Question

Consider the system of inequalities and its graph. y ≤ –0.75x y ≤ 3x – 2 In which section of the graph does the actual solution to the system lie? 1 2 3 4

Answer 1

Answer:

The section shows below represents the solution of the system.

Step-by-step explanation:

Given system of inequalities,

y ≤ -0.75x,

y ≤ 3x - 2

Graphing y ≤ -0.75x :

Since, the related equation of y ≤ -0.75x is y = 0.75x

If x = 0, 1,

y = 0, 0.75,

Thus, join the points (0, 0) and (1, 0.75) in coordinate plane,

0 ≤ -0.75(0) ( true )

So, the shaded region will be contain the origin.

Graphing y ≤ 3x - 2 :

Since, the related equation of y ≤ 3x - 2 is y = 3x - 2

x = 0, y = - 2

y = 0, x = 2/3,

Thus, join the points (0, -2) and (2/3, 0) in coordinate plane,

0 ≤ 3(0) - 2 ( false )

So, the shaded region won't contain the origin.

Also, ' ≤ ' represents the solid line,

By the above explanation we can make the feasible region that shows the solution of the system ( shown below )

Answer 2

We need to assign a value for x to check the possible values of y. 1st inequality: y < -0.75x X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT 2nd inequality: y < 3x -2 X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT X = 1 ; y < 3(1) – 2 ; y < 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT The actual solution to the system lies on the 4TH QUADRANT.