**Graphing linear inequalities in two variables and find common region :**

Here we are going to see some examples of finding solution of linear inequalities in two variables by graphing.

To obtain common region for both the given linear inequalities, we have to follow the steps given below.

**Step 1 :**

By choosing any two or three points on the line, first we have to draw the graph of the given lines.

For that we have to write the given inequality as y = ,

**Step 2 :**

If inequality contains greater than (or equal to) sign, then portion above the line denotes the inequality and hence is shaded. While, if inequality has less than (or equal to) sign, the part below the line represents the inequality and thus it has to be shaded.

**Step 3 :**

Some points (in the form (x,y)) are to be chosen and substituted in the inequality. If they satisfy inequality, that means they are included in it, otherwise not. Common test point is (0, 0). It is to be cross checked that these points justify our graph or not.

**Step 3 :**

The region that is common to both shaded portions is the solution to this system

Let us look into some examples based on the concept.

**Example 1 :**

Determine the region in the plane determined by the inequalities:

x ≤ 3y, x ≥ y

**Solution :**

Given that : x ≤ 3y 3y ≥ x 3y = x y = x/3 If x = -3, then y = -1 If x = 3, then y = 1 |
Given that : x ≥ y y ≤ x y = x If x = -1, then y = -1 If x = 1, then y = 1 |

The points on the line y = x/3 are (-3, -1) and (3, 1)

The points on the line y = x are (-1, -1) and (1, 1)

Test point : 3y ≥ x x = 0, and y = 0 Satisfies the inequality. Since we have ≥, shade the portion above the line. |
Test point : y ≤ x x = 0, and y = 0 Satisfies the inequality. |

**Example 2 :**

Determine the region in the plane determined by the inequalities:

y ≥ 2x, −2x + 3y ≤ 6.

**Solution :**

Given that : y ≥ 2x y = 2x If x = -1, then y = -2 If x = 1, then y = 2 |
Given that : −2x + 3y ≤ 6 3y = 2x + 6 y = (2/3)x + (6/3) y = (2x/3) + 2 If x = -3, then y = 0 If x = 3, then y = 4 |

The points on the line y ≥ 2x are (-1, -2) and (1, 2)

The points on the line −2x + 3y ≤ 6 are (-3, 0) and (3, 4)

Test point : y ≥ 2x x = 0, and y = 0 Satisfies the inequality. |
Test point : −2x + 3y ≤ 6 x = 0, and y = 0 Satisfies the inequality. |

**Question 3 :**

Determine the region in the plane determined by the inequalities:

3x + 5y ≥ 45, x ≥ 0, y ≥ 0

**Solution :**

3x + 5y ≥ 45

5y ≥ 45 - 3x

y ≥ (45 - 3x)/5

y ≥ 9 - (3x/5)

Let y = 9 - (3x/5)

If x = 5 y = 9 - 3 y = 6 |
If x = 10 y = 9 - 6 y = 3 |
If x = 15 y = 9 - 9 y = 0 |

By joining the points (5, 6) (10, 3) and (15, 0).

**Question 4 :**

Determine the region in the plane determined by the inequalities:

2 x + 3 y ≤ 35, y ≥ 2, x ≥ 5

**Solution :**

2 x + 3 y ≤ 35

After having gone through the stuff given above, we hope that the students would have understood "Graphing linear inequalities in two variables and find common region".

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