Q. 35.0( 1 Vote )

# Solve the following system of equation by elimination method.

Answer :

The given equations are

x + = 4 … (1)

+ 2y = 5 … (2)

(1) becomes 2x + y = 8

(2) becomes x + 6y = 15

Now, (2) × 2 – (1)

⇒ 2x + 12y – (2x + y) = 30 – 8

⇒ 2x + 12y – 2x – y = 22

⇒ 11y = 22

⇒ y = 2

Substituting y = 2 in (1),

⇒ 2x + 2 = 8

⇒ 2x = 6

⇒ x = 3

∴ (3, 2) is the solution to the given system.

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