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Pair Of Linear Equations In Two Variables
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Solve the following pair of equations by substitution method: 7x – 15y = 2 (1) x + 2y = 3 (2)

Step-by-Step Solution

Key Concept: In the substitution method, one variable is expressed from one equation and substituted into the other equation, reducing the system to a single‑variable linear equation.
1. Express one variable from the simpler equation (2):
$$x + 2y = 3 \;\Rightarrow\; x = 3 - 2y.$$

2. Substitute this expression for \(x\) into equation (1):
$$7x - 15y = 2 \;\Rightarrow\; 7(3 - 2y) - 15y = 2.$$

3. Simplify the resulting equation:
$$7\times3 - 7\times2y - 15y = 2$$
$$21 - 14y - 15y = 2$$
$$21 - 29y = 2.$$

4. Solve for \(y\):
$$-29y = 2 - 21$$
$$-29y = -19$$
$$y = \frac{-19}{-29} = \frac{19}{29}.$$

5. Find \(x\) using the expression from step 1:
$$x = 3 - 2y = 3 - 2\left(\frac{19}{29}\right)$$
$$x = \frac{87}{29} - \frac{38}{29} = \frac{49}{29}.$$

6. Write the ordered pair:
$$\boxed{\left( x, y \right) = \left( \frac{49}{29}, \; \frac{19}{29} \right)}.$$

7. Verification (optional):
$$7\left(\frac{49}{29}\right) - 15\left(\frac{19}{29}\right) = \frac{343 - 285}{29} = \frac{58}{29} = 2,$$
$$\left(\frac{49}{29}\right) + 2\left(\frac{19}{29}\right) = \frac{49 + 38}{29} = \frac{87}{29} = 3.$$
Both equations are satisfied, confirming the solution.

Correct Answer: (x, y) = (49/29 , 19/29)
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