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Class 7th Chapters
1. Integers	2. Fractions and Decimals	3. Data Handling
4. Simple Equations	5. Lines and Angles	6. The Triangle and its Properties
7. Congruence of Triangles	8. Comparing Quantities	9. Rational Numbers
10. Practical Geometry	11. Perimeter and Area	12. Algebraic Expressions
13. Exponents and Powers	14. Symmetry	15. Visualising Solid Shapes

Content On This Page
Linear Equations and Solving a Linear Equation in One Variable	Rule of Transposition	Solving Word Problems of Linear Equation

Chapter 4 Simple Equations (Concepts)

Welcome to this foundational exploration within the fascinating world of algebra. This section serves as a crucial introduction, meticulously designed to build your understanding of one of mathematics' most fundamental concepts: the equation. We will delve into what constitutes an equation and, more importantly, develop the systematic techniques required to solve them effectively. Think of this as learning the grammar of a new mathematical language, enabling you to express and resolve relationships between quantities.

At its heart, an equation is defined as a precise mathematical statement asserting the equality between two distinct expressions. These expressions typically involve a combination of variables (often represented by letters such as $x$, $y$, $z$, or $p$, signifying unknown values we aim to discover) and constants (which are fixed, known numerical values like $5$, $-12$, or $\frac{3}{4}$). These components are interconnected through various mathematical operations: addition, subtraction, multiplication, and division. The absolute core principle underpinning any equation is the balance signified by the equals sign ($=$). This symbol dictates that the value computed from the expression on the left-hand side (LHS) must be exactly identical to the value computed from the expression on the right-hand side (RHS). For instance, in the equation $2x + 5 = 15$, the term $2x + 5$ represents the LHS, while $15$ is the RHS.

It is vital to distinguish equations from mere mathematical expressions. An expression, like $7p - 10$, represents a value or a calculation waiting to be performed, but it doesn't assert equality with anything else. An equation, such as $7p - 10 = 4$, makes a definitive statement that must hold true for specific values of the variable $p$. Our primary objective when faced with an equation is to uncover the precise numerical value(s) of the variable(s) that validate this statement of equality. This specific value is known as the solution or, sometimes, the root of the equation. Finding the solution means determining which number makes the LHS truly equal to the RHS.

The cornerstone technique for solving equations revolves around the principle of balancing. Imagine an old-fashioned scale; to keep it balanced, whatever you do to one side, you must replicate exactly on the other. Similarly, for equations:

You can add the same quantity to both the LHS and RHS.
You can subtract the same quantity from both the LHS and RHS.
You can multiply both the LHS and RHS by the same non-zero quantity.
You can divide both the LHS and RHS by the same non-zero quantity.

Performing any of these operations consistently on both sides ensures the equality remains intact. A closely related and often faster method is transposition. This technique allows moving a term from one side of the equation to the other by reversing its operation (e.g., adding $a$ on the LHS becomes subtracting $a$ on the RHS, so $x + a = b$ can be seen as $x = b - a$). We will practice solving various forms of simple linear equations involving a single variable, progressing from basic forms like $x + a = b$ or $ax = b$ and $\frac{x}{a} = b$ to slightly more involved ones requiring multiple steps, such as $3p - 10 = 5$.

A significant practical application explored here is the translation of real-world scenarios, often presented as word problems, into the language of algebraic equations. This involves identifying the unknown element, assigning it a variable (like $y$), and carefully constructing an equation that mathematically models the relationships described in the problem. Once formulated, the equation can be solved using the balancing or transposition methods learned. Finally, we strongly emphasize the indispensable step of checking the solution. This involves substituting the value found for the variable back into the original equation to meticulously verify that it indeed makes the LHS equal to the RHS, confirming the correctness of our result. Mastering these concepts builds essential algebraic thinking, a prerequisite for tackling more advanced mathematical topics.

Linear Equations and Solving a Linear Equation in One Variable

In the previous chapter on Algebra, we were introduced to variables, constants, and algebraic expressions. We learned that an Equation is a mathematical statement that shows the equality between two expressions, indicated by an equals sign ($=$). Equations are powerful tools used to represent relationships between quantities and, most importantly, to find the value of unknown quantities represented by variables. In this chapter, we will focus on a specific type of equation called a Simple Equation or a Linear Equation in One Variable and learn methods to solve them.

What is an Equation?

An equation is like a balanced scale. It is a mathematical statement asserting that the expression on one side of the equality sign has the exact same value as the expression on the other side.

An equation always consists of:

An expression on the left side of the equals sign, called the Left Hand Side (LHS).
An equals sign ($=$).
An expression or value on the right side of the equals sign, called the Right Hand Side (RHS).

