The number of non-negative integer solutions | vedclass

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(C) Given equations are:$6x + 4y + z = 200$ $(i)$$x + y + z = 100$ $(ii)$Subtracting equation $(ii)$ from $(i)$:$(6x - x) + (4y - y) + (z - z) = 200 -

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$7$

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(C) Given equations are:$6x + 4y + z = 200$ $(i)$$x + y + z = 100$ $(ii)$Subtracting equation $(ii)$ from $(i)$:$(6x - x) + (4y - y) + (z - z) = 200 - 100$$5x + 3y = 100$Since $x$ and $y$ must be non-negative integers,we can express $y$ as $y = \frac{100 - 5x}{3} = \frac{5(20 - x)}{3}$.For $y$ to be an integer,$(20 - x)$ must be a multiple of $3$. Let $20 - x = 3k$,where $k$ is an integer.Then $x = 20 - 3k$. Since $x \ge 0$,$20 - 3k \ge 0 \implies 3k \le 20 \implies k \le 6.66$.Also,since $y \ge 0$,$5(20 - x) \ge 0 \implies x \le 20$. Since $y = 5k$,$k$ must be $\ge 0$.Possible values for $k$ are $0, 1, 2, 3, 4, 5, 6$.For each $k$,we find $x = 20 - 3k$ and $y = 5k$. Then $z = 100 - x - y = 100 - (20 - 3k) - 5k = 80 - 2k$.Since $k \in \{0, 1, 2, 3, 4, 5, 6\}$,$z$ will always be non-negative $(80, 78, 76, 74, 72, 70, 68)$.There are $7$ such solutions.

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

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