Class 11 Mathematics Chapter 5 Linear Inequalities

Key Concept: Basic Concept

b) Less than or equal to

[Solution Description]

The symbol $\leq$ is used to denote “less than or equal to” in inequalities. It indicates that one value is either less than or exactly equal to another.

Key Concept: Inequality Symbols

c) Greater than or equal to

[Solution Description] The symbol $\geq$ represents “greater than or equal to”. It means that the value on the left side of the inequality is either greater than or exactly equal to the one on the right.

Key Concept: Solving Absolute Value Inequality

c) $-3 < x < 7$ [Solution Description] To solve the inequality $|x - 2| < 5$, we consider two cases: Case 1: $x - 2 < 5$ $$x < 5 + 2$$ $$x < 7$$ Case 2: $-(x - 2) < 5$ $$-x + 2 < 5$$ $$-x < 3 x > -3$$

Combining both conditions, the solution set is $-3 < x < 7$.

Key Concept: Solving Literal Inequalities

c) $x > -3$

[Solution Description] Start by subtracting $3x$ from both sides:

$$3x – 4 – 3x < 5x + 2 - 3x$$ which simplifies to: $$-4 < 2x + 2$$ Next, subtract 2 from both sides: $$-4 - 2 < 2x + 2 - 2$$ this gives: $$-6 < 2x$$ Divide both sides by 2: $$\frac{-6}{2} < x$$ Simplifying gives: $-3 < x$ Therefore, the solution is $x > -3$.

Key Concept: Quadratic Inequalities

b) $(-\infty, -1) \cup (5, \infty)$

[Solution Description] Begin by factoring the quadratic expression $x^2 – 4x – 5$. We set it equal to zero to find the roots:

$$x^2 – 4x – 5 = (x – 5)(x + 1)$$

The roots are $x = 5$ and $x = -1$. The inequality is positive outside the roots. Hence, the intervals satisfying $x^2 – 4x – 5 > 0$ are:

$$(-\infty, -1) \cup (5, \infty)$$

Key Concept: Double Inequality

c) (1, 3)

[Solution Description] To find the solution set for the double inequality $-2 < 3x - 5 < 4$, we begin with the left part: $$-2 < 3x - 5$$ Add 5 to both sides: $$3 < 3x$$ Divide by 3: $$1 < x$$ Now for the right part: $$3x - 5 < 4$$ Add 5 to both sides: $$3x < 9$$ Divide by 3: $$x < 3$$ Combining, the range is $1 < x < 3$, or in interval notation: $(1, 3)$.

Key Concept: Simple Calculation

a) $x < -2$ [Solution Description] To solve for $x$, divide both sides of the inequality by -4, remembering to reverse the inequality sign: $$x < -2$$

Key Concept: Basic Inequality

c) $x < 2$ [Solution Description] To solve the inequality $5x - 7 < 3$, we first add 7 to both sides of the inequality to isolate the term with x: $$5x - 7 + 7 < 3 + 7$$ Simplifying, we have: $$5x < 10$$ Next, divide both sides by 5 to isolate x: $$x < \frac{10}{5}$$ Simplifying the right side gives: $$x < 2$$ Therefore, the solution set is all values of x that are less than 2.

Key Concept: Basic Inequality

d) $>$

[Solution Description] The symbols for inequalities that do not include equality are ‘<' and '>‘.

Key Concept: Integer and Real Solutions

b) {…, 4, 5, 6}

[Solution Description] Start by isolating $x$: Subtract $x$ from both sides of $2x – 3 < x + 4$ to get $x - 3 < 4$. Adding 3 to both sides, we have $x < 7$. Thus, for integers, $x$ can be {..., 4, 5, 6}.

Key Concept: Graphical Representation

c) Open circle at $-1$, line to the right

[Solution Description] To graphically represent $x > -1$, we use an open circle at $x = -1$ and draw a line to the right, indicating all numbers greater than $-1$. An open circle means that $-1$ is not included in the solution set.

Key Concept: Quadratic Inequality Solution

b) $x \in [2, 3]$

[Solution Description] First, factorize the quadratic expression: $x^2 – 5x + 6 = (x – 2)(x – 3)$. The roots of the equation are $x = 2$ and $x = 3$. The solution set is found by checking intervals around the roots: for $x \in [2, 3]$, $(x – 2)(x – 3) \leq 0$. Hence, the solution set is $x \in [2, 3]$.

Key Concept: Application of Inequalities

b) 72

[Solution Description]

We want to set up the inequality for the average of four exams:

$$\frac{55 + 60 + 65 + x}{4} \geq 63$$

Clearing the fraction by multiplying by 4:

$$55 + 60 + 65 + x \geq 252$$

Solving for $x$:

$$x \geq 252 – 180$$

Thus,

$$x \geq 72$$

So, the minimum score required in the fourth exam is 72.

Key Concept: Strict vs Slack Inequality

b) $n \geq 5$

[Solution Description] A slack inequality includes $\leq$ or $\geq$ symbols, indicating less restrictive conditions than strict inequalities ($<$ or $>$).

