| Classwise Concept with Examples | ||||||
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| 6th | 7th | 8th | 9th | 10th | 11th | 12th |
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| Linear Equation and Solution or Root of Equation | Solving Linear Equations in One Variable | Application of Linear Equations |
Chapter 2 Linear Equations in One Variable (Concepts)
Welcome back to the dynamic world of algebra! Building upon the foundational understanding of simple equations acquired in Class 7, this chapter significantly elevates our skills by tackling more intricate Linear Equations in One Variable. While we continue to focus on equations involving a single unknown quantity (the variable), the complexity increases as we encounter scenarios where this variable might appear on both sides of the equals sign. Our goal is to develop robust, systematic methods to solve these equations accurately and apply them confidently to a variety of real-world situations.
Let's first solidify our understanding of the core subject: a linear equation in one variable is fundamentally an equation that can be rearranged into the standard form $ax + b = 0$. Here, $x$ represents the variable we aim to find, while $a$ and $b$ are constant numbers (coefficients or fixed terms), with the crucial condition that $a \neq 0$. The term 'linear' specifically signifies that the highest power of the variable $x$ appearing in the equation is exactly 1 (i.e., just $x$, not $x^2$, $x^3$, etc.). Our journey begins by revisiting and reinforcing the essential techniques learned previously for solving simpler equations. These cornerstone methods include:
- The Balancing Method: Performing the exact same mathematical operation (addition, subtraction, multiplication, or division by a non-zero number) on both the left-hand side (LHS) and the right-hand side (RHS) of the equation to maintain the equality while progressively isolating the variable.
- Transposition: A related and often quicker technique where a term is moved from one side of the equation to the other by changing its sign (e.g., adding $k$ on one side becomes subtracting $k$ on the other).
The principal new challenge introduced here is solving equations where the variable terms appear on both sides of the equality, such as in the equation $3x + 5 = x - 7$. The strategy involves systematically manipulating the equation to collect all terms containing the variable on one side (typically the LHS) and all constant terms on the other side (usually the RHS). For $3x + 5 = x - 7$, we might subtract $x$ from both sides ($2x + 5 = -7$) and then subtract $5$ from both sides ($2x = -12$), before finally dividing by $2$ to find $x = -6$. We also tackle equations containing brackets, which necessitates applying the distributive property (e.g., $a(b+c) = ab + ac$) to remove the brackets as a first step before proceeding with collection and solving. Furthermore, we encounter equations incorporating rational numbers, either as coefficients or constants. Solving these often involves a preliminary step of 'clearing the fractions' by multiplying the entire equation by the Least Common Multiple (LCM) of all the denominators present, transforming it into an equivalent equation with only integer coefficients and constants, which is generally easier to handle.
Perhaps the most significant aspect of this chapter is the heavy emphasis placed on applying these equation-solving skills to real-world scenarios presented as word problems. This translation process is a critical mathematical skill, requiring careful reading and interpretation to:
- Identify the unknown quantity or quantities.
- Assign a variable (like $y$ or $t$) to the primary unknown.
- Express the relationships and conditions described in the problem using an algebraic equation.
Linear Equation and Solution or Root of Equation
In your previous classes, you have been introduced to algebraic expressions. These expressions are built using variables and constants, combined through mathematical operations like addition, subtraction, multiplication, and division.
Algebraic Expressions and Equations
Recall some examples of algebraic expressions:
- $x+3$
- $y-5$
- $2z$
- $3m+7$
- $\frac{a}{4} - 2$
In these expressions, $x, y, z, m, a$ are variables (symbols that can take various numerical values), and 3, -5, 2, 7, 4, -2 are constants (symbols with fixed numerical values).
Now, consider what happens when we put an 'equals to' sign (=) between two algebraic expressions or between an expression and a constant. This creates an equation.
An equation is a statement that says the value of one expression is equal to the value of another expression.
Examples of equations:
- $x + 3 = 10$
- $y - 5 = 2y$
- $2p + 1 = 5$
- $3k - 7 = k + 3$
In any equation, the expression on the left side of the equality sign is called the Left Hand Side (LHS), and the expression on the right side is called the Right Hand Side (RHS).
For example, in the equation $x+3 = 10$:
- LHS is $x+3$
- RHS is $10$
Equations are powerful tools because they allow us to represent relationships between quantities and solve for unknown values.
