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,\frac{x}{3} + \frac{x}{10} + \frac{x}{12} + 58 = x,

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

20x+6x+5x+5860=60x,20 x + 6 x + 5 x + 58 \cdot 60 = 60 x,

31x=3480,31 x = 3480,

x=120.x = 120.