NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 6

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 6

Question:

In a shop the cost of 2 pencils and 3 erasers is ₹9 and the cost of 4 pencils and 6 erasers is ₹18. Find the cost of each pencil and each eraser.

Given:

  • The cost of 2 pencils and 3 erasers is ₹9.
  • The cost of 4 pencils and 6 erasers is ₹18.

To Find:

  • The cost of one pencil.
  • The cost of one eraser.

Formula:

  • Let the cost of one pencil be ₹\( x \).
  • Let the cost of one eraser be ₹\( y \).
  • Form equations using the given information:

$$ \begin{align*} 2x + 3y &= 9 \\ 4x + 6y &= 18 \end{align*} $$

  • Solve the pair of linear equations to find \( x \) and \( y \).

Solution:

Step 1: Let the cost of one pencil be ₹\( x \) and the cost of one eraser be ₹\( y \). Write the equations from the given statements.

$$ \begin{align*} 2x + 3y &= 9 \quad \cdots (1) \\ 4x + 6y &= 18 \quad \cdots (2) \end{align*} $$

Step 2: Observe that equation (2) is a multiple of equation (1).

$$ 4x + 6y = 2 \times (2x + 3y) = 2 \times 9 = 18 $$

So, both equations represent the same line and have infinitely many solutions.

Step 3: Express \( x \) in terms of \( y \) using equation (1).

$$ \begin{align*} 2x + 3y &= 9 \\ 2x &= 9 - 3y \\ x &= \frac{9 - 3y}{2} \end{align*} $$

Step 4: Assign a value to \( y \) to find one possible solution. Let \( y = 1 \) (cost of one eraser = ₹1).

$$ x = \frac{9 - 3 \times 1}{2} = \frac{9 - 3}{2} = \frac{6}{2} = 3 $$

So, the cost of one pencil is ₹3.

Step 5: Verify the solution in both equations.

$$ \begin{align*} 2x + 3y &= 2 \times 3 + 3 \times 1 = 6 + 3 = 9 \\ 4x + 6y &= 4 \times 3 + 6 \times 1 = 12 + 6 = 18 \end{align*} $$

Both equations are satisfied.

Result:

  • The cost of one pencil is ₹3.
  • The cost of one eraser is ₹1.
  • There are infinitely many solutions, but one possible answer is: Pencil = ₹3, Eraser = ₹1.
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