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Complete Course of Mathematics
Topic 1: Numbers & Numerical Applications Topic 2: Algebra Topic 3: Quantitative Aptitude
Topic 4: Geometry Topic 5: Construction Topic 6: Coordinate Geometry
Topic 7: Mensuration Topic 8: Trigonometry Topic 9: Sets, Relations & Functions
Topic 10: Calculus Topic 11: Mathematical Reasoning Topic 12: Vectors & Three-Dimensional Geometry
Topic 13: Linear Programming Topic 14: Index Numbers & Time-Based Data Topic 15: Financial Mathematics
Topic 16: Statistics & Probability


Content On This Page
Definition of an Equation Linear Equation in One Variable: Introduction and Solution/Root Solving Linear Equations in One Variable (Methods, Rule of Transposition)
Solving Word Problems based on Linear Equations in One Variable


Linear Equations in One Variable



Definition of an Equation

Mathematical Statements of Equality

In mathematics, an equation is a fundamental concept used to express the equality between two mathematical expressions. It is a statement that asserts that the expression on one side of the equals sign represents the same value as the expression on the other side.

An equation always contains an equals sign ($=$) in the middle, separating the two expressions. The expression to the left of the equals sign is called the Left Hand Side (LHS), and the expression to the right is called the Right Hand Side (RHS).

LHS = RHS

[General form of an equation]

The expressions in an equation can be purely numerical (e.g., $2 + 3 = 5$) or algebraic (e.g., $x + 5 = 10$), involving variables, constants, and mathematical operations. When an equation contains variables, the truth of the statement (whether the LHS is indeed equal to the RHS) often depends on the specific numerical values assigned to those variables.


Types of Equations based on Truth Value

Equations can be classified based on whether the equality holds true for some, all, or no values of the variables:

  1. Conditional Equation:

    A conditional equation is an equation that is true only for certain specific value(s) of the variable(s) and false for other values. These are the most common types of equations encountered when we are asked to "solve" an equation; our goal is to find the values of the variable(s) that make the equation true.

    Example:

    $x + 5 = 10$

    This equation is true only when $x = 5$. If we substitute any other value for $x$, like $x=3$, we get $3+5=8$, which is not equal to $10$, so the statement is false. The set of values for which the equation is true $\{5\}$ is called the solution set.

    Example:

    $y^2 = 4$

    This equation is true only when $y=2$ or $y=-2$. For any other value of $y$, the equality does not hold. The solution set is $\{-2, 2\}$.

  2. Identity:

    An identity is an equation that is true for all possible values of the variable(s) for which both sides of the equation are defined. Identities represent fundamental equivalences between two algebraic expressions.

    Example:

    $2(x - 3) = 2x - 6$

    This equation is true for any real value of $x$. If $x=5$, LHS $= 2(5-3) = 2(2) = 4$, RHS $= 2(5)-6 = 10-6 = 4$. If $x=0$, LHS $= 2(0-3) = 2(-3) = -6$, RHS $= 2(0)-6 = 0-6 = -6$. This identity is based on the distributive property.

    Example:

    $(a+b)^2 = a^2 + 2ab + b^2$

    This is true for all real values of $a$ and $b$.

  3. Contradiction:

    A contradiction is an equation that is never true, no matter what values are substituted for the variable(s). It represents a false statement for all possible variable values (within the domain where both sides are defined).

    Example:

    $x + 1 = x + 2$

    If we subtract $x$ from both sides, we get $1 = 2$, which is a false statement. No value of $x$ can make this equation true. This equation has no solutions, and its solution set is empty.

    Example:

    $y^2 = -5$

    In the set of real numbers, the square of any real number is non-negative ($y^2 \ge 0$). Therefore, there is no real number $y$ such that $y^2 = -5$. This equation has no real solutions (it has complex solutions, which are discussed later).

While identities and contradictions are specific types of equations, in the context of learning to "solve equations," we are primarily focused on identifying the conditions (values of variables) under which a conditional equation becomes a true statement.


Components of an Equation

An equation is composed of several key elements:

Understanding what constitutes an equation is the foundational step for learning how to manipulate them, classify them by the types of expressions they contain (e.g., linear, quadratic), and develop methods to find their solutions.



Linear Equation in One Variable: Introduction and Solution/Root

Defining a Linear Equation in One Variable

An equation is a mathematical statement asserting that two expressions are equal. Equations can be classified based on the type of expressions they involve and the number of variables they contain. A particularly important type is the linear equation.