Example: In the equation $x + 5 = 12$,

LHS is the expression $x + 5$.
RHS is the value $12$.
The variable is $x$, and the constants are 5 and 12.

An equation involving a variable is true only for specific value(s) of that variable. This value that makes the LHS equal to the RHS is called the Solution or Root of the equation.

Linear Equation in One Variable

A Linear Equation in One Variable is an equation that has only one variable, and the highest power of that variable is 1. This means the variable is not squared ($x^2$), cubed ($x^3$), or raised to any higher power. These are often called 'simple equations' because they are the most basic type of algebraic equations.

A linear equation in one variable can always be written in the standard form:

$$ax + b = 0$$

where:

$x$ is the single variable.
$a$ and $b$ are constants (numerical values).
The constant $a$ (the coefficient of the variable) cannot be zero ($a \neq 0$). If $a$ were 0, the term $ax$ would be $0 \times x = 0$, and the equation would simplify to $b=0$, which is either a true statement (if $b$ is 0) or a false statement (if $b$ is not 0), but it wouldn't be an equation involving the variable $x$ anymore.

(Note: The standard form can also be given as $ax + b = c$, where $a, b, c$ are constants and $a \neq 0$. This is equivalent to $ax + (b-c) = 0$, where a new constant $(b-c)$ replaces the original $b$).

Examples of Linear Equations in One Variable:

$x + 7 = 10$: Here, $a=1, b=7, c=10$. (Equivalent to $x - 3 = 0$, where $a=1, b=-3$). Highest power of $x$ is 1. Only one variable ($x$).
$2y - 3 = 7$: Here, the variable is $y$. $a=2, b=-3, c=7$. Highest power of $y$ is 1. Only one variable ($y$).
$\frac{p}{4} + 1 = 5$: This is $\frac{1}{4}p + 1 = 5$. Here, the variable is $p$. $a=\frac{1}{4}, b=1, c=5$. Highest power of $p$ is 1. Only one variable ($p$).
$6z = 30$: This can be written as $6z + 0 = 30$. Here, the variable is $z$. $a=6, b=0, c=30$. Highest power of $z$ is 1. Only one variable ($z$).
$m + 10 = 0$: Here, the variable is $m$. $a=1, b=10, c=0$. Highest power of $m$ is 1. Only one variable ($m$).

Examples of Equations that are NOT Linear in One Variable:

$x^2 + 5 = 14$: This is an equation in one variable ($x$), but the highest power of $x$ is 2, so it is a quadratic equation, not a linear equation.
$xy + 3 = 10$: This is an equation, but it has two variables ($x$ and $y$), so it is not a linear equation in one variable.
$2x + 3 > 7$: This has one variable ($x$) with power 1, but it contains an inequality sign ($>$) instead of an equals sign ($=$), so it is an inequality, not an equation.
$5x - 9$: This involves a variable ($x$) and operations, but it does not have an equals sign ($=$). It is an algebraic expression, not an equation.

Solution of an Equation

The Solution (or Root) of an equation involving a variable is the specific value of the variable that makes the equation true. That is, it is the value that, when substituted into the equation, makes the LHS equal to the RHS.

Example: Consider the equation $x + 5 = 12$. We want to find the value of $x$ that makes the left side equal to the right side.

Let's try a value for $x$, say $x = 3$.
Substitute $x=3$ into the equation: $3 + 5 = 12 \Rightarrow 8 = 12$.

Since $8 \neq 12$, the equation is false when $x=3$. So, $x=3$ is not the solution.
Let's try another value for $x$, say $x = 7$.
Substitute $x=7$ into the equation: $7 + 5 = 12 \Rightarrow 12 = 12$.

Since $12 = 12$, the equation is true when $x=7$. So, $x=7$ is the solution to the equation $x + 5 = 12$.

The process of finding the solution(s) of an equation is called Solving the Equation.

Methods for Solving a Linear Equation in One Variable

The main goal in solving a linear equation is to isolate the variable on one side of the equation (either LHS or RHS) so that the other side contains only a constant, which will be the value of the variable. We achieve this isolation by performing inverse operations.

Remember the idea of an equation as a balanced scale. To maintain the balance, whatever operation we perform on one side of the equation, we must perform the exact same operation on the other side.

1. Trial and Error Method (or Inspection)

This method involves substituting different values for the variable into the equation and checking if the LHS becomes equal to the RHS. We keep trying values until we find the one that satisfies the equation. This method is only practical for very simple equations where the solution is likely to be a small whole number or a simple fraction.

Example: Solve $m - 2 = 3$ using trial and error.

