Question

How do you simplify $\dfrac{1}{2}-\left( \dfrac{1}{8}+\dfrac{1}{8} \right)$ ?

Answer 1

Hint: We can see there are 2 fractions inside the bracket, so first we have to solve the expression inside the bracket. So, we will add $\dfrac{1}{8}+\dfrac{1}{8}$ then we will subtract then result from $\dfrac{1}{2}$ . While adding 2 fractions, if the denominator is the same for 2 fractions, we can simply add 2 numerators keeping the denominator unchanged. If the denominator is different then we have to make it the same.

Complete step by step solution:
The given term is $\dfrac{1}{2}-\left( \dfrac{1}{8}+\dfrac{1}{8} \right)$
So we will add $\dfrac{1}{8}+\dfrac{1}{8}$ first , we can see that the denominator is same for both the fraction. So we can simply add the 2 fractions.
We can write
$\Rightarrow $ $\dfrac{1}{8}+\dfrac{1}{8}$ = $\dfrac{2}{8}$
In the fraction $\dfrac{2}{8}$ we can cancel out 2 from numerator and denominator
So we can write
$\Rightarrow \dfrac{2}{8}=\dfrac{1}{4}$
The term $\dfrac{1}{2}-\left( \dfrac{1}{8}+\dfrac{1}{8} \right)$ is equal to $\dfrac{1}{2}-\dfrac{1}{4}$
Now we can see the denominators in the 2 fractions are different, we have to change the denominator of $\dfrac{1}{2}$ to 4.
$\Rightarrow \dfrac{1}{2}=\dfrac{2}{4}$
So we can say
 $\Rightarrow \dfrac{1}{2}-\dfrac{1}{4}=\dfrac{2}{4}-\dfrac{1}{4}$
Now we can simply subtract 1 from 2.
$\Rightarrow \dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}$
$\Rightarrow \dfrac{1}{2}-\left( \dfrac{1}{8}+\dfrac{1}{8} \right)=\dfrac{1}{4}$

Note: In the multiplication of 2 fractions we can simply multiply the 2 numerators and 2 denominators. For example, the product of $\dfrac{a}{b}$ and $\dfrac{x}{y}$ is equal to $\dfrac{ax}{by}$ . In the division of 2 fractions, we find the reciprocal of divisor and multiply with the dividend. For example, in $\dfrac{a}{b}\div \dfrac{x}{y}$ we find the reciprocal of $\dfrac{x}{y}$ which is $\dfrac{y}{x}$. Product of $\dfrac{a}{b}$ and $\dfrac{y}{x}$ is equal to $\dfrac{ay}{bx}$ . So $\dfrac{a}{b}\div \dfrac{x}{y}$ is equal to $\dfrac{ay}{bx}$.