numerators but NOT the denominators.

$\Rightarrow \frac{5 x}{6} - \frac{3}{10} = \frac{25 x}{30} - \frac{9}{30} = \frac{25 x - 9}{30}$ $rArr(5x)/6-3/10=(25x)/30-9/30=(25x-9)/30$

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Question #3b032

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