SOLUTION: Help! Pls answer all with solution and explanation. Thankyou! 1.Edwin is twice as old as Ruel. In three years, Edwin will be 13 less than twice as old as Jill. Find their prese

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: Help! Pls answer all with solution and explanation. Thankyou! 1.Edwin is twice as old as Ruel. In three years, Edwin will be 13 less than twice as old as Jill. Find their prese      Log On

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Question 929840: Help! Pls answer all with solution and explanation.
Thankyou!
1.Edwin is twice as old as Ruel. In three years, Edwin will be 13 less
than twice as old as Jill. Find their present ages.
2. Olan is 3 years older than his sister Mary. One-third of Olan's age
is two years less than one-half his sister's ageone years ago.
Find their present ages.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1.Edwin is twice as old as Ruel. In three years, Edwin will be 13 less
than twice as old as Ruel. Find their present ages.
Equations:
E = 2R
E+3 = 2(R+3)-13
---
Substitute for "E" and solve for "R"::
2R + 3 = 2R +6 - 13
2R + 3 = 2R - 7
That is a contradiction.
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2. Olan is 3 years older than his sister Mary. One-third of Olan's age
is two years less than one-half his sister's ageone years ago.
Find their present ages.
Equations:
O = M + 3
(1/3)O = (1/2)(M-1)-2
---
(1/3)(M+3) = (1/2)M - (3/2)
Multiply thru by 6 to get:
2M+6 = 3M + 9
M = -3
That is a contradiction.
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Cheers,
Stan H.