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 claims that if the product of two numbers is positive, then one must be positive and the other negative. This is incorrect because a positive product requires both numbers to have the same sign. The statement is False.

The core idea here revolves around understanding how the signs of two numbers determine the sign of their product. This is a fundamental concept in algebra.

When we multiply two real numbers, $x$ and $y$ , the sign of their product $xy$ follows these rules:

If $x$ is positive and $y$ is positive, then $xy$ is positive.
If $x$ is negative and $y$ is negative, then $xy$ is positive.
If $x$ is positive and $y$ is negative, then $xy$ is negative.
If $x$ is negative and $y$ is positive, then $xy$ is negative.

The statement given is an "If... then..." statement: "If $xy > 0$ , then $x > 0$ and $y < 0$ ." For such a statement to be true, the conclusion must always hold whenever the initial condition is met. If we can find even one case where the initial condition is true but the conclusion is false, then the entire statement is false.

Analyze the initial condition: The statement begins with the condition $xy > 0$ .

This means the product of $x$ and $y$ is a positive number. Based on the rules of multiplication, for $xy$ to be positive, $x$ and $y$ must have the same sign.

There are two possibilities for this:
- Case 1: Both $x$ and $y$ are positive. That is, $x > 0$ and $y > 0$ .
- Case 2: Both $x$ and $y$ are negative. That is, $x < 0$ and $y < 0$ .
Analyze the proposed conclusion: The statement then claims "then $x > 0$ and $y < 0$ ."

This conclusion describes a situation where $x$ is a positive number and $y$ is a negative number.
Evaluate the product for the conclusion: Let's consider what the product $xy$ would be if the conclusion ( $x > 0$ and $y < 0$ ) were true.

If $x > 0$ (positive) and $y < 0$ (negative), then their product $xy$ would be a negative number. That is, $xy < 0$ .
Compare the initial condition and the conclusion:

The initial condition states $xy > 0$ .

The proposed conclusion, if true, would imply $xy < 0$ . …

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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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