Question

Which of the following systems of inequalities has point D as a solution?

Two linear functions f of x equals 3 times x plus 4 and g of x equals negative one half times x minus 5 intersecting at one point, forming an X on the page. A point above the intersection is labeled A. A point to the left of the intersection is labeled B. A point below the intersection is labeled C. A point to the right of the intersections is labeled D.

A. f(x) ≤ 3x + 4
g of x is less than or equal to negative one half times x minus 5
B. f(x) ≥ 3x + 4
g of x is less than or equal to negative one half times x minus 5
C. f(x) ≤ 3x + 4
g of x is greater than or equal to negative one half times x minus 5
D. f(x) ≥ 3x + 4
g of x is greater than or equal to negative one half times x minus 5

Answer 1

The point labeled D is to the right of the intersection of the two linear functions. This means that its x-coordinate is greater than the x-coordinate of the point of intersection.

We can find the point of intersection by setting the two functions equal to each other:

3x + 4 = (-1/2)x - 5

Solving for x, we get:

(7/2)x = -9

x = -18/7

So the point of intersection is (-18/7, -29/7).

Since the x-coordinate of point D is greater than -18/7, we can eliminate options A and C.

Now we need to check whether option B or option D includes point D as a solution. To do this, we can simply plug in the coordinates of D into the two inequalities and see which one holds true.

Option B:

f(x) ≥ 3x + 4

2 ≥ 3(D) + 4

2 ≥ 3D + 4

-2 ≥ 3D

D ≤ -2/3

g(x) ≤ (-1/2)x - 5

2 ≤ (-1/2)(D) - 5

7 ≤ -D

D ≥ -7

Since -2/3 is less than -7, option B does not include point D as a solution.

Option D:

f(x) ≥ 3x + 4

2 ≥ 3(D) + 42 ≥ 3D + 4

-2 ≥ 3D

D ≤ -2/3

g(x) ≥ (-1/2)x - 5

2 ≥ (-1/2)(D) - 5

7 ≥ -D

D ≤ -7

Since -2/3 is less than -7, option D does not include point D as a solution either.

Therefore, neither option B nor option D includes point D as a solution. The correct answer is that neither system of inequalities has point D as a solution.