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

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Question 697895: Find a number that is 58 more than the sum of its third, tenth, and twelfth parts?
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

a number x that is 58+more than the sum of its third=x%2F3, tenth=x%2F10, and twelfth=x%2F12 parts;
so, we have
x=x%2F3%2Bx%2F10%2Bx%2F12%2B58....LCD for 3,10, and 12 is 60
x=20x%2F%2820%2A3%29%2B6x%2F%286%2A10%29%2B5x%2F%285%2A12%29%2B58
x=20x%2F60%2B6x%2F60%2B5x%2F60%2B58...both sides multiply by 60
60x=20x%2B6x%2B5x%2B58%2A60
60x=31x%2B3480
60x-31x=3480
29x=3480
highlight%28x=120%29

check:
the sum of x%2F3%2Bx%2F10%2Bx%2F12=120%2F3%2B120%2F10%2B120%2F12=40%2B12%2B10=62
a number x=120%7D%7D++is+%7B%7B%7B58 more than the sum:
120-62=58 which shows us it's true