A linear equation is an equation where each term is either a constant or the product of a constant and a single variable raised to the power of 1. This means that variables do not appear with exponents other than 1 (e.g., no $x^2, y^3, \ldots$), they are not multiplied together (e.g., no $xy$), they are not in the denominator (e.g., no $1/x$), and they are not inside radicals (e.g., no $\sqrt{y}$).

A linear equation in one variable is a linear equation that involves only one type of variable. Such an equation can always be rearranged and written in the standard form:

$ax + b = 0$

[Standard Form of a Linear Equation in One Variable]

In this standard form:

Examples of linear equations in one variable:

Examples of equations that are NOT linear equations in one variable:


Solution or Root of a Linear Equation

A solution (or sometimes called a root) of a linear equation in one variable is the specific numerical value that, when substituted for the variable in the equation, makes the equation a true statement (i.e., makes the LHS equal to the RHS). For a linear equation in one variable $ax + b = 0$ where $a \neq 0$, there is always exactly one unique solution.

Finding the solution to a linear equation is the primary goal when solving such an equation. It involves using inverse operations and the properties of equality to isolate the variable on one side of the equation.

Let's find the solution for the standard linear equation $ax + b = 0$, assuming $a \neq 0$:

$ax + b = 0$

[Original equation]

To isolate the term with $x$, subtract $b$ from both sides of the equation (using the subtraction property of equality):

$ax + b - b = 0 - b$

$ax = -b$

Now, to isolate $x$, divide both sides of the equation by $a$ (using the division property of equality, which is allowed since $a \neq 0$):

$\frac{ax}{a} = \frac{-b}{a}$

$x = -\frac{b}{a}$

[Unique Solution]

This formula $x = -\frac{b}{a}$ gives the unique solution for any linear equation in one variable $ax + b = 0$ (with $a \neq 0$).

Example: Find the solution of $2x + 3 = 0$.

Comparing with $ax+b=0$, we have $a=2$ and $b=3$.

Using the formula, $x = -\frac{b}{a} = -\frac{3}{2}$.

The solution is $\textbf{x = -\frac{3}{2}}$.

Check: Substitute $x = -3/2$ back into the original equation: $2\left(-\frac{3}{2}\right) + 3 = -3 + 3 = 0$. LHS = RHS. The solution is correct.

Example: Find the solution of $y - 5 = 0$.

Comparing with $ay+b=0$, we have $a=1$ and $b=-5$.

Using the formula, $y = -\frac{b}{a} = -\frac{(-5)}{1} = -(-5) = 5$.

The solution is $\textbf{y = 5}$.

Check: Substitute $y=5$ back into the original equation: $5 - 5 = 0$. LHS = RHS. The solution is correct.

The solution of the linear equation $ax + b = 0$ is precisely the same as the zero (or root) of the linear polynomial $P(x) = ax + b$. Geometrically, it is the x-intercept of the graph of the linear function $y = ax + b$.



Solving Linear Equations in One Variable (Methods, Rule of Transposition)

The Goal of Solving a Linear Equation

Solving a linear equation in one variable means finding the specific value of the variable that makes the equation a true statement. This value is called the solution or root of the equation. For a linear equation in one variable of the standard form $ax + b = 0$ with $a \neq 0$, there is always exactly one unique solution.

The process of solving involves manipulating the equation algebraically to isolate the variable on one side of the equals sign. All valid manipulations are based on maintaining the balance of the equality.


Principles of Solving Equations

To maintain the equality of an equation while manipulating it, whatever operation is performed on one side of the equation must also be performed on the other side. These principles are derived from the basic properties of equality:

  1. Addition Property of Equality:

    If the same quantity is added to both sides of an equation, the equation remains balanced (true).

    If $A = B$, then $A + k = B + k$.

    [Addition Property]

  2. Subtraction Property of Equality:

    If the same quantity is subtracted from both sides of an equation, the equation remains balanced (true).

    If $A = B$, then $A - k = B - k$.

    [Subtraction Property]

  3. Multiplication Property of Equality:

    If both sides of an equation are multiplied by the same non-zero quantity, the equation remains balanced (true).

    If $A = B$, then $A \times k = B \times k$ (where $k \neq 0$).

    [Multiplication Property]

  4. Division Property of Equality:

    If both sides of an equation are divided by the same non-zero quantity, the equation remains balanced (true).

    If $A = B$, then $A / k = B / k$ (where $k \neq 0$).

    [Division Property]

The process of solving a linear equation involves applying these principles to move terms and coefficients away from the variable, effectively isolating it.


