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Question

Two years ago, a father was five times as old as his son. Two years later, his age will be $$8$$ more than three times the age of the son. Find the present ages of father and son.


Solution

Let father age be $$x$$, his son age be $$y$$.
Before $$2$$ years,

$$\Rightarrow$$$$(x-2)=5(y-2)-------(1)$$

After 2 years 
$$\Rightarrow$$$$(x+2)=8+3(y+2)--------(2)$$

From (1) $$x-5y+8=0$$

From (2) 
$$\Rightarrow$$$$x+2-3y-6-8=0$$

$$\Rightarrow$$$$ x-3y-12=0$$

Solving them 
$$\Rightarrow$$$$-2y+20=0$$

$$\Rightarrow$$$$y=10$$

When $$y=10 , x=42$$

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

Maths

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