SOLUTION: The sum of James’ age and David’s age is 34. Five years ago, the sum of twice James’ age and three times David’s age was 61. What are their present ages

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Question 1109968: The sum of James’ age and David’s age is 34. Five years ago, the sum of twice James’ age and three times David’s age was 61. What are their present ages
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616)   (Show Source): You can put this solution on YOUR website!

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
The condition gives you two equations

J + D = 34,                  (1)
2*(J-5) + 3*(D-5) = 61.      (2)


Simplify (2):

2J - 10 + 3D - 15 = 61,

2J + 3D = 61 + 10 + 15,

2J + 3D = 86.                (3)


From (1),  J = 34-D.  Substitute it into (3). You will get the single equation for only one unknown D:

2*(34-D) + 3D = 86,         (It is how the Substitution method works).

68 - 2D + 3D = 86  ====>  D = 86 - 68 = 18.


Answer.  David is 18 years old.   James is 34-18 = 16 years old.


Check.   2*(16-5) + 3*(18-5) = 2*11 + 3*13 = 22 + 39 = 61.

Solved.   //   On the way,  you learned on how the Substitution method works.

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There is a bunch of lessons on age word problems
    - Age problems and their solutions
    - A fresh formulation of a traditional age problem
    - Really intricate age word problem
    - Selected age word problems from the archive
    - Age problems for mental solution
in this site.

Read them and become an expert in solving age problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Age word problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

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