Ex.3.1 Q1 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

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Question

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

 Video Solution
Pair Of Linear Equations In Two Variables
Ex 3.1 | Question 1

Text Solution

What is Known?

\(7\) years ago, Aftab was \(7\) times as old as his daughter then and \(3\) years from now, Aftab shall be \(3\) times as old as his daughter will be.

What is Unknown?

Represent the situation algebraically and graphically.

Reasoning:

Assume the present age of Aftab be \(x\)  years and his daughter be \(y\) years then represent their ages \(7\) years later and \(3\) years ago in term of \(x\) and \(y.\) Two linear equations can be formed to represent the above situation algebraically.

Using algebraic equation and truth table they can be graphically represented.

Steps:

(i) Present age of Aftab \( = x\) years and his daughter \( = y\) years

Therefore, 7 years ago, age of Aftab \( = \left( {x - 7} \right)\) years and his daughter \( = \left( {y - 7} \right)\)years

Using this information and applying the known condition that \(7\) years ago, Aftab was \(7\) times as old as his daughter then:

\[\begin{align}x - 7 &= 7(y - 7)\\x - 7 &= 7y - 49\\x - 7y - 7 + 49 &= 0\\x - 7y + 42 &= 0\end{align}\]

After \(3\) years from now, age of Aftab \( = \left( {x + 3} \right)\) years and his daughter \( = \left( {y + 3} \right)\) years and also Aftab will be \(3\) times as old as his daughter. Then mathematically,

\[\begin{align}x + 3 &= 3\left( {y + 3} \right)\\x + 3 &= 3y + 9\\x - 3y + 3 - 9 &= 0\\x - 3y - 6 &= 0\end{align}\]

Algebraic representations, where \(x\) and \(y\) are present ages of Aftab and his daughter respectively:

\[\begin{align}x - 7y + 42 &= 0\qquad(1)\\x - 3y-6 &= 0\qquad(2)\end{align}\]

Therefore, the algebraic representation is for equation \(1\) is:

\[\begin{align}x - 7y + 42 &= 0\\- 7y &= - x - 42\\7y &= x + 42\\y &= \frac{{x + 42}}{7}\end{align}\]

And, algebraic representation for equation \((2)\) is:

\[\begin{align}x - 3y - 6 &= 0\\- 3y &= - x + 6\\3y &= x - 6\\y &= \frac{{x - 6}}{3}\end{align}\]

Let us represent these equations graphically. For this, we need at least two solutions for each equation. We give these solutions in table shown below.

For equation \((1)\)

\(x\)

\(21\)

\(28\)

\(y = \frac{{x + 42}}{7}\)

\(9\)

\(10\)

For equation (2)

\(x\) \(30\) \(21\)
\(y = \frac{{x - 6}}{3}\) \(8\) \(5\)

The graphical representation is as follows.

Unit: \(1 \,\rm{cm}=5\) years.

The answer is \((42, \;12)\)

  
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