# Ex.3.7 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

## Question

One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

\( \begin{bmatrix} \textbf{Hint:} \, x + 100 = 2 ({y - 100}),\\ y + 10 = 6 (x - 10) \end{bmatrix} \)

## Text Solution

**Reasoning:**

Assume the friends have ₹ *\(x\)* and ₹ *\(y\)* with them. Then based on given conditions, two linear equations can be formed which can be easily solved.

**Steps:**

Let the first friend has ₹ *\(x\)*

And second friend has ₹ *\(y\)*

Using the information given in the question,

When second friend gives ₹ \(100\) to first friend;

\[\begin{align}x + 100 &= 2\left( {y - 100} \right)\\x + 100 &= 2y - 200\\x - 2y &= - 300 \qquad \qquad \left( 1 \right)\end{align}\]

When first friend gives ₹ \(10\) to second friend;

\[\begin{align}y + 10 &= 6\left( {x - 10} \right)\\y + 10 &= 6x - 60\\6x - y &= 70 \qquad \qquad \left( 2 \right)\end{align}\]

Multiplying equation \((2)\) by \(2,\) we obtain

\[12x - 2y = 140 \qquad {\rm{ }}\left( 3 \right)\]

Subtracting equation \((1)\) from equation \((3),\) we obtain

\[\begin{align}11x &= 440\\x &= \frac{{440}}{{11}}\\x &= 40\end{align}\]

Substituting \(x = 40\) in equation \((1)\), we obtain

\[\begin{align}40 - 2y &= - 300\\2y &= 40 + 300\\y &= \frac{{340}}{2}\\y &= 170\end{align}\]

Therefore, first friend has ₹ \(40\) and second friend has ₹ \(170\) with them.