Answer:
A) see below for the graph
B) no; it does not satisfy the second inequality
Step-by-step explanation:
It is useful to start by graphing the boundary line. Replace the inequality symbol with an equal sign, and you have the equation of that line.
When the equations are in slope-intercept form, as here, it is useful to start by plotting the y-intercept, then identifying additional points based on the slope.
y > 2x +3
The boundary line is dashed, because the > symbol does not include the "or equal to" case. The boundary line is y = 2x +3, a line with a y-intercept of 3 and a slope of 2. The slope of 2 means the line rises 2 units for each unit it goes to the right. Similarly, it falls 2 units for each unit to the left. Using this idea, you can plot the additional points (-2, -1) and (2, 7) on this (dashed) line.
The comparison symbol is >, and the location of y is to its left, meaning y > ... is the area you want to shade--the area above (greater than) the line.
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y < -3/2x -4
As above, the boundary line is dashed, because the symbol is not ≤ or ≥. The y-intercept is -4, and the slope is -3/2. This slope means the line falls 3 units for each 2 units it goes to the right. Similarly, it rises 3 units for each 2 units to the left. Therefore a couple of additional points are (-2,-1) and (2, -7).
You may notice that the two lines have point (-2, -1) in common. This is where they intersect.
The comparison symbol is <, and y is to its left (y < ...) so the area you want to shade is y-values below (less than) the (dashed) line.
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A) The graph is two dashed diagonal lines that intersect at (-2, -1). The solution area is to the left of their point of intersection, the doubly-shaded area between the diagonal lines. It does not include the lines themselves.
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B) To show mathematically whether the point (-4, 6) is in the solution area, we can see if it satisfies both inequalities.
y > 2x +3 ⇒ 6 > 2(-4) +3 ⇒ 6 > -5 TRUE
y < -3/2x -4 ⇒ 6 < -3/2(-4) -4 ⇒6 < 2 FALSE
The point is NOT included in the solution area. (The graph shows this as well.)
