Bunuel wrote:
In the rectangular coordinate system shown above, which quadrant, if any, contains no point (x,y) that satisfies the inequality 2x – 3y ≤ –6?
(A) None
(B) I
(C) II
(D) III
(E) IV
Attachment:
2017-12-18_1011_001.png
Attachment:
quadrants.png
Rewrite the inequality in slope-intercept form for the inequality's boundary line\(2x – 3y ≤ –6\)
\(-3y ≤ -2x - 6\)
Because dividing by -3, switch the sign:
\(y \geq \frac{2}{3} x + 2\)
Slope is positive.
Positive slope
always runs through Quadrants I and III
Eliminate answers B and D
Find the intercepts to graph the line and/or to assess signsSet x equal to 0 to find y-intercept of the inequality's boundary line
\(y \geq \frac{2}{3} (0) + 2\)
\(y \geq 2\)
Set y equal to 0 to find x-intercept of boundary line
\(0 \geq \frac{2}{3} x + 2\)
\(-2 \geq \frac{2}{3} x\)
\((\frac{3}{2})(-2) \geq{x}\)
\(x ≤ - 3\)
Intercepts of the inequality's boundary line are (0,2) and (-3,0)
AssessI graphed the intercepts, connected the points and extended the line. See diagram.
No point in Quadrant IV satisfies the inequality
You don't have to graph.
(I graph often. For me, it is quick. This problem took just under a minute.)
Quadrants II has positive y-values and negative x-values.
-- y-intercept is positive
-- x-intercept is negative
Those signs fit: The boundary line runs through Quadrant II.
Eliminate Answer C
At this point, we know the graphed inequality's boundary line runs through Quadrants I, II, and III.
All straight lines pass through at least two, and at most three, quadrants.
(Most straight lines pass through three quadrants. Never four.)
By POE, Quadrant IV contains no points that satisfy the inequality.
AND/OR
Quadrant IV has no positive y-values and no negative x-values.
-- y-intercept is positive. The x-intercept is negative.
-- Quadrant IV's (x, y) values are exactly the opposite: (+x, -y)
No point in Quadrant IV satisfies this inequality
Answer
The question asks if any (x,y) of the 4-Quadrants will not satisfy \(\frac{-x}{3} + \frac{y}{2} ≤ 1\) equation. In IV quadrant (6,-4) will definitely satisfy the equation. Can you please clarify my doubt or am I missing something here??