Linear Equation in One Variable
Among various types of equations, we will focus on a specific type called a linear equation in one variable.
A linear equation in one variable is an equation that has the following characteristics:
- It involves only one variable (like only $x$, or only $y$, or only $z$, etc., throughout the equation).
- The highest power of the variable in the equation is 1.
A general form of a linear equation in one variable can be written as $ax + b = 0$, where $a$ and $b$ are constants and $a \neq 0$. However, often they appear in slightly different forms like $ax + b = c$, $ax = b$, $ax + b = cx + d$, etc., where $a, b, c, d$ are constants and the coefficient of the variable ($a$ or $a-c$ if variables are on both sides) is not zero.
Let's look at some examples:
- $x + 5 = 12$: This equation has only one variable ($x$), and the highest power of $x$ is 1 ($x^1$). This is a linear equation in one variable.
- $2y - 3 = 7$: This equation has only one variable ($y$), and the highest power of $y$ is 1 ($y^1$). This is a linear equation in one variable.
- $4z = 20$: This equation has only one variable ($z$), and the highest power of $z$ is 1 ($z^1$). This is a linear equation in one variable.
- $\frac{m}{2} + 1 = 5$: This equation has only one variable ($m$), and the highest power of $m$ is 1 ($m^1$, as $\frac{m}{2} = \frac{1}{2}m$). This is a linear equation in one variable.
- $2x + 3 = x - 5$: This equation has only one variable ($x$). If we rewrite it as $2x - x = -5 - 3$, we get $x = -8$. The highest power of $x$ is 1. This is a linear equation in one variable.
Let's look at some examples that are NOT linear equations in one variable and understand why:
- $x^2 + 2x = 5$: This equation has only one variable ($x$), but the highest power of $x$ is 2 ($x^2$). So, it is not a linear equation. It is a quadratic equation.
- $xy + y = 10$: This equation involves two variables ($x$ and $y$). So, it is not an equation in one variable.
- $\frac{1}{x} = 4$: This equation involves only one variable ($x$). If we rewrite it by multiplying both sides by $x$ (assuming $x \neq 0$), we get $1 = 4x$, or $4x - 1 = 0$. This rearranged form IS a linear equation in one variable. However, the original expression $\frac{1}{x}$ is usually considered to involve $x$ with power -1 ($x^{-1}$), making it not linear in terms of polynomial degree. At Class 8 level, we usually deal with variables having positive integer powers in the expressions that form linear equations. So, $\frac{1}{x}=4$ as presented might not fit the typical scope of a linear equation in one variable for this level, but its equivalent form $4x=1$ certainly does. It's important to clarify the context or focus on the standard forms. Let's stick to the rule: the highest power of the variable must be 1, and the variable should not be in the denominator in a form like $\frac{1}{x}$.
- $x^3 - 4 = 0$: Highest power of $x$ is 3. Not linear.
Solution or Root of an Equation
A linear equation in one variable has only one possible value for the variable that makes the equation true. This value is called the solution or the root of the equation.
The solution is the specific number that, when substituted in place of the variable, makes the LHS of the equation exactly equal to the RHS of the equation. Finding the solution is the main goal when we are asked to "solve" a linear equation.
To check if a particular value of the variable is a solution, substitute that value into the equation and evaluate both the LHS and the RHS. If LHS = RHS, the value is a solution.
Example 1. Is $x=4$ a solution to the equation $3x - 2 = 10$?
Answer:
Given equation: $3x - 2 = 10$.
We need to check if $x=4$ satisfies the equation.
Substitute $x=4$ into the LHS of the equation:
LHS $= 3x - 2$
$= 3(4) - 2$
$= 12 - 2$
$= 10$
Now, compare the calculated LHS value with the RHS of the given equation.
RHS $= 10$
Since LHS = RHS ($10 = 10$), the value $x=4$ makes the equation true.
Therefore, $x=4$ is a solution to the equation $3x - 2 = 10$.
Example 2. Check if $y=5$ is a solution to the equation $2y + 1 = 8$.
Answer:
Given equation: $2y + 1 = 8$.
We need to check if $y=5$ satisfies the equation.
Substitute $y=5$ into the LHS of the equation:
LHS $= 2y + 1$
$= 2(5) + 1$
$= 10 + 1$
$= 11$
Now, compare the calculated LHS value with the RHS of the given equation.
