Question

How do you solve \[\dfrac{1}{3}y + \dfrac{1}{4} = \dfrac{5}{{12}}\] ?

Answer 1

Hint: The equation is an algebraic equation and here in this question we have to solve this equation. By transforming or by shifting the terms we have to find the value for the variable y. Since the equation is in the form of fraction, all terms which are in the fraction are proper fraction.

Complete step-by-step answer:
The fraction has three kinds namely, proper fraction, improper fraction and the mixed fraction. In proper fraction the value of the numerator is less than the value of the denominator. In an improper fraction the value of the numerator is greater than the value of the denominator. In the mixed fraction, it is a combination of whole number and proper fraction or improper fraction.
Now consider the given equation \[\dfrac{1}{3}y + \dfrac{1}{4} = \dfrac{5}{{12}}\]
All the terms are proper fractions.
Now take \[\dfrac{1}{4}\] to the RHS we have
 \[ \Rightarrow \dfrac{1}{3}y = \dfrac{5}{{12}} - \dfrac{1}{4}\]
To solve this, we have to take LCM for the numbers 12 and 4
The LCM of 12 and 4 are

3	12, 4
2	4, 4
2	2,2
	1,1

Therefore, the LCM of 12 and 4 is \[3 \times 2 \times 2 = 12\]
 \[ \Rightarrow \dfrac{1}{3}y = \dfrac{{\dfrac{5}{{12}} \times 12 - \dfrac{1}{4} \times 12}}{{12}}\]
On simplifying we get
 \[ \Rightarrow \dfrac{1}{3}y = \dfrac{{5 - 3}}{{12}}\]
 \[ \Rightarrow \dfrac{1}{3}y = \dfrac{2}{{12}}\]
Now we multiply the above equation by 3 so we have
 \[ \Rightarrow \dfrac{1}{3} \times 3y = \dfrac{2}{{12}} \times 3\]
In the LHS of the above equation the number 3 will get cancel. The RHS of the above equation we have simplified with the help of tables of multiplication.
So we have
 \[ \Rightarrow y = \dfrac{1}{2}\]
Hence, we have solved the equation and found the value for y.
So, the correct answer is “$ y = \dfrac{1}{2}$”.

Note: To simplify the numbers which are in the form of fraction we must know about the concept of LCM, which means least common factor. While adding or subtracting the fraction the value of the denominator should be the same. If it is different then we solve with the help of LCM concept