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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 age of father and son.

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Hint- Assume father’s present age and son’s present age then make equations using given information. Then solve Pair of linear equations in two variables.

Let, father’s present age $ = x$ years
son’s present age $ = y$ years
Considering $2$ years ago,
$ \Rightarrow \left( {x - 2} \right) = 5\left( {y - 2} \right){\text{ - - - - - - - (1)}}$
Considering $2$ years later,
$ \Rightarrow \left( {x + 2} \right) = 8 + 3\left( {y + 2} \right){\text{ - - - - - - - (2)}}$
Rewriting equation $\left( 1 \right)$as:
$
   \Rightarrow x - 2 - 5y + 10 = 0 \\
   \Rightarrow x - 5y + 8 = 0{\text{ - - - - - - - - (3)}} \\
$
Rewriting equation$\left( 2 \right)$as:
$
   \Rightarrow x + 2 - 3y - 6 - 8 = 0 \\
   \Rightarrow x - 3y - 12 = 0{\text{ - - - - - - - - (4)}} \\
$
Now, we have two equations and two variables.
By using subtraction method,
Subtracting $\left( 4 \right)$from$\left( 3 \right)$, we get:
$
   \Rightarrow x - 5y + 8 - \left( {x - 3y - 12} \right) = 0 - 0 \\
   \Rightarrow - 2y + 20 = 0 \\
   \Rightarrow y = \dfrac{{20}}{2} \\
   \Rightarrow y = 10 \\
$
Putting $y = 10$ in $\left( 4 \right)$, we get:
$
   \Rightarrow x - 3\left( {10} \right) - 12 = 0 \\
   \Rightarrow x - 30 - 12 = 0 \\
   \Rightarrow x - 42 = 0 \\
   \Rightarrow x = 42 \\
$
Therefore, father’s present age $ = 42$ years and son’s present age $ = 10$ years

Note- Always let the present ages be some unknown variable and try to write the information given in the problem in form of equations. The three methods most commonly used to solve systems of equations are substitution, elimination and augmented matrices.