# Ex.3.1 Q2 Pair of Linear Equations in Two Variables Solution - NCERT Maths Class 10

## Question

The coach of a cricket team buys \(3\) bats and 6 balls for \(₹\,3900.\) Later, she buys another bat and \(3\) more balls of the same kind for ₹ \(1300.\) Represent this situation algebraically and geometrically.

## Text Solution

**What is Known?**

(i) Three bats and six balls for ₹ \(3900\)

(ii) One bat and three balls for ₹ \(1300\)

**What is Unknown?**

Represent the situation geometrically and algebraically

**Reasoning:**

Assuming the cost of one bat as ₹ \(x\) and the cost of one ball as ₹ \(y,\) two linear equations can be formed for the above situation.

**Steps:**

The cost of \(3\) bats and 6 balls is ₹ \(3900.\)

Mathematically:

\[\begin{align}3x + 6y&= 3900\\3(x + 2y)& = 3900\\x + 2y& = 1300\end{align}\]

Also, the cost of \(1\) bat and \(3\) balls is ₹ \(1300.\)

Mathematically:

\[x + 3y = 1300\]

Algebraic representation where *\(x\)* and *\(y\)* are cost of bat and ball respectively.

\[\begin{align}x + 2y &= 1300 \qquad(1)\\x + 3y &= 1300\qquad(2)\end{align}\]

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

\[\begin{align}x + 2y &= 1300\\2y &= 1300-x\\y &= \frac{{1300 - x}}{2}\end{align}\]

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

\[\begin{align}x + 3y &= 1300\\3y &= 1300-x\\y &= \frac{{1300 - x}}{3}\end{align}\]

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.

\(x\) | \(700\) | \(500\) |

\(y = \frac{{1300 - x}}{2}\) | \(300\) | \(400\) |

\(x\) | \(400\) | \(700\) |

\(y = \frac{{1300 - x}}{3}\) | \(300\) | \(200\) |

The graphical representation is as follows.

Unit: \(1\,\rm{cm} =\) ₹ \(100.\)

The answer is \((1300, \,0)\)