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

## 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.

## 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 \((42, \;12)\)