Question

How do you calculate the subtraction of fraction $6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)?$

Answer 1

Hint: We consider each of these compound fractions or mixed fractions individually. We convert the compound fractions into the simple improper fractions or vulgar fractions. Then we find the difference of these fractions.

Complete step by step solution:
Consider the difference of fractions given here, $6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right).$
Now, we consider each of the compound fractions individually so that they can be evaluated before we go to the main subtraction part.
So, let us consider the first compound fraction in the given subtraction problem.
That is, $6\dfrac{1}{8}.$
We convert this compound fraction into the simple improper fraction by cross multiplying the whole number $6$ with the denominator $8.$ Then, we add the product to the numerator and we divide this sum by the denominator \[8.\]
So, we get,
$\Rightarrow 6\dfrac{1}{8}=\dfrac{6\times 8+1}{8}.$
Therefore, we will get the following,
$\Rightarrow 6\dfrac{1}{8}=\dfrac{48+1}{8}.$
Thus, we will get the following simple improper fraction from the compound fraction,
$\Rightarrow 6\dfrac{1}{8}=\dfrac{49}{8}.$
 Next, we are considering the next compound fraction.
That is, $1\dfrac{2}{3}.$
We convert this compound fraction into the simple fraction.
So, the calculation is as follows:
We, first, find the product of the whole number and the denominator. Then we add this product to the numerator. Finally, we divide the sum by the denominator,
$\Rightarrow 1\dfrac{2}{3}=\dfrac{3\times 1+2}{3}.$
This calculation will give us,
$\Rightarrow 1\dfrac{2}{3}=\dfrac{3+2}{3}=\dfrac{5}{3}.$
Now we consider the given subtraction problem, $6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right).$
When we substitute the corresponding simple fractions in this problem, we get,
$\Rightarrow 6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)=\dfrac{49}{8}-\left( 5-\dfrac{5}{3} \right)$
Consider the terms inside the bracket, $5-\dfrac{5}{3}.$
The difference is,
$\Rightarrow 5-\dfrac{5}{3}=\dfrac{3\times 5-5}{3}$
That is,
$\Rightarrow 5\dfrac{5}{3}=\dfrac{15-5}{3}=\dfrac{10}{3}.$
Now we substitute this in the given problem.
We will get,
$\Rightarrow 6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)=\dfrac{49}{8}-\dfrac{10}{3}.$
We have to cross multiply between the terms to get the difference,
$\Rightarrow 6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)=\dfrac{49}{8}-\dfrac{10}{3}=\dfrac{3\times 49-10\times 8}{8\times 3}$
So, we will get
$\Rightarrow 6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)=\dfrac{147-80}{24}=\dfrac{67}{24}$
Hence the difference is given as $6\dfrac{1}{8}-\left( 5-1\dfrac{2}{3} \right)=2\dfrac{19}{24}.$

Note: We can do this as follows:
$\Rightarrow 6\dfrac{1}{8}=6+\dfrac{1}{8}=\dfrac{48+1}{8}=\dfrac{49}{8}$
\[\Rightarrow \left( 5-1\dfrac{2}{3} \right)=5-\left( 1+\dfrac{2}{3} \right)=5-1-\dfrac{2}{3}=4-\dfrac{2}{3}=\dfrac{12-2}{3}=\dfrac{10}{3}.\]
And, $\dfrac{49}{8}-\dfrac{10}{3}=\dfrac{67}{24}.$