Solve $3x + 8 > 2$, when (i) $x$ is an integer. (ii) $x$ is a real number.

> [!TLDR] > The inequality $3x + 8 > 2$ simplifies to $x > -2$. For integer $x$, the solution is $\{-1, 0, 1, 2, \dots\}$; for real $x$, it is $(-2, \infty)$. The core idea here is simple: a linear inequality behaves almost exactly like a linear equation, with one crucial exception — multiplying or dividing by a negative number flips the inequality sign. Since we only add, subtract, and divide by a positive number in this problem, the sign stays unchanged. The twist comes only in how we *present* the answer, depending on whether $x$ is an integer or a real number. Let’s work through it. 1. **Isolate $x$.** Start with $3x + 8 > 2$. Subtract 8 from both sides: $$3x > 2 - 8$$ $$3x > -6$$ Now divide both sides by 3 (positive, so inequality direction stays the same): $$x > -2$$ That’s the core condition. Every $x$ greater than $-2$ satisfies the inequality. 2. **Case (i): $x$ is an integer.** Integers are whole numbers (…, -3, -2, -1, 0, 1, 2, …). The condition $x > -2$ means $x$ must be strictly greater than $-2$. So $-2$ itself is **not** included. The smallest integer that works is $-1$, then $0$, $1$, $2$, and so on forever. In set notation: $\{-1, 0, 1, 2, 3, \dots\}$. > [!WARNING] > A common mistake is to include $-2$ because “greater than or equal” feels close. But $x > -2$ means $-2$ is excluded — check: $3(-2) + 8 = 2$, which is not greater than 2. 3. **Case (ii): $x$ is a real number.** …

Solve 3x + 8 > 2, when (i) x is an integer. (ii) x – Class 11 Mathematics

Exercise 5.1 · Q4

3x + 8 > 2

(i)

x

is an integer.

(ii)

x

is a real number.

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Concept understanding — Linear Inequality Solutions

Linear Inequality Solutions – A First Look

Imagine you're standing on a number line. You know exactly where the number 5 is. But what if I asked you to stand on "all numbers greater than 5"? You can't stand on all of them at once — they stretch infinitely to the right. That's the core idea of an inequality: instead of one exact point, you get a whole region of possible values.

A linear inequality is just like a linear equation ( $ax + b = 0$ ), but instead of an equals sign, you have one of these: $<$ , $>$ , $\le$ , or $\ge$ . The solution is not a single number — it's an interval (or a union of intervals) on the number line.

From Equation to Inequality

Start with a simple equation:

2x + 3 = 7

Solve it: $2x = 4 \implies x = 2$ . One point.

Now change it to an inequality:

2x + 3 > 7

Solve it the same way — but the meaning changes. Subtract 3: $2x > 4$ . Divide by 2: $x > 2$ .

The solution is all numbers greater than 2. On a number line, you draw an open circle at 2 (because 2 itself is not included) and shade everything to the right.

Watch out

If you multiply or divide both sides of an inequality by a negative number, the inequality sign reverses. For example: $-x < 5$ becomes $x > -5$ . This is the single most common mistake students make.

The Four Types of Solutions

Inequality	Meaning	Number line representation
$x > a$	All numbers strictly greater than $a$	Open circle at $a$ , shade right
$x \ge a$	All numbers greater than or equal to $a$	Closed (filled) circle at $a$ , shade right
$x < a$	All numbers strictly less than $a$	Open circle at $a$ , shade left
$x \le a$	All numbers less than or equal to $a$	Closed circle at $a$ , shade left

The solution set is usually written in interval notation:

$x > 2$ → $(2, \infty)$
$x \le -3$ → $(-\infty, -3]$

Parentheses $($ or $)$ mean the endpoint is not included. Brackets $[$ or $]$ mean it is included.

Solving a Linear Inequality: Step by Step

Solve $3x - 5 \le 7x + 3$ .

Bring variable terms to one side:

$3x - 5 - 7x \le 3$

$-4x - 5 \le 3$
Isolate the variable term:

$-4x \le 8$
Divide by the coefficient (here it's $-4$ , so reverse the sign):

$x \ge -2$

The solution is $x \ge -2$ , or in interval notation: $[-2, \infty)$ .

Tip

Always check your answer by testing a number from the solution set. For $x \ge -2$ , test $x = 0$ : $3(0)-5 = -5$ and $7(0)+3 = 3$ . Is $-5 \le 3$ ? Yes. Now test a number outside, say $x = -3$ : $3(-3)-5 = -14$ and $7(-3)+3 = -18$ . Is $-14 \le -18$ ? No — so the inequality fails, confirming our solution is correct.

Why This Matters …

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The inequality $3x + 8 > 2$ simplifies to $x > -2$ . For integer $x$ , the solution is $\{-1, 0, 1, 2, \dots\}$ ; for real $x$ , it is $(-2, \infty)$ .

The core idea here is simple: a linear inequality behaves almost exactly like a linear equation, with one crucial exception — multiplying or dividing by a negative number flips the inequality sign. Since we only add, subtract, and divide by a positive number in this problem, the sign stays unchanged. The twist comes only in how we present the answer, depending on whether $x$ is an integer or a real number.

Let’s work through it.

Isolate $x$ . Start with $3x + 8 > 2$ . Subtract 8 from both sides:

3x > 2 - 8

3x > -6

Now divide both sides by 3 (positive, so inequality direction stays the same):

x > -2

That’s the core condition. Every $x$ greater than $-2$ satisfies the inequality.

Case (i): $x$ is an integer.

Integers are whole numbers (…, -3, -2, -1, 0, 1, 2, …). The condition $x > -2$ means $x$ must be strictly greater than $-2$ . So $-2$ itself is not included. The smallest integer that works is $-1$ , then $0$ , $1$ , $2$ , and so on forever.

In set notation: $\{-1, 0, 1, 2, 3, \dots\}$ .

Watch out

A common mistake is to include $-2$ because “greater than or equal” feels close. But $x > -2$ means $-2$ is excluded — check: $3(-2) + 8 = 2$ , which is not greater than 2.
Case (ii): $x$ is a real number. …

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