Method of Balancing the Equation

This method directly applies the principles of equality by performing the inverse operation on both sides of the equation to "undo" operations being applied to the variable. The typical strategy is to first isolate the term containing the variable, and then isolate the variable itself.

Example 1. Solve the equation $3x - 7 = 8$ using the balancing method.

Answer:

The goal is to isolate $x$. The term with $x$ is $3x$. The constant term $-7$ is with $3x$. To undo the subtraction of $7$, we add $7$ to both sides of the equation:

$3x - 7 + 7 = 8 + 7$

(Add 7 to both sides)

Simplify both sides:

$3x = 15$

Now the variable term $3x$ is isolated. To isolate $x$, we need to undo the multiplication by $3$. We do this by dividing both sides of the equation by $3$:

$\frac{3x}{3} = \frac{15}{3}$

(Divide both sides by 3)

Simplify both sides:

$x = 5$

The solution is $\textbf{x = 5}$.

Check: Substitute $x=5$ back into the original equation: $3(5) - 7 = 15 - 7 = 8$. The LHS is $8$, which equals the RHS. The solution is correct.


Example 2. Solve $\frac{y}{2} + 5 = 11$ using the balancing method.

Answer:

The goal is to isolate $y$. The term with $y$ is $\frac{y}{2}$. The constant term $+5$ is with $\frac{y}{2}$. To undo the addition of $5$, subtract $5$ from both sides:

$\frac{y}{2} + 5 - 5 = 11 - 5$

(Subtract 5 from both sides)

Simplify both sides:

$\frac{y}{2} = 6$

Now the variable term $\frac{y}{2}$ is isolated. To isolate $y$, we need to undo the division by $2$. We do this by multiplying both sides of the equation by $2$:

$\frac{y}{2} \times 2 = 6 \times 2$

(Multiply both sides by 2)

Simplify both sides:

$y = 12$

The solution is $\textbf{y = 12}$.

Check: Substitute $y=12$ back into the original equation: $\frac{12}{2} + 5 = 6 + 5 = 11$. The LHS is $11$, which equals the RHS. The solution is correct.


Rule of Transposition

The Rule of Transposition is a convenient shortcut that stems directly from the balancing method. Instead of explicitly showing the same operation being done on both sides, we can conceptually "transpose" or move a term from one side of the equation to the other by changing the sign of the operation associated with that term.

Here's how transposition works:

The rule of transposition streamlines the steps of solving linear equations by allowing you to rearrange terms more fluidly.

Example 3. Solve the equation $5y + 10 = 3y - 6$ using the transposition method.

Answer:

The equation is $5y + 10 = 3y - 6$. We want to gather all terms with the variable ($y$) on one side and all constant terms on the other side.

Let's move the variable term $3y$ from the RHS to the LHS. Since it's $+3y$ on the RHS, it becomes $-3y$ on the LHS:

$5y - 3y + 10 = -6$

[Transpose 3y]

Now, let's move the constant term $+10$ from the LHS to the RHS. Since it's $+10$ on the LHS, it becomes $-10$ on the RHS:

$5y - 3y = -6 - 10$

[Transpose 10]

Combine the like terms on each side of the equation:

$(5 - 3)y = (-6 - 10)$

$2y = -16$

Finally, isolate $y$. The coefficient $2$ is multiplying $y$. Transpose $2$ to the RHS, where it will divide $-16$:

$y = \frac{-16}{2}$

[Transpose 2]

Simplify the fraction:

$y = -8$

The solution is $\textbf{y = -8}$.

Check: Substitute $y=-8$ into the original equation: LHS $= 5(-8) + 10 = -40 + 10 = -30$. RHS $= 3(-8) - 6 = -24 - 6 = -30$. LHS = RHS. The solution is correct.


Solving Equations with Brackets

If a linear equation contains expressions within parentheses (brackets), the first step is generally to remove these brackets using the distributive property. After removing the brackets, the equation will be in a form where you can apply the balancing method or transposition to solve for the variable.

Example 4. Solve the equation $2(x - 3) = 5 - 3(x + 1)$.

Answer:

The equation is $2(x - 3) = 5 - 3(x + 1)$.

Step 1: Remove brackets using the distributive property.

LHS: $2(x - 3) = 2 \times x - 2 \times 3 = 2x - 6$.

RHS: $5 - 3(x + 1) = 5 - (3 \times x + 3 \times 1) = 5 - (3x + 3)$. Now, distribute the negative sign before the parenthesis: $5 - 3x - 3$.

