Question

How do you solve \[\dfrac{x}{3} + \dfrac{x}{2} = 15\]?

Answer 1

Hint: We solve for the value of unknown variable x by taking LCM on the left side of the equation and then cross multiplying the values from both sides of the equation. Shift the required terms to opposite sides and divide with a suitable number to calculate the value of x.
* LCM is the least common multiple of two or more numbers. We write each number in the form of its prime factors and write the LCM of the numbers as multiplication of prime factors with their highest powers.

Complete step-by-step solution:
We are given the equation \[\dfrac{x}{3} + \dfrac{x}{2} = 15\]................… (1)
Since the equation has only one variable i.e. x we will calculate the value of the unknown variable ‘x’.
We will take LCM on the left side of the equation.
The fractions on the left hand side have denominators 3 and 2, so we will calculate LCM of 3 and 2.
Since we know LCM means the least common multiple, we can say that LCM of 3 and 2 is 6 as 6 is the least number that is a multiple of 3 and a multiple of 2 as well.
Take LCM as 6 and write equation (1) as
\[ \Rightarrow \dfrac{{2 \times x + 3 \times x}}{6} = 15\]
Multiply the values in the numerator of left hand side of the equation
\[ \Rightarrow \dfrac{{2x + 3x}}{6} = 15\]
Since both terms in numerator are having variable associated with them as x, we can add the terms
\[ \Rightarrow \dfrac{{5x}}{6} = 15\]
Cancel same factors from both sides of the equation i.e. 5
\[ \Rightarrow \dfrac{x}{6} = 3\]
Cross multiply the values from left side of the equation to right side of the equation
\[ \Rightarrow x = 6 \times 3\]
\[ \Rightarrow x = 18\]

\[\therefore \]The solution of the equation \[\dfrac{x}{3} + \dfrac{x}{2} = 15\] is \[x = 18\].

Note: Many students make the mistake of writing the LCM as 6 but forget to multiply the terms in the numerator with the respective factors of the multiple ( factor multiplied by denominator gives LCM ). Keep in mind we can transform the fractions into fractions with the same denominator by multiplying required value to both numerator and denominator.