NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.3 Question 1(iii)

NCERT Class X Chapter 3: Pair of Linear Equations In Two Variables Exercise 3.3 Question 1(iii)

Question:

Solve the following pair of linear equations by the elimination method and the substitution method: 3x – 5y – 4 = 0 and 9x = 2y + 7

Given:

3x – 5y – 4 = 0
9x = 2y + 7

To Find:

Solve the given pair of linear equations using elimination and substitution methods.

Formula:

No specific formula, but we'll use the principles of elimination and substitution.

Solution:

Elimination Method

Rewrite the equations as:
3x – 5y = 4 (1)
9x – 2y = 7 (2)
Multiply (1) by 3: 9x – 15y = 12 (3)
Subtract (2) from (3): (9x – 15y) – (9x – 2y) = 12 – 7 ⇒ -13y = 5 ⇒ y = -513
Substitute y = -513 in (1): 3x – 5(-513) = 4 ⇒ 3x + 2513 = 4 ⇒ 3x = 4 - 2513 = -713 ⇒ x = -739

Substitution Method

From (1), 3x = 5y + 4 ⇒ x = 5y + 43 (3)
Substitute (3) in (2): 9(5y + 43) = 2y + 7 ⇒ 15y + 12 = 2y + 7 ⇒ 13y = -5 ⇒ y = -513
Substitute y = -513 in (3): x = 5(-513) + 43 = -2513 + 43 = -7133 = -739

Result:

There is a mistake in the calculation above. The correct solution is x=1 and y=-1/5. Let's correct the solution: **Elimination Method:** 3x - 5y = 4 (1) 9x - 2y = 7 (2) Multiply (1) by 3: 9x -15y =12 (3) Subtract (2) from (3): -13y = 5 => y = -5/13 Substitute y = -5/13 in (1): 3x -5(-5/13) = 4 => 3x + 25/13 = 4 => 3x = 27/13 => x = 9/13 **Substitution Method:** From (1): 3x = 5y+4 => x = (5y+4)/3 Substitute in (2): 9((5y+4)/3) = 2y+7 => 15y+12 = 2y+7 => 13y = -5 => y = -5/13 Substitute y = -5/13 in x = (5y+4)/3: x = (5(-5/13)+4)/3 = (27/13)/3 = 9/13 Therefore, the solution is x = 9/13 and y = -5/13
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