Key Concept: Numerical Inequality

d) 9 ≥ 5

[Solution Description] To identify the true numerical inequality, compare each pair of numbers using the given inequality symbols.

Key Concept: Inequality Transformation

c) $x > -4$

[Solution Description] To solve this inequality, we start by isolating the variable x.

$$5x + 7 > 3x – 1$$

Subtract $3x$ from both sides:

$$5x + 7 – 3x > -1$$

Simplify to get:

$$2x + 7 > -1$$

Now subtract 7 from both sides:

$$2x > -8$$

Divide both sides by 2:

$$x > -4$$

So, the solution set is $x > -4$.

Key Concept: Solution Set Identification

b) $[-\infty, 4]$

[Solution Description] The solution set for $x \leq 4$ includes all real numbers less than or equal to 4, which can be written as $(-\infty, 4]$.

Key Concept: Symbol Identification

c) ≤

[Solution Description] The inequality $7 \leq 10$ uses the “less than or equal to” symbol, which is represented by $\leq$.

Key Concept: Inequality Understanding

c) 68°F < F < 77°F [Solution Description] Use $$F = \frac{9}{5}C + 32$$ to convert each endpoint of the Celsius interval to Fahrenheit. For 20°C: $$F_1 = \frac{9}{5}\times20 + 32 = 36 + 32 = 68^\circ F$$ For 25°C: $$F_2 = \frac{9}{5}\times25 + 32 = 45 + 32 = 77^\circ F$$ Hence, the inequality is 68°F < F < 77°F.

Key Concept: Graphical Representation

c) Solid circle on 4 with line to the left

[Solution Description] For $x \leq 4$, we use a solid (dark) circle on 4 with a line extending to the left, indicating that all numbers less than or equal to 4 satisfy the inequality.

Key Concept: Quadratic Inequality Basics

a) $ax^2 + bx + c \leq 0$

[Solution Description] A quadratic inequality involves an expression where the highest degree of the variable is two. In this question, we look for an option that includes terms with $ax^2$.

Option a), $ax^2 + bx + c \leq 0$, fits the criteria for a quadratic inequality as it includes a term with $x^2$.

Key Concept: Basic Inequality

c) $\{5, 6, 7, \ldots\}$

[Solution Description]

To solve the inequality $5x > 20$, we divide both sides by 5:

$$\frac{5x}{5} > \frac{20}{5}$$

Simplifying this, we get:

$x > 4$

Since $x$ is a whole number, the solution set is $\{5, 6, 7, \ldots\}$.

Key Concept: Graphical Solution

b) Line starting at 2 and going to the right

[Solution Description]

First, solve the inequality for $x$:

$$4x – 3 \geq 5$$

Add 3 to both sides:

$$4x \geq 8$$

Next, divide by 4:

$$x \geq 2$$

On a number line, this would be represented as a line extending to the right from 2 with a closed circle at 2.

Key Concept: Multi-variable Inequality, Transformation

b) $x \leq 1800, y \leq 5$

[Solution Description] Divide the entire inequality by $5$ to simplify:

$$\frac{5x + 200y}{5} \leq \frac{10000}{5}$$

Simplifying gives:

$$x + 40y \leq 2000$$

Now, evaluate options based on this inequality:

Key Concept: Basic Concept

d) $x > 2$

[Solution Description] Dividing both sides of the inequality by a positive number does not change the direction of the inequality. Therefore, dividing both sides by 7 gives $x > 2$.

Key Concept: Basic Definition

d) $\geq$

[Solution Description] The symbol for ‘greater than or equal to’ is $\geq$. This means that one quantity is either greater than or exactly equal to another.

Key Concept: Inequality with Fractions

a) $x > \frac{30}{7}$

[Solution Description]

To solve the inequality $\frac{5}{3}x – 4 > \frac{1}{2}x + 1$, follow these steps:

First, clear the fractions by finding a common denominator, which is 6.

Multiply every term by 6:

$$6 \left( \frac{5}{3}x \right) – 6 \cdot 4 > 6 \left( \frac{1}{2}x \right) + 6 \cdot 1$$

Simplify:

$$10x – 24 > 3x + 6$$

Now, move all terms involving $x$ to one side:

$$10x – 3x > 6 + 24$$

$$7x > 30$$

Divide both sides by 7:

$$x > \frac{30}{7}$$

The solution is $x > \frac{30}{7}$.

Key Concept: Double Inequality

a) $$1 < x < 3$$ [Solution Description] A double inequality shows that one value falls between two others using inequalities. In this case, $1 < x < 3$ indicates that x lies between 1 and 3.

