Question

How do you subtract $10-6\dfrac{3}{8}$?

Answer 1

Hint: We first convert the mixed fraction of $6\dfrac{3}{8}$ into its mixed fraction form. Then we subtract it from 10. We take the LCM for the integer and the fraction. We find the solution from the subtraction.

Complete step-by-step answer:
Improper fractions are those fractions who have greater value in numerator than the denominator.
We need to convert the mixed fraction which is representation in the form sum of an integer and a proper fraction. We express the process in the form of variables.
Let the mixed fraction be $x\dfrac{c}{b}$. The condition is $c < b$. $x\dfrac{c}{b}$ can be expressed as $x+\dfrac{c}{b}$. Now we express it in the form of improper fraction. Let’s assume the improper fraction is $\dfrac{a}{b}$ where $a>b$.
Then the equational condition will be $\dfrac{a}{b}=x+\dfrac{c}{b}$.
The solution of the equation $x+\dfrac{c}{b}=\dfrac{bx+c}{b}$.
This means $\dfrac{a}{b}=\dfrac{bx+c}{b}$. Therefore, the final condition for the fraction will be $a=bx+c$.
Therefore, $6\dfrac{3}{8}=\dfrac{6\times 8+3}{8}=\dfrac{51}{8}$.
We need to subtract the fraction from the value of 10.
Now the LCM for the subtraction will be 8.
Therefore, $10-6\dfrac{3}{8}=10-\dfrac{51}{8}=\dfrac{10\times 8-51}{8}=\dfrac{29}{8}$.
The subtracted value for $10-6\dfrac{3}{8}$ is $\dfrac{29}{8}$.

Note: We need to remember that the denominator in both cases of improper fraction and the mixed fraction will be the same. The only change happens in the numerator. The relation being the equational representation of $a=bx+c$.