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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