State whether the following statement is True or F – Class 11 Mathematics

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xy < 0

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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 statement is False. The condition $xy < 0$ means $x$ and $y$ have opposite signs — one is positive and the other negative — not that both are negative.

The core idea here is the sign rule for multiplication. When you multiply two real numbers, the sign of the product tells you whether the signs of the two numbers are the same or different. If the product is positive, both numbers have the same sign (both positive or both negative). If the product is negative, the numbers have opposite signs — one is positive and the other negative.

The statement given says: "If $xy < 0$ , then $x < 0$ and $y < 0$ ." This claims that a negative product forces both numbers to be negative. But that’s only half the story — it ignores the case where one is positive and the other negative.

Let’s break it down.

What $xy < 0$ actually tells us

A negative product means the two numbers have opposite signs. So either:
- $x > 0$ and $y < 0$ , or
- $x < 0$ and $y > 0$ .
The statement only allows the second case ( $x < 0$ and $y < 0$ ), which would actually give $xy > 0$ , not $xy < 0$ .
Test with a concrete example

Take $x = 2$ and $y = -3$ . Then $xy = -6 < 0$ , so the condition holds. But $x = 2$ is not less than $0$ — it’s greater than $0$ . So the conclusion " $x < 0$ and $y < 0$ " is false here.
Why the statement is false …

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Q.State whether the following statement is True or False: If xy<0xy < 0xy<0, then x<0x < 0x<0 and y<0y < 0y<0.

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Q.State whether the following statement is True or False: If $xy < 0$ , then $x < 0$ and $y < 0$ .

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