The equation becomes:

$$ 2x - 6 = 5 - 3x - 3 $$

Step 2: Simplify the RHS by combining constant terms:

$$ 2x - 6 = (5 - 3) - 3x $$ $$ 2x - 6 = 2 - 3x $$

Step 3: Use transposition to group variable terms on one side (say LHS) and constant terms on the other side (say RHS).

Transpose $-3x$ from RHS to LHS: it becomes $+3x$.

Transpose $-6$ from LHS to RHS: it becomes $+6$.

$$ 2x + 3x = 2 + 6 $$

Step 4: Combine like terms on both sides:

$$ 5x = 8 $$

Step 5: Isolate $x$. Transpose the coefficient $5$ (which is multiplying $x$) to the RHS, where it will divide $8$:

$$ x = \frac{8}{5} $$

The solution is $\textbf{x = \frac{8}{5}}$.

Check: Substitute $x = 8/5$ into the original equation:

LHS = $2\left(\frac{8}{5} - 3\right) = 2\left(\frac{8}{5} - \frac{15}{5}\right) = 2\left(\frac{8-15}{5}\right) = 2\left(\frac{-7}{5}\right) = -\frac{14}{5}$.

RHS = $5 - 3\left(\frac{8}{5} + 1\right) = 5 - 3\left(\frac{8}{5} + \frac{5}{5}\right) = 5 - 3\left(\frac{8+5}{5}\right) = 5 - 3\left(\frac{13}{5}\right) = 5 - \frac{39}{5}$.

To subtract on the RHS, find a common denominator:

$$ 5 - \frac{39}{5} = \frac{5 \times 5}{1 \times 5} - \frac{39}{5} = \frac{25}{5} - \frac{39}{5} = \frac{25-39}{5} = \frac{-14}{5} $$

LHS $= -14/5$ and RHS $= -14/5$. LHS = RHS. The solution is correct.

The ability to solve linear equations in one variable is a foundational skill in algebra. Whether using the explicit balancing method or the shortcut of transposition, the underlying principle is always to perform equivalent operations on both sides of the equation to isolate the variable. This skill is vital for solving more complex problems in mathematics and various applied fields.



Solving Word Problems based on Linear Equations in One Variable

Translating Words into Algebra

Word problems are narratives describing situations, often from real life, that pose a question requiring a mathematical answer. A crucial skill in algebra is the ability to translate the information given in a word problem into a mathematical model, such as a linear equation. Once the problem is expressed as a linear equation in one variable, we can use algebraic techniques to solve the equation and find the answer to the original problem.


Steps for Solving Word Problems

A systematic approach can greatly simplify the process of solving word problems using linear equations:

  1. Read and Understand the Problem:

    Read the problem carefully, perhaps several times, to fully grasp the situation. Identify exactly what information is given (the knowns) and precisely what question is being asked (the unknown quantity you need to find). Look for relationships between different quantities.
  2. Assign a Variable:

    Choose a letter (like $x, y, n, w$, etc.) to represent the single unknown quantity that you are asked to find. Write down a clear definition of what this variable stands for (e.g., "Let $x$ be the unknown number," or "Let $w$ represent the width of the rectangle in centimetres").
  3. Translate the Problem into an Equation:

    Convert the verbal description of the relationships between the quantities into an algebraic equation. Use the variable(s) you defined, constants, and mathematical operation symbols ($+, -, \times, \div$). Pay close attention to keywords and phrases:
    • "sum", "total", "more than", "increased by" suggest addition ($+$).
    • "difference", "less than", "subtracted from", "decreased by" suggest subtraction ($-$). Be careful with the order in "less than" or "subtracted from".
    • "product", "times", "multiplied by", "twice", "thrice" suggest multiplication ($\times$).
    • "quotient", "divided by" suggest division ($\div$ or fraction bar).
    • "is", "equals", "the result is", "gives" translate to the equals sign ($=$).

    Build the equation step-by-step based on the wording of the problem.

  4. Solve the Equation:

    Solve the linear equation you have formulated for the variable using the algebraic methods learned (balancing method or transposition).
  5. Check the Solution and State the Answer:

    Verify if the numerical solution you obtained for the variable makes sense in the context of the original word problem. Substitute the value back into the original problem description (not just the equation) to see if all conditions are met. Finally, write your answer in a clear sentence, making sure to include any units (like cm, kg, Rupees, years, etc.) if applicable.

