Q3 of 227 Page 73

Solve the following system of equation by elimination method.

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.


More from this chapter

All 227 →
1

Solve the following system of equation by elimination method.

x + 2y = 7, x – 2y = 1

 2

Solve the following system of equation by elimination method.

3x + y = 8, 5x + y = 10

 4

Solve the following system of equation by elimination method.

11x – 7y = xy, 9x – 4y = 6xy

 5

Solve the following system of equation by elimination method.