RHS $= 8$
Since LHS $\neq$ RHS ($11 \neq 8$), the value $y=5$ does not make the equation true.
Therefore, $y=5$ is not a solution to the equation $2y + 1 = 8$.
Solving Linear Equations in One Variable
Solving a linear equation means finding the specific value of the variable that makes the equation true. This value, as we learned earlier, is called the solution or the root of the equation. To find this value, we perform a series of inverse mathematical operations on the equation until the variable is isolated on one side of the equality sign.
The fundamental principle behind solving equations is maintaining balance. Whatever mathematical operation you do on the Left Hand Side (LHS) of the equation, you must also do the exact same operation on the Right Hand Side (RHS) to keep the equality valid.
Basic Principles for Solving Equations (Balancing Method)
Think of an equation as a balanced scale. If you add weight to one side, you must add the same weight to the other side to keep it balanced. Similarly, if you remove weight from one side, you must remove the same weight from the other side. In algebra, these weights are numbers or terms involving the variable.
The basic operations we use to manipulate equations are the inverse operations:
- To undo addition, we subtract.
- To undo subtraction, we add.
- To undo multiplication, we divide.
- To undo division, we multiply.
When applying these operations to both sides of an equation:
- Addition: If you add a number to one side, add the same number to the other side.
- Subtraction: If you subtract a number from one side, subtract the same number from the other side.
- Multiplication: If you multiply one side by a non-zero number, multiply the other side by the same non-zero number.
- Division: If you divide one side by a non-zero number, divide the other side by the same non-zero number.
If $a = b$, then $a+c = b+c$.
If $a = b$, then $a-c = b-c$.
If $a = b$, then $a \times c = b \times c$ (where $c \neq 0$).
If $a = b$, then $a \div c = b \div c$ (where $c \neq 0$).
By applying these operations strategically, we can move terms around and isolate the variable.
Example 1. Solve $x + 5 = 12$ using the balancing method.
Answer:
We want to find the value of $x$ that makes the equation $x+5=12$ true.
$x + 5 = 12$
... (i)
The variable $x$ is on the LHS, and it is being added with 5. To isolate $x$, we need to undo the addition of 5. The inverse operation of adding 5 is subtracting 5.
Subtract 5 from both sides of the equation to maintain the balance:
$x + 5 - 5 = 12 - 5$
(Subtracting 5 from both sides)
Simplify both sides:
$x + 0 = 7$
$x = 7$
The solution is $x=7$.
Check: Substitute $x=7$ back into the original equation:
LHS $= x+5 = 7+5 = 12$
RHS $= 12$
Since LHS = RHS, the solution $x=7$ is correct.
Example 2. Solve $2m = 18$ using the balancing method.
Answer:
We want to find the value of $m$ that makes the equation $2m=18$ true.
$2m = 18$
... (i)
The variable $m$ is on the LHS, and it is being multiplied by 2 ($2m$ means $2 \times m$). To isolate $m$, we need to undo the multiplication by 2. The inverse operation is dividing by 2.
Divide both sides of the equation by 2:
$\frac{2m}{2} = \frac{18}{2}$
(Dividing both sides by 2)
Simplify both sides:
$\frac{\cancel{2}m}{\cancel{2}} = \frac{\cancel{18}^{\normalsize 9}}{\cancel{2}_{\normalsize 1}}$
$m = 9$
The solution is $m=9$.
Check: Substitute $m=9$ back into the original equation:
LHS $= 2m = 2(9) = 18$
RHS $= 18$
Since LHS = RHS, the solution $m=9$ is correct.
Method 2: Transposing Method
The transposing method is a shortcut derived from the balancing method. Instead of explicitly showing the same operation on both sides, we "transpose" or move a term from one side of the equation to the other by changing its sign or inverse operation.
Rules for transposing:
- If a term is being added on one side (e.g., $+c$), it can be moved to the other side by subtracting it (e.g., $-c$).
- If a term is being subtracted on one side (e.g., $-c$), it can be moved to the other side by adding it (e.g., $+c$).
- If a term is multiplying the variable on one side (e.g., $c \times x$), it can be moved to the other side by dividing (e.g., $\div c$), provided $c \neq 0$.
