Linear Inequalities Questions

Linear inequalities are mathematical expressions involving variables, constants, and inequality signs (such as ≤, ≥, <, >). They describe relationships where one side of the inequality is not necessarily equal to the other. Solving linear inequalities involves isolating the variable, similar to solving linear equations, but the direction of the inequality sign may change when multiplying or dividing by negative numbers. The solution is often expressed as a range of values or represented on a number line.

1.0What are Linear Inequalities?

Linear inequalities are expressions where two algebraic expressions are compared using inequality signs, such as:

• > (greater than)
• < (less than)
• ≥ (greater than or equal to)
• ≤ (less than or equal to)

For example:

2x + 3 > 5

This inequality represents all values of x that make the expression 2x + 3 greater than 5.

2.0Graphing Linear Inequalities Questions

Graphing linear inequalities is a visual way of representing the set of solutions. A linear inequality with two variables can be represented on a coordinate plane, where the solution is shown as a shaded region.

Steps to Graph Linear Inequalities:

1. Rewrite the inequality in slope-intercept form (y = mx + b) if necessary. For example:

$2x+y\le 4\Rightarrow y\le -2x+4$

2. Graph the corresponding linear equation (ignore the inequality for now). For $y\le -2x+4$, graph the line y = –2x + 4.
3. Determine whether the line is solid or dashed:
4. • If the inequality is ≤ or ≥, the line is solid (the points on the line are included in the solution).
   • If the inequality is < or >, the line is dashed (the points on the line are not included).
4. Shade the correct region:
5. • If the inequality is y ≤ or x ≤ , shade below the line.
   • If the inequality is y ≥ or x ≥ , shade above the line.

Example of Graphing Linear Inequalities:

Consider the inequality y ≥ 2x + 1.

1. Graph the line y = 2x + 1 (a solid line since ≥ is used).
2. Since the inequality is y ≥ 2x + 1, shade the region above the line.

The shaded area represents all the possible solutions to this inequality.

3.0Simultaneous Linear Inequalities Questions

Simultaneous linear inequalities refer to a set of linear inequalities that must all be satisfied at the same time. These problems involve finding the region where all the inequalities overlap on a graph.

Steps to Solve Simultaneous Linear Inequalities:

1. Graph each inequality separately. For each inequality, follow the same steps as you would for graphing a single inequality.
2. Identify the common region. The solution to the system of inequalities is the area where all the shaded regions overlap.
3. Check the boundary lines. If the inequalities involve ≤ or ≥, include the boundary line. If they involve < or >, exclude the boundary line (use a dashed line).

4.0Linear Inequalities Solved Questions

Example 1. Solve 4x – 5 ≥ 7 

Solution:

4x ≥ 7 + 5 

4x ≥ 12 

x ≥ 3  

Example 2. Solve 2x + 3 < 11

Solution:

2x < 11 - 3 

2x < 8 

x < 4 

Example 3. Solve 3(x − 4) > 2(x + 1) + 5 

Solution:

3x - 12 > 2x + 2 + 5 

3x - 12 > 2x + 7 

3x - 2x > 7 + 12 

x > 19 

Example 4. Solve 7x – 4 ≥ 3x + 8 

Solution:

7x − 3x ≥ 8 + 4 

4x ≥ 12  

x ≥ 3  

Example 5. Solve 5(x + 2) ≤ 3(x − 4) + 6 

Solution:

5x + 10 ≤ 3x – 12 + 6

5x + 10 ≤ 3x − 6

5x − 3x ≤ −6 − 10

2x ≤ −16

x ≤ −8

Example 6. Solve for x: |x + 1| + |x| > 3 

Solution:
We break this into cases based on the values of x.

• Case 1: x ≥ −1

For x ≥ −1, |x + 1| = x + 1 and |x| = x, so the inequality becomes:

(x + 1) + x > 3 

2x + 1 > 3 

2x > 2 

x > 1 

• Case 2: –1 < x < 0

For −1 < x < 0, |x + 1| = x + 1 and |x| = –x, so the inequality becomes:

(x + 1) - x > 3 

1 > 3 

This is a contradiction, so there is no solution for this range.

