Solving Equations
Algebraic Properties of Real Numbers
By the time you reach class X, you've already practiced and become familiar with these fundamental algebraic properties of the set of real numbers (\(\mathbb{R}\)). This foundational knowledge not only helps in solving algebraic expressions and equations but also lays the groundwork for understanding more advanced mathematical concepts in higher classes and subjects like calculus and real analysis. These properties ensure the predictability and stability needed for mathematical operations and proofs, forming an essential part of your mathematical toolkit.

Closure Property:
 Addition: For any two real numbers \(a\) and \(b\), their sum \(a + b\) is also a real number.
 Multiplication: For any two real numbers \(a\) and \(b\), their product \(a \cdot b\) is also a real number.
 This means you can add or multiply any two real numbers and always get another real number as a result.

Associativity:
 Addition: \((a + b) + c = a + (b + c)\) for any real numbers \(a\), \(b\), and \(c\).
 Multiplication: \((a \cdot b) \cdot c = a \cdot (b \cdot c)\) for any real numbers \(a\), \(b\), and \(c\).
 This property allows you to change the grouping of numbers in addition or multiplication without changing the result.

Commutativity:
 Addition: \(a + b = b + a\) for any real numbers \(a\) and \(b\).
 Multiplication: \(a \cdot b = b \cdot a\) for any real numbers \(a\) and \(b\).
 This means the order in which you add or multiply two numbers does not affect the result.

Distributivity of Multiplication over Addition:
 \(a \cdot (b + c) = a \cdot b + a \cdot c\) for any real numbers \(a\), \(b\), and \(c\).
 This property links multiplication and addition, showing how multiplication distributes over addition.

Identity Elements:
 Addition: There exists an additive identity, 0, such that \(a + 0 = a\) for any real number \(a\).
 Multiplication: There exists a multiplicative identity, 1, such that \(a \cdot 1 = a\) for any real number \(a\).
 These identities are unique numbers that leave another number unchanged when added or multiplied.

Inverse Elements:
 Addition: For every real number \(a\), there exists an additive inverse \(a\) such that \(a + (a) = 0\).
 Multiplication: For every real number \(a\), except 0, there exists a multiplicative inverse \(1/a\) such that \(a \cdot (1/a) = 1\).
An important algebraic property related to the real numbers (\(\mathbb{R}\)) that you've likely encountered by class X is related to multiplication:
 Zero Product Property: If the product of two real numbers is zero, then at least one of the numbers must be zero. Mathematically, if \(a \cdot b = 0\), then either \(a = 0\) or \(b = 0\) (or both).
Order properties
In mathematics, "order" refers to the arrangement of elements in a structured sequence based on a specific criterion that determines how each element compares to others in the set.
The order properties of real numbers, particularly the trichotomy property, play a crucial role in understanding the relationships within the set of real numbers (\(\mathbb{R}\)). Here's a simplified look:
Trichotomy Property
For any two real numbers \(a\) and \(b\), exactly one of the following is true:
 \(a < b\), meaning \(a\) is less than \(b\),
 \(a = b\), meaning \(a\) and \(b\) are equal,
 \(a > b\), meaning \(a\) is greater than \(b\).
This foundational property establishes a comprehensive comparison framework for real numbers, affirming their ordered structure.
As a direct consequence of the trichotomy property, we can understand that:
 If \(a \geq b\) and \(b \geq a\), then we must have \(a = b\). This situation arises because, according to the trichotomy property, \(a\) and \(b\) can either be equal or one can be greater than the other. If each number is both not less than and not greater than the other, the only remaining possibility is that they are equal. This consequence helps in determining equality through comparative reasoning in various mathematical contexts.
Notation
The notation \(a < b < c\) is a concise way to state that \(a\) is less than \(b\) and that \(b\) is less than \(c\). This shorthand notation implies two separate inequalities:
 \(a < b\): \(a\) is less than \(b\),
 \(b < c\): \(b\) is less than \(c\).
Using this notation simplifies expressions and mathematical arguments, making it clear that there's an ordered relationship among three elements, with \(a\) being the smallest and \(c\) being the largest in this specific arrangement.
Transitivity
Transitivity is a fundamental property in the context of order relations, especially within the set of real numbers. It states that if we have three elements \(a\), \(b\), and \(c\), and the relations \(a < b\) and \(b < c\) hold true, then it logically follows that \(a < c\). In simpler terms, if \(a\) is less than \(b\) and \(b\) is less than \(c\), then \(a\) must also be less than \(c\).
This rule helps us figure out and keep track of how things are arranged in order. It makes sure that when we compare things one by one, we can still see the big picture of how everything lines up from smallest to largest.
More properties