Key Concept: Complex Algebraic Solution, Inequality Transformation

b) $x \leq \frac{11}{2}$

[Solution Description] To solve this inequality, first clear the fractions by finding a common denominator, which is 15:

$$15 \left(\frac{3x – 4}{5}\right) \leq 15 \left(\frac{x + 2}{3}\right)$$

This simplifies to:

$$3(3x – 4) \leq 5(x + 2)$$

Expanding both sides gives:

$$9x – 12 \leq 5x + 10$$

Next, subtract $5x$ from both sides:

$$4x – 12 \leq 10$$

Add 12 to both sides:

$$4x \leq 22$$

Finally, divide by 4:

$$x \leq \frac{22}{4} = \frac{11}{2}$$

So, the solution is $x \leq \frac{11}{2}$.

Key Concept: Simple Inequality

c) $z < 12$ [Solution Description] Starting with the inequality $z + 3 < 15$, we apply Rule 1, which allows us to subtract the same number from both sides. Subtracting 3 from both sides, we get: $$z + 3 - 3 < 15 - 3$$ This simplifies to: $$z < 12$$

Key Concept: Basic Inequality

c) $x < 3$ [Solution Description] To solve the inequality $5x + 3 < 18$, we first subtract 3 from both sides of the inequality: $$5x + 3 - 3 < 18 - 3$$ Simplifying this gives: $$5x < 15$$ Next, divide both sides by 5 to isolate $x$: $$x < 3$$ Therefore, the solution to the inequality is $x < 3$.

Key Concept: Perimeter Inequality

c) 9 cm

[Solution Description]

Let the shortest side be $x$. Then the longest side is $4x$, and the third side is $4x – 3$.

The perimeter condition states:

$$x + 4x + (4x – 3) \geq 70$$

Simplifying this inequality gives:

$$9x – 3 \geq 70$$

Add 3 to both sides:

$$9x \geq 73$$

Divide by 9:

$$x \geq \frac{73}{9} \approx 8.11$$

Therefore, the shortest integer length satisfying this condition is 9 cm.

Key Concept: Basic Inequality

b) $7 – x > 10$

[Solution Description] An inequality involves a relationship between two expressions using symbols like ‘<', '>‘, ‘≤’, or ‘≥’. In this case, we look for such a symbol in each option.

Key Concept: Basic Inequality

a) $\{ x | x \leq 7 \}$

[Solution Description] We begin by solving the inequality $2x – 5 \leq 9$.

Add 5 to both sides to get: $2x \leq 14$.

Divide both sides by 2 to solve for $x$: $x \leq 7$.

Therefore, the solution set includes all real numbers less than or equal to 7.

Key Concept: Basic Inequality

b) $(-\infty, 2)$

[Solution Description]

To solve the inequality, we subtract 5 from both sides:

$3x + 5 – 5 < 11 - 5$, which simplifies to $3x < 6$. Then, divide both sides by 3: $\frac{3x}{3} < \frac{6}{3}$, resulting in $x < 2$. The solution set is $(-\infty, 2)$ on a number line.

Key Concept: Quadratic Inequality

b) $ax^2 + bx + c \leq 0$

[Solution Description]

A quadratic inequality involves a term with the variable raised to the second power, such as $ax^2 + bx + c > 0$ or $ax^2 + bx + c \leq 0$.

Option B has a quadratic expression since it includes $ax^2 + bx + c \leq 0$. The rest of the options involve linear terms only.

Key Concept: Numerical Inequality

a) $7 > 3$

[Solution Description]

Comparing the numbers given in each option:

– For option A, $7 > 3$ is true because 7 is greater than 3.

– Option B states $5 = 9$, which is false because 5 is not equal to 9.

– Option C says $6 < 4$, which is false because 6 is greater than 4. - Option D states $8 \leq 7$, which is false because 8 is greater than 7. Therefore, the correct answer is option A: $7 > 3$.

Key Concept: Solving Inequality

a) $x > \frac{1}{2}$

[Solution Description]

Step 1: Distribute 5 into the expression: $5 \cdot 2x + 5 \cdot 3 > 20$.

Simplifies to $10x + 15 > 20$.

Step 2: Subtract 15 from both sides: $10x > 5$.

Step 3: Divide both sides by 10: $x > \frac{1}{2}$.

Thus, the solution set for $x$ is $(\frac{1}{2}, \infty)$.

Key Concept: Basic Definition

c) $5 > 3$

[Solution Description] An inequality is represented when two expressions are related using symbols such as ‘<', '>‘, ‘≤’, or ‘≥’. Here, option c) 5 > 3 uses the symbol ‘>’, indicating it is an inequality.

Key Concept: Combined Mixture Problem

c) 200 ml

[Solution Description]

To find out how much pure water is needed, we first calculate the amount of saline in the original solution: $10%\ of\ 400$ ml = $40$ ml. Since we want a 5% saline solution in 400 ml total, this means $$5\%\ of\ 400$$ ml = $20$ ml will be saline. Therefore, the remaining $20$ ml needs to be diluted with pure water.

Let $$x$$ be the volume of pure water that needs to be added.

The equation for the new concentration is:

$$(40\ \text{ml saline})/(400 + x)\ \text{ml} = 0.05$$

Solving for $x$:

$$400 – 20 = x$$

$$x = 200\ \text{ml}$$

Thus, 200 ml of pure water should be added.