Question

The point (−2, 2) is a solution to which of the following systems?

y > −2x + 2 and y > x + 5
y x − 1
y < 2x + 8 and y ≥ −x − 3
y < 2x + 3 and y ≥ −2x − 5

Answer 1

Answer:

quadrant IV has no solution

Step-by-step explanation:

given, y≥2x+1 and  

y>  

2

1

​  

x−1

first, draw the graph for equations  y=2x+1 and y=  

2

1

​  

x−1

for y=2x+1  

substitute y=0 we get, 2x+1=0⟹x=−0.5

substitute x=0 we get, y=1

therefore, y=2x+1 line passes through (0.5,0) and (0,1) as shown in fig.

Hence,  y≥2x+1 includes the region above the line.

for y=  

2

1

​  

x−1  

substitute y=0 we get,  

2

1

​  

x−1=0⟹x=2

substitute x=0 we get, y=−1

therefore, y=2x+1 line passes through (2,0) and (0,-1) as shown in fig.

Hence,  y>  

2

1

​  

x−1 includes the region above the line.

the intersection region is the shaded region as shown in above figure which includes I, II and III quadrants.

Therefore, quadrant IV has no solution

Answer 2

Given:

• point (-2, 2)
• y > −2x + 2 and y > x + 5
• y x − 1
• y < 2x + 8 and y ≥ −x − 3
• y < 2x + 3 and y ≥ −2x − 5

To find:

point (-2, 2) is a solution of which equation among options.

Solution:

We will put the value of point (-2, 2) in each of the options.

1. y > −2x + 2 and y > x + 5

replacing x and y with -2 and 2 respectively we get,

2 > −2(-2) + 2 and 2 > (-2) + 5

2 > 4 + 2 and 2 > 5 - 2

2 > 6 and 2 > 3

now since we have a 'and' in between we need both the equations to be true but clearly they aren't true as 2 is less than 6.

Hence (-2, 2) is not a solution for this.

2. This equation doesn't have a logical sign.

3. y < 2x + 8 and y ≥ −x − 3

replacing x and y with -2 and 2 respectively we get,

2 < 2(-2) + 8 and 2 ≥ −(-2) − 3

2 < -4 +8 and 2 ≥ 2-3

2 < 4 and 2 ≥ -1

now since we have a 'and' in between, we need both the equations to be true and for a fact we know, 2 is smaller than 4 and greater than -1.

Hence (-2, 2) is a solution for this.

Since we got the answer already let's stop here.