Addition Property: If \(a < b\), then adding the same number \(c\) to both sides keeps the inequality the same. So, \(a + c < b + c\). This means if one number is smaller than another, adding the same amount to both doesn't change which one is bigger.

Multiplication Property for Positive Numbers: If \(a < b\) and \(c > 0\), then multiplying both sides by \(c\) keeps the inequality direction the same, so \(ac < bc\). This tells us that if one number is smaller than another, and you multiply both by the same positive number, the order doesn't change—the first number is still smaller.

Multiplication Property for Negative Numbers: If \(a < b\) and \(c < 0\), then when you multiply both sides by \(c\), the inequality flips direction, meaning \(ac > bc\). This happens because multiplying by a negative number reverses the order. So, if one number was smaller before, after multiplying both by a negative number, it becomes larger.
Example
Solve for all real values of \(c\) which satisfy the inequality \( \frac{x}{2} + 1 \geq 5x + 4 \)
Solution:
To solve the inequality \( \frac{x}{2} + 1 \geq 5x + 4 \):
Subtract 1 from both sides:
\[ \frac{x}{2} \geq 5x + 3 \]Multiply both sides by 2:
\[ x \geq 10x + 6 \]Subtract \( x \) from both sides:
\[ 0 \geq 9x + 6 \]Subtract 6 from both sides:
\[ 6 \geq 9x \]Divide both sides by 9, noting that dividing by a positive number will not change the inequality's direction:
\[ \frac{6}{9} \geq x \]This simplifies to:
\[ \frac{2}{3} \geq x \]So, \( x \) is less than or equal to \(\frac{2}{3}\). The solution set can be written in interval notation as:
\[ x \in (\infty, \frac{2}{3}] \]Example
Find all real values of \( x \) which satisfy \( 9x + 11 \leq x  10 \).
Solution:
\[ \begin{align*} 9x + 11 &\leq x  10 \\ 9x  x &\leq 10  11 \\ 8x &\leq 21 \\ x &\leq \frac{21}{8} \end{align*} \]\( x \in (\infty, \frac{21}{8}] \)

