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# Simplify: $\dfrac{2x}{5}+\dfrac{3x}{4}-\dfrac{3x}{5}$

Last updated date: 22nd Feb 2024
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Hint: LCM (least common factor) is the smallest positive integer that is divisible by all the denominators of all the given fractions. A fraction represents a part of a whole or more generally, any number of equal parts whereas in mathematics a fraction is a numerical quantity that is not a whole number. For e.g. $\dfrac{2}{3},\dfrac{5}{9}$ etc.

To simplify $\dfrac{2x}{5}+\dfrac{3x}{4}-\dfrac{3x}{5}$ first we have to find the LCM of their denominators.
So, we have to find the LCM of 5, 4 and 5.
LCM of 5,4 and 5 would be the same as the LCM of 4 and 5. Now we have to write the prime factors of each number. So,
5,4 = 5*2*2
5,4 = 20
So, 20 is the LCM of 4 and 5.
As 20 is the smallest positive integer that is divisible by 5, 4 and 5. So 20 is the LCM.
So $\dfrac{2x}{5}+\dfrac{3x}{4}-\dfrac{3x}{5}$ can be written as
$\dfrac{\{2x(4)+3x(5)-3x(4)\}}{20}$
Now by using the BODMAS (Brackets, order, division, multiplication, addition and subtraction) rule we would solve the brackets first.
So $\dfrac{\{2x(4)+3x(5)-3x(4)\}}{20}$ would become
$\dfrac{(8x+15x-12x)}{20}$
Now again we have to use the BODMAS rule according to which we have to solve the values between the brackets first.
We also have to use the BODMAS rule on the terms inside the bracket, so we would do addition first.
So, $\dfrac{(8x+15x-12x)}{20}$ becomes
$\dfrac{(23x-12x)}{20}$
Now we have to do subtraction
So, $\dfrac{(23x-12x)}{20}$ would become
$\dfrac{11x}{20}$
Hence by simplifying $\dfrac{2x}{5}+\dfrac{3x}{4}-\dfrac{3x}{5}$ it becomes $\dfrac{11x}{20}$ .
So, the correct answer is “ $\dfrac{11x}{20}$ ”.

Note: Use the BODMAS (Brackets, order, division, multiplication, addition and subtraction) rule to simplify step by step and without making any calculation mistakes. To find LCM first find all the prime factors of each given number, then list all the prime numbers found, as many times as they occur most often for each number, then multiply the list of prime factors together to finally find the LCM.