In the system $8y − 3x ≤ 16, 3x + 8y ≥ −18$ for which solution $(x, y)$ is $x + y$ least? In the system $8y − 3x ≤ 16, 3x + 8y ≥ −18$ for which solution $(x, y)$ is $x + y$ least?
The correct answer was $(-6,-4)$, but I don't know why. I tried graphing it and solving for the variables but that didn't seem to help much.
 A: The region defined by the two lines is unbounded in the first and fourth quadrants, where $x>0.$ Clearly, the sum $x+y$ increases here as $x\to+\infty.$ To see this, write the first line as $$y=\frac{16+3x}{8}$$ and the second as $$y=\frac{-18-3x}{8}.$$ Clearly as $x\to+\infty,$ we have $y\to+\infty$ in the first equation, so that the sum must go to $+\infty$ too. For the second, compute $$x+y= \frac{-18}{8}-\frac{3x}{8}+x=\frac{-18}{8}+\frac{5x}{8},$$ which also goes to $+\infty$ as $x\to+\infty.$
Now, first note that these lines have slopes that are negatives of each other; also, they meet at the point $(-17/3,-1/8),$ the sharp point of the region, as it were. I shall now show is that for any point in the region, as $x\to+\infty,$ the sum also goes to positive infinity. We do this by considering the pencil of lines pivoted at the sharp point of the region, and with slopes varying between $-3/8$ and $3/8.$ Such lines have the form $$y+\frac18=m\left(x+\frac{17}{3}\right),$$ where $m$ ranges in the interval $[-3/8,3/8].$ All these however are similar to the case for the boundary lines; it is obvious for $m\ge 0$ that the claim holds. For $m<0,$ it is obvious once you calculate the sum $x+y.$ Thus, the minimum must exist as $x\to-\infty.$ In particular, it is in the third quadrant, and somewhere along the boundary line $8y+3x=-18$ in the interval $-17/3\le x\le 0.$ From the equation the sum $$x+y=\frac{-18}{8}+\frac{5x}{8},$$ as before. As $x$ increases, this increases, and decreases with decreasing $x.$ But the least value of $x$ in this interval is $-17/3.$ Thus, the least sum $x+y$ is given by $$\frac{-18}{8}+\frac{5\left(\frac{-17}{3}\right)}{8}=-\frac{139}{24},$$ which is just a little bit over $-6.$ Thus, the given answer (being $-10$) is outright wrong. Another way to see this is that the point $(-6,-4)$ does not satisfy the system. Who gave this answer?
A: 
From the graph above, the red line represents $8y-3x=16$, the blue line represents $3x+8y=-18$. 
From the inequality above, we are looking at the region below the red line and above the blue line. Assume that we are only considering integer coordinates. From the graph, the important point is $(-5,0)$. Since the slope of $3x+8y=-16$ is greater than the slope of $y=x$. This means if we take points down the blue line, $y+x$ will be increasing.
Therefore the solution is $(-5,0)$ for least $x+y$
