Name: Linear Equations With One or More Variables
Uploaded: 2019-07-19
Duration: 11 min 20 s
Description: Let's learn the concept of linear equations, the introduction of linear algebra, the formation of linear equations in one, two and three variables, etc.

Question

Four years ago, a father was six times as old as his son. Ten years later, the father will be two and a half times as old as his son. Determine the present age of father and his son.

Accepted Answer

Let $x$ and $y$ be the present age of son and father respectively.

Now according to the question-

Condition I:- Four years ago father's age was $6$ times that of his son.

$4$ years ago son's age $= x - 4$

$4$ years ago the father's age $= y - 4$

$\Rightarrow y - 4 = 6 (x - 4)$

$\Rightarrow y - 4 = 6 x - 24$

$\Rightarrow 6 x - y = 24 - 4 = 20$ ..........(i)

Condition II:- Ten years later, the father will be two and a half times as old as his son.

Ten years later, son's age $= x + 10$

Ten years later, father's age $= y + 10$

$\Rightarrow y + 10 = 2 \frac{1}{2} (x + 10)$

$\Rightarrow y + 10 = \frac{5}{2} (x + 10)$

$\Rightarrow 2 (y + 10) = 5 (x + 10)$

$\Rightarrow 2 y + 20 = 5 x + 50$

$\Rightarrow 5 x - 2 y = 20 - 50$

$\Rightarrow 5 x - 2 y = - 30$ ..........(ii)

Multiply eq. (i) by $2$ , we get

$12 x - 2 y = 40$ ..........(iii)

Subtract eq. (ii) from (iii), we get

$12 x - 2 y - (5 x - 2 y) = 40 - (- 30)$

$\Rightarrow 12 x - 2 y - 5 x + 2 y = 40 + 30$

$\Rightarrow 7 x = 70$

$\Rightarrow x = 10 years$

Substitute value of $x$ in eq. (i), we get

$6 (10) - y = 20$

$\Rightarrow 60 - y = 20$

$y = 40 years$

Hence, the present age of father and son is $10 years$ and $40 years$ respectively.

Four years ago, a father was six times as old as his son. Ten years later, the father will be two and a half times as old as his son. Determine the present age of father and his son.

Important Questions on Pair of Linear Equations in Two Variables

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