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Mr. Jain is 4 times as old as his son. After 10 years he will be twice as old as his son. Find Mr. Jain and his son's present ages.

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    $\begingroup$ $$\dfrac{4x+10}{x+10}=2$$ $\endgroup$ – lab bhattacharjee Mar 2 '17 at 11:24
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A direct retranscript of the statement is $$j=4s,\\j+10=2(s+10).$$

By substitution,

$$4s+10=2s+20,$$ giving $5$ and $20$ (and later $15$ and $30$).

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    $\begingroup$ That's a young parent. Not medically impossibly young, but perhaps quite irresponsibly young. At least in modern western society. $\endgroup$ – Arthur Mar 2 '17 at 11:41
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If we set $j$ as the current age of Mr. Jain and $s$ as the current age of his son, we can then write the statements as two linear equations

\begin{align} j&=4s\\ (j+10)&=2(s+10) \end{align}

We can then solve this by substituting $j=4s$ into the second equation

\begin{align} (j+10) &= 2(s+10) \\ (4s+10)&=2(s+10) \\ 4s+10&=2s+20 \\ 2s&=10\\ s &= 5 \end{align}

We can then say that \begin{align}j&=4s \\ j &= 4\times 5 \\ j &= 20 \end{align}

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How to model the statements:

Mr. Jain is 4 times as old as his son.

$$ J = 4 S \quad (1) $$

After 10 years he will be twice as old as his son.

$$ J + 10 = 2 (S + 10) \quad (2) $$

These are linear equations. One way to solve this system is inserting equation $(1)$ into equation $(2$): $$ 4S + 10 = 2 (S + 10) \iff \\ 2 S = 10 \iff \\ S = 5 $$ which then is used with equation $(1)$ to give $$ J = 4\cdot 5 = 20 $$

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