2200.21 – An Equation with Fractions


Find a number that is 58 more than the sum of its third, tenth, and twelfth parts.


Solution

Let xx represent the target number. Then,

x3+x10+x12+58=x\dfrac{x}{3} + \dfrac{x}{10} + \dfrac{x}{12} + 58 = x

The least common denominator is 60. To clear fractions, we multiply by it:

20x+6x+5x+5860=60x31x=3480x=120\begin{align*}20 x + 6 x + 5 x + 58 \cdot 60 &= 60 x \\31 x &= 3480 \\x &= 120\end{align*}