- If a term is dividing the variable on one side (e.g., $x \div c$ or $\frac{x}{c}$), it can be moved to the other side by multiplying (e.g., $\times c$).
This method is faster once you understand the underlying principle of balancing.
Example 3. Solve $y - 3 = 10$ using the transposing method.
Answer:
Given equation: $y - 3 = 10$.
We want to isolate $y$ on the LHS. The term -3 is being subtracted from $y$ on the LHS.
Transpose the term -3 from the LHS to the RHS. When we transpose a term being subtracted, it becomes added on the other side.
$y = 10 + 3$
(Transposing -3 to RHS)
Simplify the RHS:
$y = 13$
The solution is $y=13$.
Check: Substitute $y=13$ back into the original equation:
LHS $= y-3 = 13-3 = 10$
RHS $= 10$
Since LHS = RHS, the solution $y=13$ is correct.
Example 4. Solve $\frac{p}{4} = 5$ using the transposing method.
Answer:
Given equation: $\frac{p}{4} = 5$.
We want to isolate $p$ on the LHS. The term 4 is dividing $p$ on the LHS ($\frac{p}{4}$ means $p \div 4$).
Transpose the term 4 from the LHS to the RHS. When we transpose a term that is dividing, it becomes multiplying on the other side.
$p = 5 \times 4$
(Transposing 4 to RHS)
Simplify the RHS:
$p = 20$
The solution is $p=20$.
Check: Substitute $p=20$ back into the original equation:
LHS $= \frac{p}{4} = \frac{20}{4} = 5$
RHS $= 5$
Since LHS = RHS, the solution $p=20$ is correct.
Solving Equations with Variables on One Side (Multiple Steps)
Many equations involve more than one operation applied to the variable. To solve these, we apply the inverse operations in reverse order of how they were applied to the variable. Think of peeling layers off an onion.
General strategy: Undo addition/subtraction first, then undo multiplication/division.
Example 5. Solve $3x - 4 = 11$.
Answer:
Given equation: $3x - 4 = 11$.
On the LHS, the variable $x$ is first multiplied by 3, and then 4 is subtracted. We need to undo these operations in reverse order: first undo subtraction of 4, then undo multiplication by 3.
Step 1: Undo subtraction of 4. Add 4 to both sides (or transpose -4 to RHS):
$3x - 4 + 4 = 11 + 4$
$3x = 15$
... (i)
Step 2: Undo multiplication by 3. Divide both sides by 3 (or transpose 3 to RHS):
$\frac{3x}{3} = \frac{15}{3}$
(Dividing both sides by 3)
$x = 5$
The solution is $x=5$.
Check: Substitute $x=5$ back into the original equation $3x - 4 = 11$:
LHS $= 3(5) - 4 = 15 - 4 = 11$
RHS $= 11$
Since LHS = RHS, the solution $x=5$ is correct.
Solving Equations with Variables on Both Sides
Equations can have variable terms on both the LHS and the RHS. The strategy here is to collect all the variable terms on one side and all the constant terms on the other side using the transposing method or balancing operations. It is usually convenient to collect the variable terms on the side where the coefficient of the variable is larger to avoid negative coefficients, but it's not mandatory.
Example 6. Solve $5x - 3 = 3x + 7$.
Answer:
Given equation: $5x - 3 = 3x + 7$.
We have variable terms ($5x$ and $3x$) and constant terms (-3 and 7) on both sides.
Step 1: Collect variable terms on one side. Let's transpose $3x$ from RHS to LHS. When transposing $+3x$, it becomes $-3x$ on the other side.
$5x - 3x - 3 = 7$
(Transposing $3x$ to LHS)
Simplify the LHS:
$2x - 3 = 7$
... (i)
Step 2: Collect constant terms on the other side. Transpose -3 from LHS to RHS. When transposing -3, it becomes +3 on the other side.
$2x = 7 + 3$
(Transposing -3 to RHS)
Simplify the RHS:
$2x = 10$
Step 3: Solve the resulting simple linear equation (undo multiplication by 2). Divide both sides by 2 (or transpose 2 to RHS):
$x = \frac{10}{2}$
(Transposing 2 to RHS)
$x = 5$
The solution is $x=5$.
Check: Substitute $x=5$ back into 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 $x=5$ is correct.
