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(D) $1$. The line passing through $A(7, 0)$ and $(0, 7)$ has the equation $\frac{x}{7} + \frac{y}{7} = 1$,which simplifies to $x + y = 7$. Since the s

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$x+y \leq 7, 2x-3y+6 \geq 0, x \geq 0, y \geq 0$

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(D) $1$. The line passing through $A(7, 0)$ and $(0, 7)$ has the equation $\frac{x}{7} + \frac{y}{7} = 1$,which simplifies to $x + y = 7$. Since the shaded region is towards the origin,the inequality is $x + y \leq 7$.$2$. The line passing through $C(0, 2)$ and $B(3, 4)$ has the slope $m = \frac{4-2}{3-0} = \frac{2}{3}$. The equation is $y - 2 = \frac{2}{3}(x - 0)$,which simplifies to $3y - 6 = 2x$,or $2x - 3y + 6 = 0$. Testing the point $(3, 0)$ which is in the shaded region: $2(3) - 3(0) + 6 = 12 \geq 0$. Thus,the inequality is $2x - 3y + 6 \geq 0$.$3$. Since the region is in the first quadrant,$x \geq 0$ and $y \geq 0$.$4$. Combining these,the system is $x + y \leq 7, 2x - 3y + 6 \geq 0, x \geq 0, y \geq 0$. Therefore,option $(D)$ is correct.

The shaded region in the figure is the solution of which of the following systems of inequations?

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