Question

The number of non-negative integer solutions of the equations $6x+4y+z=200$ and $x+y+z=100$ is

• A. $3$
• B. $5$
• C. $7$
• D. Infinite

Answer 1

The correct option is C $7$

Given that $6x+4y+z=200$ and $x+y+z=100$
Subtracting the second equaiton from the first
$5x+3y=100$
The total number of ordered pairs of $(x,y)=7\left[\right(2,30),(5,25),(8,20),(11,15),(14,10),(17,5),(20,0\left)\right]$

We can choose the value of $z$ accordingly.