Question

A father is now 38 years older than his son. Ten years ago , he was twice as old as his son . How old is the father and son now?

Answer 1

Given:

“A father is now 38 years older than his son”

y=x+38

“Ten years ago, he was twice as old as his son.”

y-10=2(x-10)

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Solve for x:

y=x+38

(subtract 38 from both sides)

y-38=x

y-10=2(x-10)

(distribute the 2 into parentheses)

y-10=2x-20

(add 20 to both sides)

y+10=2x

(divide both sides by 2)

(y/2)+5=x

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Set x values equal to each other and solve for y:

x=y-38

x=(y/2)+5

y-38=(y/2)+5

(multiply both sides by 2)

2y-76=y+10

(add 76 to both sides)

2y=y+86

(subtract y from both sides)

y=86

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If y (father’s age) is 86, then x (son’s age) is 48.

This can be checked by rewriting the equations with x and y values inputted.

86=48+38

86-10=2(48-10)

The father is 86 and his son is 48.

Answer 2

Answer:

age of son = 10

age of father = 48

Step-by-step explanation:

father is 38 year older than his son

y= 38+x

solve for x

so, x=y-38

after 10 year ago,

y-10=2(x-10)

add 20 in both side

y+10=2x

x=(y+10)/2

x= (y/2) +5

therfore .

y-38=(y/2)+5

2y-76=y+10

2y=y+86

subtarct the value of 86 in both side

we get y= 48

so

age of father =48+38=86

and age of son = 86-10=2(48-10)

= 48