NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 4

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Example 4

Question:

Solve the following pair of equations by substitution method: \(7x - 15y = 2\) and \(x + 2y = 3\)

Given:

The given pair of equations is:

\(7x - 15y = 2 \)    ...(1)
\(x + 2y = 3 \)    ...(2)

To Find:

Find the values of \(x\) and \(y\) using the substitution method.

Formula:

In the substitution method:

  • Solve one equation for one variable in terms of the other.
  • Substitute this value in the second equation to get the value of one variable.
  • Substitute back to get the other variable.

Solution:

Step 1: Express \(x\) in terms of \(y\) using equation (2).

$$ x + 2y = 3 \\ \Rightarrow x = 3 - 2y \quad ...(3) $$

Step 2: Substitute the value of \(x\) from (3) into equation (1).

$$ 7x - 15y = 2 \\ 7(3 - 2y) - 15y = 2 \\ 21 - 14y - 15y = 2 \\ 21 - 29y = 2 $$

Step 3: Solve for \(y\).

$$ 21 - 29y = 2 \\ -29y = 2 - 21 \\ -29y = -19 \\ y = \frac{-19}{-29} \\ y = \frac{19}{29} $$

Step 4: Substitute \(y = \frac{19}{29}\) into equation (3) to find \(x\).

$$ x = 3 - 2y \\ x = 3 - 2 \left(\frac{19}{29}\right) \\ x = 3 - \frac{38}{29} \\ x = \frac{87}{29} - \frac{38}{29} \\ x = \frac{49}{29} $$

Result:

Therefore, the solution to the given pair of equations is:
\[ x = \frac{49}{29}, \quad y = \frac{19}{29} \]

© Kaliyuga Ekalavya. All rights reserved.

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