Solving Equations Involving Fractions
When the equation contains fractions with the variable or constants in the numerator, it can seem more complex. There are two main approaches:
- Work with the fractions directly using the rules of addition, subtraction, etc.
- Clear the denominators by multiplying both sides of the equation by a common multiple of all the denominators. The LCM of the denominators is the most efficient choice.
The second method often simplifies the equation into one without fractions, which is usually easier to solve.
Example 7. Solve $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$.
Answer:
Given equation: $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$.
The denominators are 2, 5, 3, and 4. Let's find their LCM.
$\text{LCM}(2, 5, 3, 4) = 60$
Step 1: Multiply both sides of the equation by the LCM (60) to eliminate the denominators.
$60 \times (\frac{x}{2} - \frac{1}{5}) = 60 \times (\frac{x}{3} + \frac{1}{4})$
(Multiplying both sides by 60)
Apply the distributive property on both sides:
$(60 \times \frac{x}{2}) - (60 \times \frac{1}{5}) = (60 \times \frac{x}{3}) + (60 \times \frac{1}{4})$
Cancel out the denominators:
$(\cancel{60}^{\normalsize 30} \times \frac{x}{\cancel{2}_{\normalsize 1}}) - (\cancel{60}^{\normalsize 12} \times \frac{1}{\cancel{5}_{\normalsize 1}}) = (\cancel{60}^{\normalsize 20} \times \frac{x}{\cancel{3}_{\normalsize 1}}) + (\cancel{60}^{\normalsize 15} \times \frac{1}{\cancel{4}_{\normalsize 1}})$
This simplifies the equation to one without fractions:
$30x - 12 = 20x + 15$
... (i)
Step 2: Solve the resulting linear equation with variables on both sides (as in Example 6).
Collect variable terms on LHS and constant terms on RHS:
$30x - 20x = 15 + 12$
(Transposing 20x to LHS and -12 to RHS)
Simplify both sides:
$10x = 27$
Step 3: Isolate $x$ by dividing both sides by 10:
$x = \frac{27}{10}$
(Dividing both sides by 10)
The solution is $x = \frac{27}{10}$. This is also equivalent to the decimal 2.7.
Check: Substitute $x = \frac{27}{10}$ back into the original equation $\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}$:
LHS $= \frac{27/10}{2} - \frac{1}{5}$
$= \frac{27}{10} \div 2 - \frac{1}{5} = \frac{27}{10} \times \frac{1}{2} - \frac{1}{5}$
$= \frac{27}{20} - \frac{1}{5}$
LCM of 20 and 5 is 20.
$= \frac{27}{20} - \frac{1 \times 4}{5 \times 4} = \frac{27}{20} - \frac{4}{20} = \frac{27-4}{20} = \frac{23}{20}$
RHS $= \frac{27/10}{3} + \frac{1}{4}$
$= \frac{27}{10} \div 3 + \frac{1}{4} = \frac{27}{10} \times \frac{1}{3} + \frac{1}{4}$
$= \frac{\cancel{27}^{\normalsize 9}}{\cancel{10}_{\normalsize 10}} \times \frac{1}{\cancel{3}_{\normalsize 1}} + \frac{1}{4} = \frac{9}{10} + \frac{1}{4}$
LCM of 10 and 4 is 20.
$= \frac{9 \times 2}{10 \times 2} + \frac{1 \times 5}{4 \times 5} = \frac{18}{20} + \frac{5}{20} = \frac{18+5}{20} = \frac{23}{20}$
Since LHS = RHS ($\frac{23}{20} = \frac{23}{20}$), the solution $x=\frac{27}{10}$ is correct.
Solving Equations Involving Brackets
When an equation contains terms inside brackets multiplied by a factor, the first step is usually to remove the brackets by applying the distributive property. This converts the equation into a simpler form (often with variables on one or both sides) that you already know how to solve.
Recall the distributive property: $a(b+c) = ab + ac$ and $a(b-c) = ab - ac$. This also applies to expressions like $a(b+c+d) = ab + ac + ad$, and so on.
Example 8. Solve $3(z - 1) = 5(2z + 3)$.
Answer:
Given equation: $3(z - 1) = 5(2z + 3)$.
Step 1: Open the brackets on both sides using the distributive property.
On the LHS: $3(z - 1) = 3 \times z - 3 \times 1 = 3z - 3$.