Examples of Word Problems

Example 1. The sum of two consecutive integers is $45$. Find the integers.

Answer:

Step 1: Understand

We are given that the sum of two integers that follow each other in order is $45$. We need to find these two specific integers.

Step 2: Assign a Variable

Let the first integer be represented by the variable $x$.

Since the integers are consecutive, the second integer is one greater than the first. So, the second integer is $x + 1$.

Step 3: Translate into an Equation

The problem states that the "sum of two consecutive integers is $45$".

Sum of integers = (First integer) + (Second integer)

The equation is: $x + (x + 1) = 45$.

Step 4: Solve the Equation

Now, solve the linear equation for $x$:

$$ x + x + 1 = 45 $$

Combine like terms on the LHS:

$$ 2x + 1 = 45 $$

Subtract $1$ from both sides (or transpose $+1$ to RHS):

$$ 2x = 45 - 1 $$ $$ 2x = 44 $$

Divide both sides by $2$ (or transpose $2$):

$$ x = \frac{44}{2} $$ $$ x = 22 $$

So, the first integer is $x = 22$.

The second consecutive integer is $x + 1 = 22 + 1 = 23$.

Step 5: Check and State the Answer

Check: The two consecutive integers are $22$ and $23$. Their sum is $22 + 23 = 45$. This matches the condition given in the problem.

Answer: The two consecutive integers are $\textbf{22}$ and $\textbf{23}$.


Example 2. A number is such that when it is multiplied by $5$ and $7$ is subtracted from the product, the result is $23$. Find the number.

Answer:

Step 1: Understand

We are given a description of an operation performed on an unknown number, yielding a specific result ($23$). We need to find this unknown number.

Step 2: Assign a Variable

Let the unknown number be represented by the variable $n$.

Step 3: Translate into an Equation

Let's break down the description:

  • "when it is multiplied by $5$": This translates to $5 \times n$, or simply $5n$.
  • "$7$ is subtracted from the product": The product is $5n$. Subtracting $7$ gives $5n - 7$.
  • "the result is $23$": This means the expression $5n - 7$ is equal to $23$.

The equation is: $5n - 7 = 23$.

Step 4: Solve the Equation

Solve the linear equation for $n$:

$$ 5n - 7 = 23 $$

Add $7$ to both sides (or transpose $-7$ to RHS):

$$ 5n = 23 + 7 $$ $$ 5n = 30 $$

Divide both sides by $5$ (or transpose $5$):

$$ n = \frac{30}{5} $$ $$ n = 6 $$

Step 5: Check and State the Answer

Check: The number is $6$. Multiply by $5$: $6 \times 5 = 30$. Subtract $7$: $30 - 7 = 23$. This matches the given result in the problem statement.

Answer: The number is $\textbf{6}$.


Example 3. The perimeter of a rectangle is $40 \text{ cm}$. If its length is $12 \text{ cm}$, find its width.

Answer:

Step 1: Understand

We are given the perimeter and the length of a rectangle. We need to find the width. We should recall the formula for the perimeter of a rectangle.

Step 2: Assign a Variable

Let the unknown width of the rectangle be $w$ centimetres.

We are given the length, $l = 12 \text{ cm}$.

We are given the perimeter, $P = 40 \text{ cm}$.

Step 3: Translate into an Equation

The formula for the perimeter of a rectangle is $P = 2 \times (\text{length} + \text{width})$.

Substitute the known values and the variable into the formula:

$$ 40 = 2 \times (12 + w) $$

The equation is $40 = 2(12 + w)$.

Step 4: Solve the Equation

Solve the linear equation for $w$. First, simplify by dividing both sides by $2$:

$$ \frac{40}{2} = 12 + w $$ $$ 20 = 12 + w $$

Now, isolate $w$ by subtracting $12$ from both sides (or transpose $12$ to LHS):

$$ 20 - 12 = w $$ $$ 8 = w $$

So, the width is $8$.

Step 5: Check and State the Answer

Check: A rectangle with length $12 \text{ cm}$ and width $8 \text{ cm}$. Perimeter $= 2 \times (12 \text{ cm} + 8 \text{ cm}) = 2 \times (20 \text{ cm}) = 40 \text{ cm}$. This matches the given perimeter.

Answer: The width of the rectangle is $\textbf{8 cm}$.

Solving word problems effectively requires strong reading comprehension, the ability to translate carefully from verbal language to algebraic language, and proficiency in solving the resulting equations. Practice with various types of problems helps develop this skill.