document.write( "Question 507474: ten years ago the age of a father was four times of his son . Ten years hence the age of the father will be twice that of his son . The present ages of the father and the son are. \n" ); document.write( "

Algebra.Com's Answer #340503 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let D = Dad's present age and S = Son's present age.
\n" ); document.write( "1) D-10 = 4(S-10) \"Ten years ago the age of a father (D-10) was four times the age of his son (4(S-10)).\"
\n" ); document.write( "2) D+10 = 2(S+10) \"Ten years hence, the age of the father (D+10) will be twice that of his son ((2(S+10)).\"
\n" ); document.write( "So now you have two equations with two unknowns (D & S). Simplify the two equations:
\n" ); document.write( "1a) D-10 = 4S-40
\n" ); document.write( "2a) D+10 = 2S+20 Subtract equation 2a) from equation 1a) to get:
\n" ); document.write( "3) -20 = 2S-60 Now add 60 to both sides.
\n" ); document.write( "3a) 40 = 2S Divide both sides by 2.
\n" ); document.write( "3b) 20 = S or S = 20 Now substitute this value of S into 1a) or 2a) and solve for D.
\n" ); document.write( "2a) D+10 = 2S+20 Substitute S = 20
\n" ); document.write( "2b) D+10 = 2(20)+20 Simplify.
\n" ); document.write( "3c) D+10 = 60 Subtract 10 from both sides.
\n" ); document.write( "3d) D = 50
\n" ); document.write( "The father is 50 years old and the son is 20 years old. \n" ); document.write( "

\n" );