On the RHS: $5(2z + 3) = 5 \times 2z + 5 \times 3 = 10z + 15$.
The equation becomes:
$3z - 3 = 10z + 15$
... (i)
Step 2: Solve the resulting linear equation with variables on both sides.
Collect variable terms on one side (let's move $3z$ to RHS to keep the coefficient positive, although moving $10z$ to LHS is also fine) and constant terms on the other side.
$-3 - 15 = 10z - 3z$
(Transposing 15 to LHS and 3z to RHS)
Simplify both sides:
$-18 = 7z$
Step 3: Isolate $z$ by dividing both sides by 7:
$\frac{-18}{7} = z$
(Dividing both sides by 7)
So, the solution is $z = -\frac{18}{7}$.
Check: Substitute $z = -\frac{18}{7}$ back into the original equation $3(z - 1) = 5(2z + 3)$:
LHS $= 3(z - 1) = 3(-\frac{18}{7} - 1)$
Rewrite 1 as $\frac{7}{7}$ to subtract fractions:
$= 3(-\frac{18}{7} - \frac{7}{7}) = 3(\frac{-18-7}{7}) = 3(\frac{-25}{7})$
$= \frac{3 \times (-25)}{7} = \frac{-75}{7}$
RHS $= 5(2z + 3) = 5(2(-\frac{18}{7}) + 3)$
$= 5(\frac{2 \times (-18)}{7} + 3) = 5(\frac{-36}{7} + 3)$
Rewrite 3 as $\frac{21}{7}$ to add fractions:
$= 5(\frac{-36}{7} + \frac{21}{7}) = 5(\frac{-36+21}{7}) = 5(\frac{-15}{7})$
$= \frac{5 \times (-15)}{7} = \frac{-75}{7}$
Since LHS = RHS ($\frac{-75}{7} = \frac{-75}{7}$), the solution $z = -\frac{18}{7}$ is correct.
Application of Linear Equations
Linear equations in one variable are not just abstract mathematical concepts; they are powerful tools used to solve a wide variety of problems encountered in our daily lives and in various fields like science, engineering, and economics. Translating real-world situations into mathematical equations is a key skill. This section will guide you through the process of applying your knowledge of solving linear equations to tackle such problems, often presented as "word problems".
Steps to Solve Word Problems using Linear Equations
Solving a word problem involves moving from the language of words to the language of algebra, solving the algebraic problem, and then interpreting the result back in the context of the original problem. Here is a systematic approach:
- Read and Understand: Read the problem carefully, perhaps multiple times. Identify what information is given and what quantity you need to find. Try to grasp the relationship between the different parts of the problem.
- Assign a Variable: Choose a variable (like $x, y, z, p, m$, etc.) to represent the unknown quantity that you need to find. Be specific about what the variable represents (e.g., let $x$ be the age in years, let $w$ be the width in metres).
- Formulate the Equation: Translate the statements and conditions given in the problem into a mathematical equation using your chosen variable and the given numbers. Look for keywords that suggest operations (e.g., 'sum' means addition, 'difference' means subtraction, 'product' means multiplication, 'quotient' means division, 'is' or 'was' or 'will be' often mean equals to).
- Solve the Equation: Use the methods you learned in the previous section (balancing or transposing) to solve the linear equation for the variable.
- Interpret the Solution: The value you found for the variable is the mathematical solution to the equation. Now, relate this value back to the original word problem. Write down the answer clearly in the context of the problem (e.g., The length of the park is $X$ metres).
- Check the Solution: Verify if your answer satisfies all the conditions mentioned in the original word problem. Substitute the value(s) you found back into the original problem statement (not just the equation you formed) to ensure it makes sense in the real-world context.
Examples of Word Problems
Let's apply these steps to solve some word problems.
Example 1. The sum of two numbers is 80. If one number is 10 more than the other, find the two numbers.
Answer:
Step 1: Understand the problem. We are given that the sum of two numbers is 80. We also know one number is 10 greater than the other. We need to find both numbers.
Step 2: Assign a variable. Let the smaller of the two numbers be $x$.
Step 3: Formulate the equation.
- One number is $x$.
- The other number is 10 more than the smaller number, so the other number is $x + 10$.
- The sum of the two numbers is 80.
So, we write the equation: (Smaller number) + (Other number) = 80
$x + (x + 10) = 80$
... (i)
Step 4: Solve the equation.
