Question

Which system of linear inequalities is represented by the
graph?
y> 2x – 1 and y < 2x + 2
y > 2x + 2 and ys 2x-1
y < 2x + 2 and y? 2x-1
y< 2x + 1 and y>2x - 2

Answer 1

Answer:

y > 2x + 2 and y < 2x-1 .

Step-by-step explanation:

The line which the blue shaded area represent has y intercept 2 and slope $\frac{2}{1} =2$

Hence equation of the line is y=2x+2.

To check the inequality for the shaded region we take any  point (-3,0) in the shaded region .Plugging the values in the given equation :

0 > 2(-3)+2  or 0 >-4.

The inequality equation represented by the blue shaded part is y > 2x+2.

The line for the red shaded region has y intercept -1 and slope 2.

Hence equation of the line is y= 2x-1 .

Taking a point (2,0) in the shaded part and substituting the values in the equation of line we have :

0< 2(2)-1 or 0< 3 .

Hence the inequality representing the red shaded region is y<2x-1 .

Answer 2

y > 2x + 2 and y < 2x - 1

• The graph of y > 2x + 2 is a dashed line that intersects the axes at points (-1, 0) and (0, 2). The origin (0, 0) is not included in the blue shaded area.
• The graph of y < 2x - 1 is a dashed line that intersects the axes at points (¹/₂, 0) and (0, -1). The origin (0, 0) is not included in the red shaded area.

Further explanation

In this problem, we will compose the system of linear inequalities is represented by the graph. Firstly, let us state each line on the graph in terms of the equation of the line.

A shortcut to form a linear equation through the intercepts of the axes at (0, a) and (b, 0) is $\boxed{\boxed{ \ ax + by = ab \ }}$.

Part-1: a dashed line that intersects the axes at points (0, 2) and (-1, 0).

Step-1: make a linear function  

• (0, 2) → (0, a)
• (-1, 0) → (b, 0)

$\boxed{ \ ax + by = ab \ } \rightarrow \boxed{ \ 2x + (-1)y = 2 \times (-1) \ }$

2x - y = -2

Add by 2 and y on both sides.

Hence, the equation of line is $\boxed{y = 2x + 2 \ }$  

Step-2: make a linear inequality  

• y = 2x + 2 is the boundary line and we draw a dashed line since the equality symbol is " > or < ".  
• Test the point (0, 0) as origin in y = 2x + 2, i.e., $\boxed{0 = 2(0) + 2}$ which is true if 0 < 2.  

Since the test point (0, 0) is not in the blue shaded area, which means the test results must be false (or 0 > 2), then linear inequality is arranged as follows:

$\boxed{\boxed{ \ y > 2x + 2 \ }}$

Part-2: a dashed line that intersects the axes at points (¹/₂, 0) and (0, -1)..

Step-1: make a linear function  

• (0, -1) → (0, a)
• (¹/₂, 0) → (b, 0)

$\boxed{ \ ax + by = ab \ } \rightarrow \boxed{ \ (-1)x + \frac{1}{2}y = -1 \times \frac{1}{2} \ }$

$\boxed{ \ -x + \frac{1}{2}y = -\frac{1}{2} \ }$

Multiply by 2 on both sides.

-2x + y = -1

Add by 2x on both sides.

Hence, the equation of line is $\boxed{y = 2x - 1 \ }$  

Step-2: make a linear inequality  

• y = 2x - 1 is the boundary line and we draw a dashed line since the equality symbol is " > or < ".  
• Test the point (0, 0) as origin in y = 2x - 1, i.e., $\boxed{0 = 2(0) - 1}$ which is true if 0 > -1.  

Since the test point (0, 0) is not in the red shaded area, which means the test results must be false (or 0 < -1), then linear inequality is arranged as follows:

$\boxed{\boxed{ \ y < 2x - 1 \ }}$

Thus the system of linear inequalities is represented by the  graph is y > 2x + 2 and y < 2x - 1.

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