1)  4x ^2< 10x -1

4x^2 -10x +1 < 0

2) 10x - 4y + 3 < 11

10x -4y +3 - 11 < 0

First inequality 10x - 4y - 8 < 0

Second inequality y >= x^2 - 3x - 4

1. Draw the coordianate plane.

2.  Since inequality 10x - 4y - 8 < 0 symbol is < , the boundary is not included in the solution set. Graph the boundary of the inequality x + y = 8 with dotted line.

3.  To determine which half-plane to be shaded use a test point in either half-plane. A simple choice is (0, 0).

Substitute and x = 0 and y = 0 in original inequality 10x - 4y - 8 < 0.

-8 < 0

The statement is true.

4. Since the statement is true, shade the region contains point (0, 0).

Second inequality y >= x^2 - 3x - 4

2) Since inequality y >= x^2 - 3x - 4 symbol is >= , the boundary is included in the solution set. Graph the boundary of the inequality y >= x^2 - 3x - 4 with solid line.

A simple choice is (2, 3)

Substitute and x = 2 and y = 3 in original inequality y >= x^2 - 3x - 4

3 >= -12

The statement is true.

Since the statement is true, the graph will be shaded inside of parabola.