document.write( "Question 821759: 5 year ago a man was 7 times of his son 5year hence he was 3 times of his son find their present age \n" ); document.write( "
Algebra.Com's Answer #494451 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
When I solve word problems I often like to set my variable to the smallest unknown (if I know what that is). This way, the other expressions I use will have additions and/or multiplications (which are often easier to work with than subtractions and/or divisions).

\n" ); document.write( "The smallest unknown in this problem would be:
\n" ); document.write( "Let x = the son's age 5 years ago

\n" ); document.write( "Also, this problem can be solved using either 1 or 2 variables. If we use 1 variable we only need one equation. If we use two variables then we need two equations. I will solve this using 1 variable.

\n" ); document.write( "From \"5 year ago a man was 7 times of his son\" we can say
\n" ); document.write( "7x = the man's age five years ago.
\n" ); document.write( "And, if the son was x years old 5 years ago, then:
\n" ); document.write( "x + 5 = son's age now
\n" ); document.write( "And, if the father was 7x years old 5 years ago, then:
\n" ); document.write( "7x + 5 = man's age right now
\n" ); document.write( "And we can rewrite \"5year hence he was 3 times of his son\" (\"hence\" means later) as: \"Now, the man's age is three times his son's age\". And from this we can write the equation:
\n" ); document.write( "7x + 5 = 3(x + 5)

\n" ); document.write( "Now we solve. Simplifying the right side:
\n" ); document.write( "7x + 5 = 3x + 15
\n" ); document.write( "Subtracting 3x from each side:
\n" ); document.write( "4x + 5 = 15
\n" ); document.write( "Subtracting 5 from each side:
\n" ); document.write( "4x = 10
\n" ); document.write( "Dividing by 4:
\n" ); document.write( "\"x+=+10%2F4+=+5%2F2+=+2.5\"

\n" ); document.write( "Looking back we can see that x is the son's age 5 years ago. So his present age would be x + 5 or 7.5. The father's present age, 7x + 5, would be 7(2.5) + 5 = 17.5 + 5 = 22.5. So their present ages are:
\n" ); document.write( "son: 7.5
\n" ); document.write( "father: 17.5

\n" ); document.write( "I strongly suspect that there was something wrong with what you posted:
  • Usually these problems don't end up with fractions/decimals for answers. (I have checked and the answers above are correct if the problem was posted accurately.)
  • The father is quite young! These answers show that the father must have been 15 years old when the son was born!?
I hope that what I've done here will help you solve the correct problem.

\n" ); document.write( "P.S. The other tutor's answer is not correct. After all, 40 is not 3 times 10!!

\n" ); document.write( "P.P.S. If you want to use two variables:
\n" ); document.write( "Let x = the son's age 5 years ago.
\n" ); document.write( "Let y = the father's age 5 years ago.
\n" ); document.write( "This makes their current ages:
\n" ); document.write( "x + 5 = son's current age
\n" ); document.write( "y + 5 = father's current age

\n" ); document.write( "From \"5 year ago a man was 7 times of his son\" we can write:
\n" ); document.write( "y = 7x
\n" ); document.write( "From \"5year hence he was 3 times of his son\" we can write:
\n" ); document.write( "y + 5 = 3(x + 5)

\n" ); document.write( "So we end up with the system:
\n" ); document.write( "y = 7(x + 5)
\n" ); document.write( "y + 5 = 3(x + 5)
\n" ); document.write( "which has the same solution as we found earlier. \n" ); document.write( "
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