If the system of inequalities given by `x+3yle12 and 2x-3yle-3` is graphed on the coordinate plane above, which quadrant of the plane contains no solutions to the system?

Options

A. Quadrant I
B. Quadrant IV
C. Quadrant I and IV
D. Quadrant III and IV

Answer

B

Explanation

Difficult:Hard:
Category: Heart of Algebra/ Inequalities
Strategic Advice: The only way to solve a system of inequalities is to graph it, so draw a sketch on the coordinate plane provided to find the answer. Remember, you are looking for the quadrant that contains no solution to the system.
Getting to the Answer: Start rewriting each inequality in slope-intercept form, `y=mx+b`,where m is the slope and b is the y-intercept. Then, use m and b to graph each inequality.
`x+3le12 to 3yle-x+12 to yle(-1)/(3)+4`
`2x-3yle-3 to -3y le-2x-3 to yge(2)/(3)x+1`
To represent the first inequality, shade below it's line `(le)`: to represent the second inequality shade above its line `(ge)`. The solution to the system of inequalities is the region where the two shaded areas overlap. A sample graph follows

Notice tha there is a some darker shading ( the overlap) in quadrant I, II and III, making quadrant IV the only one that does not contain solutions to the systemt. Choice (B) is correct.