Table of Contents

Last modified on June 8th, 2024

Linear inequalities are algebraic expressions where the power of the unknown variable is no more than one, and the variable is connected with an inequality sign (>, <, â‰¤, or â‰¥).

7x – 12 > 16 and 5x + 11 < 2 are examples of linear inequalities

Adding or subtracting a number on or from both sides of the inequality does not change its direction.

If a < b, then a + c < b + c

If a > b, then a + c > b + c

Similarly,

If a < b, then a – c < b – c

If a > b, then a – c > b – c

Like adding and subtracting, multiplying, or dividing an inequality by the same positive number also does not change the direction of the inequality.

If a < b and c > 0, then ac < bc

If a > b and c > 0, then ac > bc

If a < b and c < 0, then ac > bc

If a > b and c < 0, then ac < bc

Similarly,

If a < b and c > 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$

If a > b and c > 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$

If a < b and c < 0, then ${\dfrac{a}{c} >\dfrac{b}{c}}$

If a > b and c < 0, then ${\dfrac{a}{c} <\dfrac{b}{c}}$

When solving a linear inequality, the solution is typically represented as an ordered pair (x, y) that satisfies the inequality, which is then graphed on a number line.

Using the above rules, we solve the inequality x + 3 > 10

**Step 1: Using the Subtraction Property**

x + 3 – 3 > 10 -3

â‡’ x > 7

**Step 3: Graphing the Solution**

**For One Variable**

Let us try solving the inequality with one variable as 4x + 3 > 23

**Step 1: Using the Subtraction Property**

4x + 3 – 3 > 23 – 3

â‡’ 4x > 20

**Step 2: Using the Division Property**

${\dfrac{4x}{4} >\dfrac{20}{4}}$

â‡’ x > 5

Thus, the solution is x > 5

**Step 3: Graphing the Solution**

**For Two Variables:**

Now, we solve the inequality 7y – 5x â‰¤ 6y – 3x + 3

**Step 1: Solve for â€˜xâ€™ Using the Addition Property**

7y – 5x + 5x â‰¤ 6y – 3x + 5x + 3

â‡’ 7y â‰¤ 6y + 2x + 3

**Step 2: Solve for â€˜yâ€™ Using the Subtraction Property**

7y – 6y â‰¤ 6y + 2x + 3 – 6y

â‡’ y â‰¤ 2x + 3

**Step 3: Graphing the Solution**

To plot a graph of the inequality y â‰¤ 2x + 3, we consider â€˜â‰¤â€™ as â€˜=â€™ sign. Now, plotting the graph y = 2x + 3, we get

Since the inequality is â€˜â‰¥,â€™ the graph of the equation is formed as a solid line.

Now, considering a point (x, y) as (-2, 0), we get

2x + 3 = 2(-2) + 3 = -4 + 3 = -1, which is not greater than 0

Thus, y â‰° 2x + 3 at (-2, 0)

Now, shading the region that does not contain (-2, 0) shows the following graph.

**Solve and Graph the inequality with one variable: 2x + 5 < 3x**

Solution:

Given, 2x + 5 < 3x**Step 1: Using the Subtraction Property**

2x + 5 – 2x < 3x – 2x

â‡’ 5 < x

â‡’ x > 5, which is graphed on the number line as shown.

**Solve and Graph the inequality with the variables on both sides: 7x – 5 > 3x + 13**

Solution:

Given, 7x – 5 > 3x + 13**Step 1: Using the Addition and Subtraction Property**

7x – 5 + 5 – 3x > 3x + 13 + 5 – 3xÂ

â‡’ 4x > 18**Step 2: Using the Division Property**

â‡’ x > ${\dfrac{18}{4}}$ (by division property)

â‡’ x > ${\dfrac{9}{2}}$

However, linear inequalities with the same solution are called **equivalent inequalities**. Thus, 7x – 5 > 3x + 13 is equivalent to x > ${\dfrac{9}{2}}$ or x > ${4\dfrac{1}{2}}$

By plotting the solution on the number line, we get the shown graph.

Sometimes, plotting the solution of linear inequalities can form a horizontal line parallel to the x-axis or a vertical line parallel to the y-axis.

**Case 1:** When we plot the graph of the linear inequality y > 1 (â€˜yâ€™ is greater than 1, excluding 1), the graph includes the entire region above the line y = 1, without the line y = 1 (shown in dotted form).

**Case 2:** When we plot the graph of the linear inequality x < 4 (â€˜xâ€™ is less than 4, excluding 4), the graph includes the entire region left of line x = 4, without the line x = 4 (shown in dotted form).

A system of linear inequalities consists of two or more linear inequalities with the same variables. Its solution includes all ordered pairs that simultaneously satisfy each inequality in the system.

Graphically, the solution set is depicted as the intersection of the regions represented by each individual inequality on the coordinate plane.

Let us consider the system of inequalities:

5y – 3x < 15 and 5x + 3y > 15

First, we graph each inequality separately, as shown.

Now, let us pick a point that is not on the two given lines and verify whether this point satisfies the inequalities.

Considering the point (x, y) as (2, 1), we get

5y – 3x = 5(1) – 3(2) = 5 – 6 = -1, which is less than 15

5x + 3y = 5(2) + 3(1) = 10 + 3 = 13, which is not greater than 15

Thus, the individual shaded regions and their overlapped region are shaded:

Here, the graphs are formed as dotted lines for the strict inequalities (â€˜<â€™ and â€˜>â€™).

**Solve and graph the linear inequality 6(x – 3) > 30**

Solution:

Here, 6(x – 3) > 30

â‡’ x – 3 > 5 (by division property)

â‡’ x > 5 + 3 (by addition property)

â‡’ x > 8

Thus, the solution is x > 8, and its graph is as follows.

**Solve the inequality 6(y – 5) < 5(4 + y)**

Solution:

Here, 6(y – 5) < 5(4 + y)

â‡’ 6y – 30 < 20 + 5y

â‡’ 6y – 5y – 30 + 30 < 20 + 5y – 5y + 30 (by addition and subtraction property)

â‡’ y < 50

Thus, the solution is y < 50

**Which linear inequality is represented by the following graph?**

Solution:

Here, the given graph represents the linear inequality x â‰¥ -2 for any variable x.

Last modified on June 8th, 2024