If \( a, b \in \mathbb{R} \) (meaning \( a \) and \( b \) are real numbers) and \( ab \geq 0 \), this implies that:
 Either both \( a \) and \( b \) are greater than or equal to zero (\( a \geq 0 \) and \( b \geq 0 \))
 Or both \( a \) and \( b \) are less than or equal to zero (\( a \leq 0 \) and \( b \leq 0 \))
Example
Find all real values of \( x \) which satisfy
\[ x^2  4x + 3 > 0 \]Sol.:
\[ x^2  4x + 3 > 0 \]\[ \Rightarrow x^2  3x  x + 3 > 0 \]\[ \Rightarrow x(x3)  (x3) > 0 \]\[ \Rightarrow (x3)(x1) > 0 \]\[ \Rightarrow x3 > 0 \quad \text{and} \quad x1 > 0 , or, x3 <0 \quad \text{and} \quad x1 < 0\]\[ \text{Case I:} \quad (x3) > 0 \quad \text{and} \quad (x1) > 0 \]\[ \text{Case II:} \quad (x3) < 0 \quad \text{and} \quad (x1) < 0 \]Case I:
\[ \Rightarrow x > 3 \quad \text{and} \quad x > 1 \]\[ \Rightarrow x > 3 \]Case II:
\[ \Rightarrow x < 3 \quad \text{and} \quad x < 1 \]\[ \Rightarrow x < 1 \]\[ x > 3 \quad \text{or} \quad x < 1 \]\[ \Rightarrow x \in (\infty,1) \cup (3,\infty) \]Example
Find all real values of \( x \) that satisfy the inequality:
\[ x^2 + 5x + 6 > 0 \]Solution:
We start by factoring the quadratic expression:
\[ x^2 + 5x + 6 > 0 \]\[ \Rightarrow x^2 + 3x + 2x + 6 > 0 \]\[ \Rightarrow x(x + 3) + 2(x + 3) > 0 \]\[ \Rightarrow (x + 3)(x + 2) > 0 \]We now have the product of two factors greater than zero. We will consider two cases based on the product of these factors.
Case I: Both factors are positive.
If both factors are positive, their product is positive. This means:
\[ x + 3 > 0 \quad \text{and} \quad x + 2 > 0 \]\[ \Rightarrow x > 3 \quad \text{and} \quad x > 2 \]Since both conditions must be satisfied simultaneously, the solution for this case is:
\[ x > 2 \]Case II: Both factors are negative.
If both factors are negative, their product is also positive. This means:
\[ x + 3 < 0 \quad \text{and} \quad x + 2 < 0 \]\[ \Rightarrow x < 3 \quad \text{and} \quad x < 2 \]However, since \( x \) cannot be less than \( 3 \) and simultaneously less than \( 2 \) (because \( 3 \) is less than \( 2 \)), there is no solution in this case, that is, \(x\in\emptyset\)
Therefore, the solution to the inequality is:
\[ x \in (\infty, 3) \cup (2, \infty) \]

If \( ab \leq 0 \), then:
 \( a \geq 0 \) and \( b \leq 0 \) or
 \( a \leq 0 \) and \( b \geq 0 \)
For the product of two numbers \( a \) and \( b \) to be nonpositive (that is, negative or zero), one of the numbers must be nonnegative (zero or positive) and the other nonpositive (zero or negative), or vice versa.
Example
Solve:
\[ x^2  5x + 6 \leq 0 \]Solution:
Starting with the inequality:
\[ x^2  5x + 6 \leq 0 \]The quadratic is factored:
\[ (x  3)(x  2) \leq 0 \]From here, we analyze two cases:
Case I:
\[ x  3 \geq 0 \quad \text{and} \quad x  2 \leq 0 \]This leads to:
\[ x \geq 3 \quad \text{and} \quad x \leq 2 \]This case has no solution because there's no number that is simultaneously greater than or equal to 3 and less than or equal to 2.
Case II:
\[ x  3 \leq 0 \quad \text{and} \quad x  2 \geq 0 \]This leads to:
\[ x \leq 3 \quad \text{and} \quad x \geq 2 \]This gives the solution interval:
\[ x \in [2, 3] \]This means the solution to the inequality \( x^2  5x + 6 \leq 0 \) is any real number \( x \) such that \( 2 \leq x \leq 3 \).

These inequalities describe the sign rules for the quotient of two numbers, indicating the conditions under which their division results in a positive, nonnegative, negative, or nonpositive value, while also highlighting the prohibition of division by zero.

If \( \frac{a}{b} > 0 \) then:
 \( a > 0 \) and \( b > 0 \) or
 \( a < 0 \) and \( b < 0 \)

If \( \frac{a}{b} \geq 0 \) then:
 \( a \geq 0 \) and \( b > 0 \) or
 \( a \leq 0 \) and \( b < 0 \)

If \( \frac{a}{b} < 0 \) then:
 \( a > 0 \) and \( b < 0 \) or
 \( a < 0 \) and \( b > 0 \)

