SOLVING SYSTEM OF INEQUALITIES IN TWO VARIABLES BY GRAPHING

Sketch the solution to each system of inequalities.

Problem 1 :

y ≤ -x-2

y ≥ -5x + 2

Solution :

y ≤ -x - 2  ----(1)

y ≥ -5x + 2

Slope and y-intercept of (1) :

Slope = -1

y-intercept = -2

Since line (1) is having negative slope, it should be a falling line. Rise = -1 and run = 1. Tracing some more points and joining them, we will the line. It should be solid line since we have ≤.

Slope and y-intercept of (2) :

Slope = -5

y-intercept = 2

Since line (2) is having negative slope, it should be a falling line. Rise = -5 and run = 1. Tracing some more points and joining them, we will get the line. It should be solid line since we have ≥.

Test point for y ≤ -x - 2 :

So, the region below y ≤ -x-2 be the solution.

Test point for y ≥ -5x + 2 :

So, the region above y ≥ -5x + 2 be the solution. So, overlapping region must be the solution region.

Sketch the solution to each system of Inequalities.

Problem 2 :

y > -x – 2

y < -5x + 2

Solution :

y > -x – 2 --- (1)

y < -5x + 2

Slope and y – intercept of (1) :

Slope = -1

y – intercept = -2

Since line (1) is having negative slope, it should be a falling line. Rise = -1 and run = 1. Tracing some more points and joining them, we will get the line, it should be dotted line since we have >. 

Slope and y – intercept of (2) :

Slope = -5

y – intercept = 2

Since line (2) is having negative slope, it should be a falling line. Rise = -5 and run = 1. Tracing some more points and joining them, we will get the line, it should be dotted line since we have <. 

Test point for y > -x – 2 :

Test point for y < -5x + 2 :

So, the region above y < -5x+2 be the solution. So, overlapping region must be the solution region.

Problem 3 :

y ≤ 1/2x + 2

y < -2x – 3

Solution :

y ≤ 1/2x + 2 --- (1)

y < -2x – 3 --- (2)

Slope and y – intercept of (1) :

Slope = 1/2

y – intercept = 2

Since line (1) is having positive slope, it should be a raising line. Rise = 1 and run = 2. Tracing some more points and joining them, we will get the line, it should be solid line since we have ≤. 

Slope and y – intercept of (2) :

Slope = -2

y – intercept = -3

Since line (2) is having negative slope, it should be a falling line. Rise = -2 and run = 1. Tracing some more points and joining them, we will get the line, it should be dotted line since we have <. 

Test point for y ≤ 1/2x + 2 :

Let us take (1, 1) below the line and apply in (1)

1 ≤ (1/2)(1) + 2

1 ≤ 5/2 (True)

Test point for y < -2x – 3 :

Let us take (-2, -2) below the line and apply in (1)

-2 < -2(-2) – 3

-2 < 4 - 3

-2 < 1 (True)

Overlapping region is the solution.

Problem 4 :

x ≤ -3

y < (5/3)x + 2

Solution :

x ≤ -3 --- (1)

y < (5/3)x + 2 --- (2)

Slope and y – intercept of (1) :

It should be the vertical line having x intercept at -3.

Slope and y – intercept of (2) :

Slope = 5/3

y – intercept = 2

Since line (2) is having negative slope, it should be a falling line. Rise = 5 and run = 3. Tracing some more points and joining them, we will get the line, it should be dotted line since we have <. 

Test point for y < (5/3)x + 2 :

Let us take (1, 2) below the line and apply in (2)

2 < (5/3)(1) + 2

2 < 11/3 (True)

Overlapping region will be the solution.

Problem 5 :

y ≤ -(5/2)x - 2

y < -(1/2)x + 2

Solution :

y ≤ -(5/2)x - 2 --- (1)

y < -(1/2)x + 2 --- (2)

Slope and y – intercept of (1) :

Slope = -5/2, y-intercept = -2

Slope and y – intercept of (2) :

Slope = -1/2, y-intercept = 2

Taking a point above the line (-1, 2) and applying in y ≤ -(5/2)x - 2

2 ≤ -(5/2)(-1) - 2

2 ≤ (5/2) - 2

2 ≤ (1/2) (False)

Taking a point below the line (1, 1) and applying in y ≤ -(1/2)x + 2

1 ≤ -(1/2)(1) + 2

1 ≤ (-1/2) + 2

1 ≤ (3/2) (True)