shrouded1 wrote:
metallicafan wrote:
In the xy-plane, region R consists of all the points (x, y) such that \(2x + 3y =< 6\) . Is the point (r,s) in region R?
(1) \(3r + 2s = 6\)
(2) \(r=< 3\) and \(s=< 2\)
Ok, the easiest way to solve this is to visualize the graph with the lines plotted on it.
2x+3y<=6, is the region below the line with X-intercept 3 and Y-intercept 2. We know it is that region, because (0,0) lies below the line and it satisfies the inequality. And all points on one side of the line satisfy the same sign of inequality. (BLUE LINE)
(1) : The line 3r+2s=6 (PURPLE LINE) represents the second line shown in the figure. It can be above or below the other line. So insufficient.
(2) : r<=3 & s<=2. Again easy to see from the graph even with that constraint, the point (r,s) may lie above or below the line in question
(1+2) : the two conditions together, only take a section of the 3r+2s=6 line as a solution, but again even with r<=3, s<=2, its not sufficient to keep solutions below the 2x + 3y =< 6 line
Answer is (e)In questions like these, once you are comfortable with graphs, you can solve in less than 30 seconds fairly easily.
Let me know if the method isn't clear
I did it a little differently.
I solved for y to get \(y <=\frac{-2}{3}x + 2\)
(1) I solved for s, so \(s = \frac{-3}{2}r + 3\)
Both x-y and r-s lines are downward sloping and have different y-intercepts. We know that they deviate in the -x region if you draw or visualize a crude graph.
Plug in an arbitrary negative x-value for r. Say, -10. Then solve for s, which is 18.
Plug the x-value of r into the x-y equation. \(y <=\frac{-2}{3}*-10\)+ 2; y <= \(\frac{26}{3}\) or 8\(\frac{2}{3}\)
If (r,s) is within the x-y region, then the y-value (s) must be <= 8\(\frac{2}{3}\). 18 <= 8\(\frac{2}{3}\) does not hold.
Inconsistent answers so insufficient.
(2) Plug given r and s values into given 2r+3s≤6 equation. 12 ≤ 6 does not hold.
Insufficient.
Both (1) and (2) together neither confirms nor eliminates the large negative x-value concern that we have, and it also neither confirms nor eliminates the positive x-values. So both together are insufficient.