Question

Solve the following pair of linear equations by substitution method:
$x+3y=4$
$2x+6y=8$

• A. $x=4,y=3$
• B. $x=1,y=1$
• C. Infinitely many solution
• D. $x=4,y=0$

Answer 1

The correct option is C Infinitely many solution

$x+3y=4$.........(1)
$2x+6y=8$ .......(2)
from equation (1),
$x+3y=4\Rightarrow x=4-3y$............(3)
Substituting $x=4-3y$ in equation (2)
$2x+6y=8\Rightarrow 2(4-3y)+6y=8$
$\Rightarrow 8-6y+6y=8$
$\Rightarrow 8=8$
So the given system of equations has infinite number of solutions:
Every solution of $x+3y=4$ is a solution of $2x+6y=8$
Some of the solutions are $(4,0),(1,1),....$