Try $m=1$: LHS = $1-2 = -1$. RHS = 3. $-1 \neq 3$.
Try $m=3$: LHS = $3-2 = 1$. RHS = 3. $1 \neq 3$.
Try $m=5$: LHS = $5-2 = 3$. RHS = 3. $3 = 3$. The equation is true.

So, by trial and error, the solution is $m=5$.

2. Systematic Method (Balancing Method)

This is a more systematic and reliable method for solving linear equations, especially as equations become more complex. The principle is to perform inverse operations on both sides of the equation to gradually isolate the variable. The inverse operations are:

Addition is the inverse of Subtraction.
Subtraction is the inverse of Addition.
Multiplication is the inverse of Division.
Division is the inverse of Multiplication.

Rules of Balancing:

Adding the same number to both sides of an equation does not change the equality.
Subtracting the same number from both sides of an equation does not change the equality.
Multiplying both sides of an equation by the same non-zero number does not change the equality.
Dividing both sides of an equation by the same non-zero number does not change the equality.

We apply these rules to 'undo' the operations performed on the variable, working backwards from the variable to isolate it.

Example 1: Solve $x + 7 = 10$.

The variable $x$ has 7 added to it. To isolate $x$, we need to undo the addition of 7 by performing the inverse operation, which is subtraction. We subtract 7 from both sides of the equation to maintain balance.

$x + 7 - 7 = 10 - 7$

(Subtracting 7 from both sides)

Simplify both sides:

$x + 0 = 3$

$x = 3$

The solution is $x=3$.

Check: Substitute $x=3$ into the original equation $x + 7 = 10$: LHS $= 3 + 7 = 10$. RHS $= 10$. Since LHS = RHS, the solution is correct.

Example 2: Solve $y - 3 = 4$.

The variable $y$ has 3 subtracted from it. To isolate $y$, we undo the subtraction of 3 by performing the inverse operation, which is addition. We add 3 to both sides.

$y - 3 + 3 = 4 + 3$

(Adding 3 to both sides)

Simplify both sides:

$y + 0 = 7$

$y = 7$

The solution is $y=7$.

Check: Substitute $y=7$ into the original equation $y - 3 = 4$: LHS $= 7 - 3 = 4$. RHS $= 4$. Since LHS = RHS, the solution is correct.

Example 3: Solve $5p = 20$.

The variable $p$ is multiplied by 5. To isolate $p$, we undo the multiplication by 5 by performing the inverse operation, which is division. We divide both sides by 5.

$\frac{5p}{5} = \frac{20}{5}$

(Dividing both sides by 5)

Simplify both sides:

$1 \cdot p = 4$

$p = 4$

The solution is $p=4$.

Check: Substitute $p=4$ into the original equation $5p = 20$: LHS $= 5 \times 4 = 20$. RHS $= 20$. Since LHS = RHS, the solution is correct.

Example 4: Solve $\frac{m}{3} = 6$.

The variable $m$ is divided by 3. To isolate $m$, we undo the division by 3 by performing the inverse operation, which is multiplication. We multiply both sides by 3.

$\frac{m}{3} \times 3 = 6 \times 3$

(Multiplying both sides by 3)

Simplify both sides:

$m \times \frac{3}{3} = 18$

$m \times 1 = 18$

$m = 18$

The solution is $m=18$.

Check: Substitute $m=18$ into the original equation $\frac{m}{3} = 6$: LHS $= \frac{18}{3} = 6$. RHS $= 6$. Since LHS = RHS, the solution is correct.

Example 5 (Solving a Two-step equation): Solve $2x + 5 = 15$.

In this equation, the variable $x$ is first multiplied by 2, and then 5 is added to the result. To isolate $x$, we need to undo these operations in reverse order.

Step 1: Undo the addition of 5. The inverse of adding 5 is subtracting 5. Subtract 5 from both sides of the equation.

$2x + 5 - 5 = 15 - 5$

(Subtracting 5 from both sides)

Simplify both sides:

$2x = 10$

Step 2: Undo the multiplication by 2. The inverse of multiplying by 2 is dividing by 2. Divide both sides by 2.

$\frac{2x}{2} = \frac{10}{2}$

(Dividing both sides by 2)

Simplify both sides:

$x = 5$

The solution is $x=5$.

Check: Substitute $x=5$ into the original equation $2x + 5 = 15$: LHS $= 2(5) + 5 = 10 + 5 = 15$. RHS $= 15$. Since LHS = RHS, the solution is correct.

Forming Equations from Statements

One of the most practical uses of equations is solving word problems. The first step in solving a word problem using algebra is to translate the problem from words into a mathematical equation. This involves identifying the unknown quantity and expressing the given conditions as a relationship involving that unknown.

