SOLUTION: Mr. Dee spent (in this order) 1/4 of his life in England, 1/3 in France, 7 years in Germany, half the remainder of his life in Russia and 1/2 as long in the U.S. as he had spent in
Algebra ->
Equations
-> SOLUTION: Mr. Dee spent (in this order) 1/4 of his life in England, 1/3 in France, 7 years in Germany, half the remainder of his life in Russia and 1/2 as long in the U.S. as he had spent in
Log On
Question 126225: Mr. Dee spent (in this order) 1/4 of his life in England, 1/3 in France, 7 years in Germany, half the remainder of his life in Russia and 1/2 as long in the U.S. as he had spent in France. If he was born in 1908, in what year did he die?
How would this be put into an equation ?
Solved it... by another method... 1992... but need to understand how to work the equation.
Thank you Answer by uma(370) (Show Source):
You can put this solution on YOUR website! Let him have lived for x years.
Then the equation would be:
(x/4)+ (x/3) + 7 + (1/2)[x - (x/4 + x/3 + 7)] + (x/6) = x
==> (1/2)[x + (x/4 + x/3 + 7)] + (x/6) = x
==> [x + (x/4 + x/3 + 7)] + (x/3) = 2x [multiplying throughout by 2]
==> 7 = 2x - x - x/4 - x/3 - x/3
==> 7 = x/12 [ on taking 12 as the common denominator for the right side ans simplifying]
==> 84 = x [ multiplying by 12 on both sides]
Thus he had lived for 84 years.
His year of birth = 1908
So his year of death = 1908 + 84
= 1992
Good luck!!!
(