• Case 3: x ≤ −1

For x ≤ −1, ∣x + 1∣ = −(x + 1) and ∣x∣ = −x, so the inequality becomes:

-(x + 1) - x > 3 

-2x - 1 > 3 

-2x > 4 

x < -2 

Final Answer:

x > 1or x < −2 

Example 7. The longest side of a triangle is 4 times the shortest side, and the third side is 5 cm shorter than the longest side. If the perimeter of the triangle is at least 80 cm, find the minimum length of the shortest side.

Solution:
Let the shortest side be x. Then:

• The longest side is 4x.
• The third side is 4x – 5.

The perimeter is the sum of all three sides, and we know the perimeter is at least 80 cm:

X + 4x + (4x − 5) ≥ 80 

9x – 5 ≥ 80  

9x ≥ 85  

$x\ge \frac{85}{9}\approx 9.44\mathrm{cm}$

Final Answer:
The minimum length of the shortest side is approximately 9.44 cm.

Example 8. Solve the inequality and graph the solution on the number line:

$\frac{3x-4}{2}\ge \frac{x+1}{4}-1$

Solution:
First, eliminate the fractions by multiplying through by 4 (the least common denominator):

$4\times \left(\frac{3x-4}{2}\right)\ge 4\times \left(\frac{x+1}{4}-1\right)$

2(3x − 4) ≥ x + 1 – 4  

6x – 8 ≥ x – 3 

Subtract x from both sides:

5x – 8 ≥ −3  

Add 8 to both sides:

5x ≥ 5 

Divide both sides by 5:

x ≥ 1  

Final Answer:
The solution is x ≥ 1. On the number line, shade from 1 to the right, including 1 (since the inequality is ≥).

Example 9. A manufacturer has 500 litres of a 15% solution of acid. How many litres of a 40% acid solution must be added to it so that the acid content in the resulting mixture is more than 20% but less than 25%?

Solution:
Let the amount of 40% solution to be added be x litres.

• The total amount of acid in the 500 litres of 15% solution is 500 × 0.15 = 75 litres.
• The amount of acid in x litres of 40% solution is x × 0.40 = 0.4x.

The total acid content after mixing is:

75 + 0.4x  

The total volume of the mixture is:

500 + x 

We want the acid content to be between 20% and 25%, so we set up two inequalities:

1. For more than 20% acid:

$\frac{75+0.4x}{500+x}>0.20$

Multiply both sides by 500 + x:

75 + 0.4x > 0.20(500 + x) 

75 + 0.4x > 100 + 0.20x 

Subtract 0.20x from both sides:

75 + 0.20x > 100 

Subtract 75 from both sides:

0.20x > 25 

x > 125 

2. For less than 25% acid:

$\frac{75+0.4x}{500+x}<0.25$

Multiply both sides by 500 + x:

75 + 0.4x < 0.25(500 + x) 

75 + 0.4x < 125 + 0.25x 

Subtract 0.25x from both sides:

75 + 0.15x < 125 

Subtract 75 from both sides:

0.15x < 50 

$x<\frac{50}{0.15}\approx 333.33$

Final Answer:
The amount of 40% solution must be between 125 litres and 333.33 litres. Therefore, 125 < x < 333.33.

Example 10. Solve the following system of linear inequalities graphically.

$2x+y\ge \mathrm{6...}(1)x-y\le \mathrm{2...}(2)$

Solution:

The graph of linear equation 2x+y=6 is drawn in fig. We note that solution of inequality (1) is represented by the shaded region above the line 2x+y=6, including the point on the line on the same set of axes; we draw a graph of the equation x-y=2 as shown in fig. Then we note that inequality (2) represents the shaded region above the line x-y=2 including the points on the line. Clearly, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Example 11. Solve the following system of inequalities graphically

$\begin{array}{rl}& x+2y\le 8\\ & 2x+y\le 8\\ & x\ge 0\\ & y\ge 0\end{array}$

Solution:

 We draw the graphs of the lines $x+2y=8\text{ and }2x+y=8$. The inequality (1) and (2) represent the region below the two lines, including the point on the respective lines.

Since x ≥ 0, y ≥ 0, every point in the shaded region in the first quadrant represents a solution of the given system of inequalities.

5.0Linear Inequalities Practice Questions 

1. Solve the inequality:

5x – 7 ≥ 3x + 5 

2. Solve the inequality:

4x + 6 < 2x + 12

3. Solve the inequality:

3(x – 2) > 4(x + 1) – 5

4. Solve the inequality:

2x – 5 ≤ 3x + 2 

5. Solve the inequality:

6x + 4 > 10x – 8