Steps to Form an Equation from a Verbal Statement:

Read the statement carefully to understand what is being asked and what information is given.
Identify the unknown quantity in the problem.
Assign a variable (like $x, y, m, n$) to represent this unknown quantity.
Translate the conditions given in the statement into an algebraic expression involving the variable and constants (using the rules for forming expressions learned in Chapter 11).
Look for the word or phrase in the statement that indicates equality (such as "is", "equals", "gives", "the result is", "is equal to"). Place the equals sign ($=$) where this phrase appears.
Write the expression or value on the other side of the equals sign, based on the statement.

Example 1. Write equations for the following statements:

(a) The sum of a number $x$ and 4 is 9.

(b) 2 subtracted from $y$ is 8.

(d) The number $b$ divided by 5 gives 6.

Answer:

(a) "The sum of a number $x$ and 4" translates to the expression $x + 4$. "is 9" translates to $= 9$.

Equation: $x + 4 = 9$.

(b) "2 subtracted from $y$" translates to the expression $y - 2$. "is 8" translates to $= 8$.

Equation: $y - 2 = 8$.

Equation: $10a = 70$.

(d) "The number $b$ divided by 5" translates to the expression $\frac{b}{5}$. "gives 6" translates to $= 6$.

Equation: $\frac{b}{5} = 6$.

Example 2. Solve the equation $3n - 2 = 46$ and check your result.

Answer:

Given equation: $3n - 2 = 46$.

We use the systematic method to isolate the variable $n$. In the expression $3n - 2$, $n$ is first multiplied by 3, and then 2 is subtracted. We undo these operations in reverse order.

Step 1: Undo the subtraction of 2 by adding 2 to both sides of the equation.

$3n - 2 + 2 = 46 + 2$

(Adding 2 to both sides)

Simplify both sides:

$3n = 48$

Step 2: Undo the multiplication by 3 by dividing both sides by 3.

$\frac{3n}{3} = \frac{48}{3}$

(Dividing both sides by 3)

Simplify both sides:

$n = 16$

The solution of the equation is $n = 16$.

Check the result:

Substitute the obtained solution $n = 16$ back into the original equation $3n - 2 = 46$.

Evaluate the Left Hand Side (LHS):

LHS $= 3n - 2 $

Substitute $n=16$:

LHS $= 3(16) - 2 $

Perform multiplication:

LHS $= 48 - 2 $

Perform subtraction:

LHS $= 46 $

Evaluate the Right Hand Side (RHS) of the original equation:

RHS $= 46 $

Compare LHS and RHS:

LHS $= 46 $

RHS $= 46 $

Since LHS = RHS, the solution $n = 16$ is correct.

Rule of Transposition

In the previous section, we learned about the systematic method for solving linear equations by performing the same inverse operations on both sides of the equation to maintain balance. While this method clearly shows the logic, writing out each step (e.g., "Subtract 5 from both sides") can become repetitive. The Rule of Transposition is a convenient shortcut derived directly from the systematic method. It simplifies the process of rearranging terms in an equation.

Understanding Transposition

Transposition literally means "changing the position" or "moving across". In the context of equations, it refers to moving a term from one side of the equals sign to the other. When a term is transposed, its associated operation changes to its inverse operation as it moves to the other side.

Here's how the operation changes during transposition:

If a term is being added (+) on one side (e.g., $x + 5$), when you move it to the other side, it becomes subtracted ($-$) (e.g., $x = 12 - 5$).
If a term is being subtracted ($-$) on one side (e.g., $x - 5$), when you move it to the other side, it becomes added (+) (e.g., $x = 12 + 5$).
If a term is multiplying ($\times$) the variable (or another term) on one side (e.g., $5x$), when you move the *factor* (the number 5) to the other side, it becomes dividing ($\div$ or fraction denominator) (e.g., $x = \frac{10}{5}$).
If a term is dividing ($\div$ or fraction denominator) the variable (or another term) on one side (e.g., $\frac{x}{5}$), when you move the *divisor* (the number 5) to the other side, it becomes multiplying ($\times$) (e.g., $x = 2 \times 5$).

Derivation of the Rule of Transposition from the Systematic Method:

The rule of transposition is simply a shorthand for applying the same operation to both sides of the equation:

1. For Addition/Subtraction:

Consider the equation $x + a = b$. Using the systematic method, we subtract $a$ from both sides: $x + a - a = b - a$, which simplifies to $x = b - a$. Notice that the term $+a$ from the LHS effectively moved to the RHS as $-a$.

Transposition: $x + a = b \implies x = b - a$.

