Question

2 years ago a father was five times as old as his son two years later from today his age will be eight years more than three times the age of his son find their present ages ​

Answer 1

Answer:

Let father age be x, his son age be y.

Before 2 years,

⇒(x−2)=5(y−2)−−−−−−−(1)

After 2 years

⇒(x+2)=8+3(y+2)−−−−−−−−(2)

From (1) x−5y+8=0

From (2)

⇒x+2−3y−6−8=0

⇒x−3y−12=0

Solving them

⇒−2y+20=0

⇒y=10

When y=10,x=42

So the present age of the father is 42 and the son age is 10.

Step-by-step explanation:

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Answer 2

SOLUTION :-

Let,

Present age of son = x

Present age of father = y

Two years ago,

Age of son = x - 2

Age of father = y - 2

After two years,

Age of son = x + 2

Age of father = y + 2

According to the first condition,

$\longrightarrow\sf 5(x-2) = y-2 \\\\\longrightarrow 5x-10=y-2\\\\\longrightarrow 5x-y = 8 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

According to the second condition,

$\longrightarrow\sf 3(x+2)+8 = y+2 \\\\\longrightarrow 3x+6+8= y+2 \\\\\longrightarrow 3x-y = -12 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(2)$

Eq(2) - Eq(1),

$\longrightarrow\sf 2x= 20\\\\\longrightarrow \bf x= 10$

Substitute the value of x in equation (1),

$\longrightarrow\sf 5\times 10-y = 8 \\\\\longrightarrow 50-y=8\\\\\longrightarrow -y= -42\\\\\longrightarrow \bf y = 42$

Present age of son = 10 years

present age of father = 42 years