If \( \frac{a}{b} \leq 0 \) then:
 \( a \geq 0 \) and \( b < 0 \) or
 \( a \leq 0 \) and \( b > 0 \)
The reason equality cannot apply to \( b \) in the inequality \( \frac{a}{b} \leq 0 \) and also in \( \frac{a}{b} \geq 0 \) is because division by zero is undefined in mathematics. If \( b \) were zero, the quotient \( \frac{a}{b} \) would be undefined, and the inequality would not hold. Thus, while \( a \) can be equal to zero (making the entire expression zero), \( b \) must be strictly positive or negative to avoid division by zero.
Example
Problem Statement:
Find all real values of \( x \) that satisfy the inequality:
\[ \frac{x+1}{2x1} \leq 0 \]Solution:
Starting with the inequality:
\[ \frac{x+1}{2x1} \leq 0 \]\[ \Rightarrow \color{red}x+1 \leq 0 \quad \text{and} \quad 2x1 > 0 \color{black} \quad \text{OR} \quad \color{blue} x+1 \geq 0 \quad \text{and} \quad 2x1 < 0 \]Case I:
\[ x + 1 \geq 0 \quad \text{and} \quad 2x  1 < 0 \]\[ \Rightarrow x \geq 1 \quad \text{and} \quad x < \frac{1}{2} \]The intersection of these two conditions gives us:
\[ x \in \left[1, \frac{1}{2}\right) \]Case II:
\[ x+1 \leq 0 \quad \text{and} \quad 2x1 > 0 \]\[ \Rightarrow x \leq 1 \quad \text{and} \quad 2x > 1 \]\[ \Rightarrow x \leq 1 \quad \text{and} \quad x > \frac{1}{2} \]This case is impossible as \( x \) cannot be both less than or equal to \( 1 \) and greater than \( \frac{1}{2} \) at the same time.
Therefore, the final solution set is:
\[ x \in \left[1, \frac{1}{2}\right) \]This represents the values of \( x \) for which the original inequality \( \frac{x+1}{2x1} \leq 0 \) holds true.
Example
Solve the inequality for real values of \(x\):
\[ \frac{x+1}{2x1} < 1 \]To find the real values of \( x \) which satisfy \( \frac{x+1}{2x1} < 1 \), the approach of crossmultiplying by \( 2x1 \) is a common method, but it can lead to an incorrect solution when dealing with inequalities.
When the inequality is manipulated by crossmultiplying, the calculation would go as follows:
\[ \frac{x+1}{2x1} < 1 \]Crossmultiplying by \( 2x1 \) gives:
\[ x+1 < 2x1 \]This simplifies to:
\[ x > 2 \]However, this solution assumes that \( 2x1 \) is positive. If \( 2x1 \) were negative, the direction of the inequality would change, and the solution would be different. This is why crossmultiplication isn't a valid step in solving inequalities unless you can be certain of the sign of the expression you're multiplying with, which in this case, you cannot.
If you were to plug in \( x = 2 \) into the original inequality:
\[ \frac{2+1}{2(2)1} = \frac{1}{5} < 1 \]This shows that \( x = 2 \) is indeed a solution to the inequality, highlighting the error that comes from improperly crossmultiplying in this context. Crossmultiplication in inequalities should only be done when you're sure the term you're multiplying with does not change sign over the domain you're considering, otherwise, it may lead to missing or extraneous solutions.

Start with the original inequality:
\[ \frac{x+1}{2x1} < 1 \] 
Bring 1 to the left side:
\[ \frac{x+1}{2x1}  1 < 0 \] 
Combine the terms over a common denominator:
\[ \frac{x+1  (2x1)}{2x1} < 0 \] 
Simplify the numerator:
\[ \frac{x+2}{2x1} < 0 \] 
Multiply by 1 (which flips the inequality sign):
\[ \frac{x2}{2x1} > 0 \] 
Now make two cases to solve the inequality:
Case 1: \( x2 > 0 \) and \( 2x1 > 0 \)
 Solve each part of the inequality:
\[ x > 2 \quad \text{and} \quad x > \frac{1}{2} \] Since \( x \) must be greater than both values, the solution for this case is \( x > 2 \).
Case 2: \( x2 < 0 \) and \( 2x1 < 0 \)
 Solve each part of the inequality:
\[ x < 2 \quad \text{and} \quad x < \frac{1}{2} \] Since \( x \) must be less than both values, the solution for this case is \( x < \frac{1}{2} \).

The final solution combines the two cases:
\[ x \in (\infty, \frac{1}{2}) \cup (2, \infty) \]