Consider the equation $x - a = b$. Using the systematic method, we add $a$ to both sides: $x - a + a = b + a$, which simplifies to $x = b + a$. Notice that the term $-a$ from the LHS effectively moved to the RHS as $+a$.

Transposition: $x - a = b \implies x = b + a$.

2. For Multiplication/Division:

Consider the equation $a \times x = b$ (or $ax = b$), where $a \neq 0$. Using the systematic method, we divide both sides by $a$: $\frac{ax}{a} = \frac{b}{a}$, which simplifies to $x = \frac{b}{a}$. Notice that the factor $a$ multiplying $x$ on the LHS effectively moved to the RHS as a divisor in the denominator.

Transposition: $ax = b \implies x = \frac{b}{a}$.

Consider the equation $\frac{x}{a} = b$, where $a \neq 0$. Using the systematic method, we multiply both sides by $a$: $\frac{x}{a} \times a = b \times a$, which simplifies to $x = ba$. Notice that the divisor $a$ on the LHS effectively moved to the RHS as a factor multiplying $b$.

Transposition: $\frac{x}{a} = b \implies x = b \times a$ (or $ab$).

The rule of transposition is a powerful shortcut because it allows us to quickly rearrange equations without explicitly writing the operation being performed on both sides at each step.

Solving Equations using Transposition

The process of solving using transposition is to move terms step-by-step away from the variable until the variable is isolated on one side of the equation and a constant is on the other side.

We usually follow a sequence where we first transpose terms that are added or subtracted to the variable term, and then transpose factors that are multiplying or dividing the variable.

Example 1: Solve $y + 6 = 10$ using transposition.

The variable $y$ is on the LHS, and $+6$ is added to it. To isolate $y$, we transpose $+6$ from the LHS to the RHS. When $+6$ moves to the RHS, its operation changes from addition to subtraction, so it becomes $-6$ on the RHS.

$y = 10 - 6$

Simplify the RHS:

$y = 4$

The solution is $y=4$.

Example 2: Solve $p - 4 = 5$ using transposition.

The variable $p$ is on the LHS, and $-4$ is subtracted from it. To isolate $p$, transpose $-4$ from the LHS to the RHS. When $-4$ moves to the RHS, its operation changes from subtraction to addition, so it becomes $+4$ on the RHS.

$p = 5 + 4$

Simplify the RHS:

$p = 9$

The solution is $p=9$.

Example 3: Solve $4m = 12$ using transposition.

The variable $m$ is on the LHS and is being multiplied by the factor 4. To isolate $m$, transpose the factor 4 from the LHS to the RHS. When 4 moves to the RHS, its operation changes from multiplication to division, so it becomes a divisor on the RHS.

$m = \frac{12}{4}$

Simplify the RHS:

$m = 3$

The solution is $m=3$.

Example 4: Solve $\frac{t}{5} = 10$ using transposition.

The variable $t$ is on the LHS and is being divided by the divisor 5. To isolate $t$, transpose the divisor 5 from the LHS to the RHS. When 5 moves to the RHS, its operation changes from division to multiplication, so it becomes a factor multiplying 10 on the RHS.

$t = 10 \times 5$

Simplify the RHS:

$t = 50$

The solution is $t=50$.

Example 5 (Solving a Multi-step equation): Solve $3x - 5 = 13$ using transposition.

In the expression $3x - 5$, the variable $x$ is first multiplied by 3, then 5 is subtracted. To isolate $x$, we undo the operations in reverse order of BODMAS applied to the variable term.

Step 1: Isolate the term containing the variable ($3x$). The term $-5$ is being subtracted from $3x$. Transpose $-5$ from the LHS to the RHS. It becomes $+5$ on the RHS.

$3x = 13 + 5$

Simplify the RHS:

$3x = 18$

Step 2: Isolate $x$. The variable $x$ is being multiplied by the factor 3 in the term $3x$. Transpose the factor 3 from the LHS to the RHS. It will divide on the RHS.

$x = \frac{18}{3}$

Simplify the RHS:

$x = 6$

The solution is $x=6$.

Check: Substitute $x=6$ into the original equation $3x - 5 = 13$: LHS $= 3(6) - 5 = 18 - 5 = 13$. RHS $= 13$. Since LHS = RHS, the solution is correct.

Example 6 (Solving an equation with variables on both sides): Solve $5x - 3 = 3x + 7$ using transposition.

The goal is to get all terms with the variable on one side (usually the LHS) and all constant terms on the other side (usually the RHS).

Step 1: Move the variable term $3x$ from the RHS to the LHS. $3x$ is being added on the RHS (even though there's no plus sign written before it, it's a positive term). Transposing it to the LHS makes it $-3x$.

