solve by linear equation of one variable.5years ago a man was 7times as old as his son.5years hence he is 3 times as old as his son.find their present ages?
2006-08-26 20:31:38 · 6 answers · asked by atul_berry 1 in Education & Reference ➔ Homework Help
let the present ages of the man and the son be y and x
the equations are
(y-5)=7(x-5)=> 7x-y=30
(y+5)=3(x+5)=>3x-y=-10
4x=40 and so x=10
plugging in y=40
the present ages of the man and the son
are 40 yrs and 10 yrs respectively
2006-08-27 00:10:27 · answer #1 · answered by raj 7 · 0⤊ 0⤋
by setting up the following equations:
1) x = 3*y
2)x-5 = 7*(y-5)
solving for x, the age of the father gives you the following ages:
Father: 22.5 years
Son: 7.5 years
thanks for the 10 points!
2006-08-26 20:57:26 · answer #2 · answered by Anonymous · 0⤊ 0⤋
dad=45
son=15
previous ages:
dad=35
son=5
2006-08-26 20:43:08 · answer #3 · answered by piece of mind 1 · 0⤊ 0⤋
Please check up whether the sum is right.If u want i can solve it in 2 variables but i wonder whether u will understand.
2006-08-26 20:43:49 · answer #4 · answered by devil'sincarnate666 2 · 0⤊ 0⤋
let the age of son 5 yrs ago be x yrs
thus, fathers age =7x
ATP,
7x +5=3*7x...................maybe
2006-08-26 20:39:01 · answer #5 · answered by Anonymous · 0⤊ 0⤋
SORRY I'M VERY YOUNG
2006-08-29 04:06:09 · answer #6 · answered by mirchi girl 3 · 0⤊ 0⤋