Question

How do you simplify $\dfrac{3}{12}-\dfrac{1}{8}?$

Answer 1

Hint: We will use the cross multiplication to find the difference of the fractions. We will find the product of the numerator of the first fraction and the denominator of the second fraction and subtract it from the product of the numerator of the second fraction and the denominator of the first fraction. Then we divide the value by the product of the denominators.

Complete step by step solution:
Let us consider the given problem where we are asked to find the difference of two fractions which have different denominators.
Consider $\dfrac{3}{12}-\dfrac{1}{8}.$
Now, let us use the cross multiplication to find the difference of the fractions $\dfrac{3}{12}$ and $\dfrac{1}{8}.$
Here, we are going to multiply the numerator of the first fraction, $3,$ with the denominator of the second fraction, $8.$ Then we will find the product of the numerator of the second fraction, $1,$ with the denominator of the first fraction, $12.$ Then, we will find the difference of these two products and we will subtract the second product from the first product. Then we divide this difference by the product of the denominators $12$ and $8.$
So, we will get, $3\times 8=24.$
And then, we will get $1\times 12=12.$
The product of the denominators is $12\times 8=96.$
So, we will get $\dfrac{3}{12}-\dfrac{1}{8}=\dfrac{24-12}{96}$
We know that $24-12=12.$
Therefore, we will get $\dfrac{3}{12}-\dfrac{1}{8}=\dfrac{12}{96}.$
Also, since $12\times 8=96,$ we will get $\dfrac{96}{12}=8$ and thus the reciprocal of this quotient will give us $\dfrac{12}{96}=\dfrac{1}{8}.$
Hence the difference is $\dfrac{3}{12}-\dfrac{1}{8}=\dfrac{1}{8}.$

Note: An alternative and relatively easy method to find this is given below:
Simplify the first fraction, $\dfrac{3}{12}=\dfrac{1}{4}.$ Then, we will get $\dfrac{1}{4}-\dfrac{1}{8}.$ We will make both the denominators the same by multiplying the numerator and the denominator of the first fraction with $2.$ We will get $\dfrac{2}{8}-\dfrac{1}{8}=\dfrac{1}{8}.$