$5x - 3x - 3 = 7$

Combine the variable terms on the LHS: $5x - 3x = 2x$.

$2x - 3 = 7$

Step 2: Move the constant term $-3$ from the LHS to the RHS. $-3$ is being subtracted on the LHS. Transposing it to the RHS makes it $+3$.

$2x = 7 + 3$

Simplify the RHS:

$2x = 10$

Step 3: Isolate the variable $x$. The factor 2 is multiplying $x$. Transpose 2 from the LHS to the RHS. It will divide on the RHS.

$x = \frac{10}{2}$

Simplify the RHS:

$x = 5$

The solution is $x=5$.

Check: Substitute $x=5$ into both sides of the original equation $5x - 3 = 3x + 7$.

LHS $= 5(5) - 3 = 25 - 3 = 22$.

RHS $= 3(5) + 7 = 15 + 7 = 22$.

Since LHS = RHS, the solution is correct.

Example 1. Solve the following equations using the rule of transposition:

(a) $x - 7 = 15$

(b) $3p + 12 = 0$

Answer:

(a) Solve $x - 7 = 15$.

Transpose $-7$ from the LHS to the RHS. It changes from subtraction to addition (+7).

$x = 15 + 7$

Simplify the RHS:

$x = 22$

(b) Solve $3p + 12 = 0$.

Step 1: Transpose the constant term $+12$ from the LHS to the RHS. It changes from addition to subtraction (-12).

$3p = 0 - 12$

Simplify the RHS:

$3p = -12$

Step 2: Transpose the factor 3 (multiplying $p$) from the LHS to the RHS. It changes from multiplication to division (dividing by 3).

$p = \frac{-12}{3}$

Simplify the RHS (divide negative by positive, result is negative):

$p = -4$

This equation can be seen as a term $2y$ being divided by 3, or as $y$ being multiplied by the fraction $\frac{2}{3}$. We can solve this in one step or two steps.

Method 1 (Two steps):

Rewrite the LHS as $(2y) \div 3 = 18$.

Step 1: Transpose the divisor 3 from the LHS to the RHS. It changes from division to multiplication (multiplying by 3).

$2y = 18 \times 3$

Simplify the RHS:

$2y = 54$

Step 2: Transpose the factor 2 (multiplying $y$) from the LHS to the RHS. It changes from multiplication to division (dividing by 2).

$y = \frac{54}{2}$

Simplify the RHS:

$y = 27$

Method 2 (One step):

Rewrite the LHS as $\frac{2}{3}y = 18$. Here, the variable $y$ is being multiplied by the fraction $\frac{2}{3}$. To isolate $y$, transpose the multiplying fraction $\frac{2}{3}$ from the LHS to the RHS. It changes from multiplication to division, which means multiplying by its reciprocal.

The reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

$y = 18 \div \frac{2}{3} $

Using the rule for dividing by a fraction (multiply by the reciprocal):

$y = 18 \times \frac{3}{2} $

Multiply (using cancellation):

$y = \frac{\cancel{18}^9 \times 3}{\cancel{2}_1} $

$y = 9 \times 3 = 27 $

Both methods give the same solution, $y=27$.

Solving Word Problems of Linear Equation

Equations are not just theoretical mathematical statements; they are incredibly useful for solving problems we encounter in real life. Many everyday situations and puzzles can be translated into the language of algebra in the form of a simple linear equation. Once we have the equation, we can use the methods we've learned (like the systematic method or transposition) to find the unknown quantity asked for in the problem.

Solving a word problem involves two main parts: first, formulating the algebraic equation that represents the problem, and second, solving that equation to find the value of the variable.

Steps to Solve Word Problems

Follow these systematic steps to successfully solve word problems using linear equations:

