How do I find a point of the vertices of three inequalities?

Question

y is greater than or equal to 3x-5

y is less or equal to -1/2 x +2

x is greater than or = to -4

 

 

 

choices( -4,20)   ( -4,-15)  (-4,-4 )( -4,4)

Answer 1

The solution is best seen graphically. The first exercise is to draw the lines (1) y=3x-5, (2) y=-½x+2, (3) x=-4 (vertical line).

So substitute x=-4 in (1) and (2): y=-17, y=4.

We get the intersections (-4,-17), (-4,4).

Next, (1) and (2) intersect when 3x-5=-½x+2, 7x/2=7, x/2=1, x=2, y=3x-5=1, so intersecting at (2,1). (None of the answer options includes this important vertex.)

We now have the intersection points. The inequalities give us the area (region) of a triangle with the above vertices: (-4,-17), (-4,4), (2,1). The space inside this triangle (including its sides) satisfies all the inequalities. The vertex (-4,4) is included in your answer options as the last option.