Read and Understand the Problem: Read the word problem carefully, perhaps multiple times. Understand the context, identify what information is given (the knowns), and clearly determine what quantity you are being asked to find (the unknown).
Define the Variable: Choose a letter (a variable, e.g., $x, y, n, m$) to represent the unknown quantity that you need to find. Write down clearly in words what this variable stands for (e.g., "Let the unknown number be $x$", "Let the breadth of the pool be $b$ metres").
Translate the Problem into an Equation: This is often the trickiest step. Translate the relationships and conditions described in the word problem into a mathematical equation involving the variable you defined. Look for keywords and phrases that indicate mathematical operations and equality:
- Words like "sum", "total", "altogether", "more than", "increased by", "added to" usually imply Addition ($+$).
- Words like "difference", "less than", "subtracted from", "decreased by", "reduced by", "remains" usually imply Subtraction ($-$). Be careful with the order in "less than" or "subtracted from".
- Words like "product", "times", "multiplied by", "of" (in contexts like "half of", "twice the number") usually imply Multiplication ($\times$).
- Words like "quotient", "divided by", "per", "ratio" usually imply Division ($\div$).
- Words or phrases like "is", "are", "was", "were", "will be", "gives", "equals", "the result is" usually imply Equality ($=$). This is where you place the equals sign to form your equation.
Write the expression on the LHS and the expression or value on the RHS as described in the problem statement.
Solve the Equation: Use the algebraic methods you have learned (the systematic method by performing inverse operations on both sides, or the transposition method) to solve the equation you formulated in Step 3 for the value of the variable.
Check the Solution: This is an important verification step. Substitute the value you found for the variable back into the original word problem (not just the equation you wrote, though checking in the equation is also good). Read the problem statement again and see if your solution makes sense and satisfies all the conditions given in the problem.
Write the Answer: State the final answer clearly in a complete sentence, ensuring you include the correct units if the problem involves measurements (like metres, kilograms, rupees, years).

Examples of Solving Word Problems

Example 1. If you subtract $\frac{1}{2}$ from a number and multiply the result by $\frac{1}{2}$, you get $\frac{1}{8}$. What is the number?

Answer:

Step 1 & 2: Understand and Define Variable

We are looking for an unknown number. Let's represent this unknown number by the variable $x$.

Let the number be $x$.

Step 3: Translate to an Equation

The statement says, "If you subtract $\frac{1}{2}$ from a number...". This translates to the expression $x - \frac{1}{2}$.

Next, "...and multiply the result by $\frac{1}{2}$...". This means we multiply the entire expression $(x - \frac{1}{2})$ by $\frac{1}{2}$. It's important to use parentheses here because we are multiplying the *entire result* of the subtraction.

This translates to $\frac{1}{2} \times (x - \frac{1}{2})$, or simply $\frac{1}{2}(x - \frac{1}{2})$.

Finally, "...you get $\frac{1}{8}$". This indicates equality. The expression we formed is equal to $\frac{1}{8}$.

So, the equation is: $\frac{1}{2} (x - \frac{1}{2}) = \frac{1}{8}$.

Step 4: Solve the Equation

Equation: $\frac{1}{2} (x - \frac{1}{2}) = \frac{1}{8}$.

We want to isolate $x$. The term $(x - \frac{1}{2})$ is being multiplied by $\frac{1}{2}$. To undo this multiplication, we can multiply both sides by the reciprocal of $\frac{1}{2}$, which is 2.

$2 \times \frac{1}{2} (x - \frac{1}{2}) = 2 \times \frac{1}{8} $

(Multiplying both sides by 2)

Simplify both sides:

$1 \times (x - \frac{1}{2}) = \frac{2}{8} $

$x - \frac{1}{2} = \frac{1}{4} $

(Simplifying $\frac{2}{8}$)

Now, we have $x$ with $\frac{1}{2}$ subtracted from it. To isolate $x$, undo the subtraction by adding $\frac{1}{2}$ to both sides (or transpose $-\frac{1}{2}$ to the RHS).

$x - \frac{1}{2} + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} $

(Adding $\frac{1}{2}$ to both sides)

Simplify the LHS and RHS. To add fractions on the RHS, find a common denominator for 4 and 2. LCM(4, 2) = 4.

$x + 0 = \frac{1}{4} + \frac{1 \times 2}{2 \times 2} $

$x = \frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} = \frac{3}{4} $

So, the solution is $x = \frac{3}{4}$.

Step 5: Check the Solution

The number we found is $\frac{3}{4}$. Let's check if it satisfies the original problem statement.

Subtract $\frac{1}{2}$ from the number: $\frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}$. (Result after subtraction)

Multiply this result by $\frac{1}{2}$: $\frac{1}{2} \times \frac{1}{4} = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}$.

The problem states that the final result should be $\frac{1}{8}$. Our calculation matches this value. So, the solution $x=\frac{3}{4}$ is correct.

Step 6: Write the Answer

The number is $\frac{3}{4}$.

Example 2. The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?

Answer:

Step 1 & 2: Understand and Define Variable

We need to find the length and breadth of the rectangular swimming pool. The problem describes the length in terms of the breadth. It's usually best to choose the quantity that the other quantity is described in relation to as your variable. Here, length is described in terms of breadth.

Let the breadth of the swimming pool be $b$ metres.

The length is given as "2 m more than twice its breadth".

Twice the breadth means $2 \times b$, or $2b$.

2 m more than twice the breadth means $2b + 2$.

So, the length $l = (2b + 2)$ metres.

The perimeter of the pool is given as 154 m.

Step 3: Translate to an Equation

We know the formula for the perimeter of a rectangle: Perimeter $P = 2 \times (\text{length} + \text{breadth})$.

Substitute the given perimeter (154 m) and the expressions for length ($2b + 2$) and breadth ($b$) into the formula:

$154 = 2 \times ((2b + 2) + b)$

Step 4: Solve the Equation

Equation: $154 = 2(2b + 2 + b)$.

Simplify the expression inside the parenthesis by combining like terms ($2b$ and $b$): $2b + b = 3b$.

$154 = 2(3b + 2) $

Now, we can solve this equation. We can either distribute the 2 on the RHS ($2 \times 3b = 6b$, $2 \times 2 = 4$, so $154 = 6b + 4$) or divide both sides by 2 (since 154 is divisible by 2). Dividing by 2 is often simpler.

$\frac{154}{2} = \frac{2(3b + 2)}{2} $

(Dividing both sides by 2)

Simplify both sides:

$77 = 3b + 2 $

Now, transpose the constant term $+2$ from the RHS to the LHS. It becomes $-2$ on the LHS.

$77 - 2 = 3b $

Simplify the LHS:

$75 = 3b $

Finally, transpose the factor 3 (multiplying $b$) from the RHS to the LHS. It will divide on the LHS.

$\frac{75}{3} = b $

Simplify the LHS:

$25 = b $

So, the breadth of the pool, $b = 25$ metres.

Now, calculate the length using the expression $l = 2b + 2$:

Length $l = 2(25) + 2 = 50 + 2 = 52 $ metres.

Step 5: Check the Solution

We found Breadth = 25 m and Length = 52 m.

Check Condition 1: Is the length 2 m more than twice the breadth?

Twice the breadth $= 2 \times 25 \text{ m} = 50 \text{ m}$.

2 m more than twice the breadth $= 50 \text{ m} + 2 \text{ m} = 52 \text{ m}$. This matches our calculated length. Condition 1 is satisfied.

Check Condition 2: Is the perimeter 154 m?

Perimeter $= 2(\text{length} + \text{breadth}) = 2(52 \text{ m} + 25 \text{ m}) = 2(77 \text{ m}) = 154 \text{ m}$. This matches the given perimeter. Condition 2 is satisfied.

The solution is correct.

Step 6: Write the Answer

The breadth of the swimming pool is 25 metres and the length is 52 metres.

Example 3. The sum of three consecutive integers is 51. What are these integers?

Answer:

Step 1 & 2: Understand and Define Variable

Consecutive integers are integers that follow each other in sequence, with a difference of 1 between them (e.g., 7, 8, 9 or -3, -2, -1). We need to find three such integers whose sum is 51.

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

Since the integers are consecutive, the next integer after $x$ is $x+1$.

The third consecutive integer after $x+1$ is $(x+1)+1 = x+2$.

Let the three consecutive integers be $x$, $x+1$, and $x+2$.

Step 3: Translate to an Equation

The problem states that the sum of these three integers is 51.

The sum of the three integers is $x + (x+1) + (x+2)$.

The statement "The sum ... is 51" translates to setting this sum equal to 51.

Equation: $x + (x+1) + (x+2) = 51$.

Step 4: Solve the Equation

Equation: $x + x + 1 + x + 2 = 51$.

Combine like terms on the LHS. There are three terms with $x$ ($x+x+x = 3x$) and two constant terms ($1+2 = 3$).

$3x + 3 = 51 $

This is a two-step linear equation. First, isolate the term with the variable ($3x$) by transposing the constant term $+3$ from the LHS to the RHS. It becomes $-3$ on the RHS.

$3x = 51 - 3 $

Simplify the RHS:

$3x = 48 $

Now, isolate $x$ by transposing the factor 3 (multiplying $x$) from the LHS to the RHS. It will divide on the RHS.

$x = \frac{48}{3} $

Simplify the RHS:

$x = 16 $

So, the first integer is $x = 16$.

The second consecutive integer is $x+1 = 16 + 1 = 17$.

The third consecutive integer is $x+2 = 16 + 2 = 18$.

The three consecutive integers are 16, 17, and 18.

Step 5: Check the Solution

Check if the numbers are consecutive integers: 16, 17, 18 are indeed consecutive integers.

Check if their sum is 51:

Sum $= 16 + 17 + 18 = 33 + 18 = 51$.

The sum matches the condition given in the problem. So, the solution is correct.

Step 6: Write the Answer

The three consecutive integers are 16, 17, and 18.