$x + x + 10 = 80$
(Removing the bracket)
$2x + 10 = 80$
(Combining like terms $x+x = 2x$)
Now, solve the linear equation $2x + 10 = 80$. Transpose the constant term 10 to the RHS:
$2x = 80 - 10$
(Transposing 10 to RHS)
$2x = 70$
Now, transpose the coefficient 2 to the RHS:
$x = \frac{70}{2}$
(Transposing 2 to RHS)
$x = 35$
Step 5: Interpret the solution.
The smaller number we represented by $x$ is 35.
The other number is $x + 10 = 35 + 10 = 45$.
The two numbers are 35 and 45.
Step 6: Check the solution.
- Are the two numbers 35 and 45? Yes.
- Is one number 10 more than the other? $45 - 35 = 10$. Yes.
- Is their sum 80? $35 + 45 = 80$. Yes.
All conditions are satisfied. The solution is correct.
Example 2. The perimeter of a rectangular park is 120 metres. If its length is 10 metres more than its width, find the length and width of the park.
Answer:
Step 1: Understand the problem. We are given the perimeter of a rectangle and a relationship between its length and width. We need to find the actual values of the length and width.
Step 2: Assign a variable. Let the width of the rectangular park be $w$ metres.
Step 3: Formulate the equation.
- Width = $w$ metres.
- The length is 10 metres more than the width, so Length, $l = w + 10$ metres.
- The perimeter of a rectangle is given by the formula: Perimeter $= 2 \times (\text{length} + \text{width})$.
- The perimeter is given as 120 metres.
Substitute the expressions for length and width and the given perimeter into the formula:
$120 = 2 \times ((w + 10) + w)$
... (i)
Step 4: Solve the equation.
Simplify the expression inside the brackets first:
$120 = 2 \times (2w + 10)$
(Combining like terms $w+w=2w$)
Divide both sides by 2:
$\frac{120}{2} = 2w + 10$
(Dividing both sides by 2)
$60 = 2w + 10$
Now, solve the linear equation $60 = 2w + 10$. Transpose the constant term 10 to the LHS:
$60 - 10 = 2w$
(Transposing 10 to LHS)
$50 = 2w$
Now, transpose the coefficient 2 to the LHS:
$\frac{50}{2} = w$
(Transposing 2 to LHS)
$25 = w$
Step 5: Interpret the solution.
The width, $w$, is 25 metres.
The length, $l$, is $w + 10 = 25 + 10 = 35$ metres.
The length of the park is 35 metres and the width is 25 metres.
Step 6: Check the solution.
- Width = 25 m, Length = 35 m.
- Is the length 10 metres more than the width? $35 - 25 = 10$. Yes.
- Is the perimeter 120 metres? Perimeter $= 2 \times (\text{length} + \text{width}) = 2 \times (35 + 25) = 2 \times 60 = 120$. Yes.
All conditions are satisfied. The solution is correct.
Example 3. The age of Rahul is 5 years more than twice the age of Rina. If the sum of their ages is 41 years, find their present ages.
Answer:
Step 1: Understand the problem. We need to find the current ages of Rahul and Rina. We are given a relationship between their ages and the sum of their ages.
Step 2: Assign a variable. Let Rina's present age be $r$ years. We choose Rina's age as the base because Rahul's age is described in terms of Rina's age.
Step 3: Formulate the equation.
- Rina's age = $r$ years.
- Twice Rina's age = $2 \times r = 2r$.
- Rahul's age is 5 years more than twice Rina's age, so Rahul's age = $2r + 5$ years.
- The sum of their ages is 41 years.
So, we write the equation: (Rina's age) + (Rahul's age) = 41
$r + (2r + 5) = 41$
... (i)
Step 4: Solve the equation.
$r + 2r + 5 = 41$
(Removing the bracket)
$3r + 5 = 41$
(Combining like terms $r+2r = 3r$)
Now, solve the linear equation $3r + 5 = 41$. Transpose the constant term 5 to the RHS:
$3r = 41 - 5$
(Transposing 5 to RHS)
$3r = 36$
Now, transpose the coefficient 3 to the RHS:
$r = \frac{36}{3}$
(Transposing 3 to RHS)
$r = 12$
Step 5: Interpret the solution.
Rina's present age, $r$, is 12 years.
Rahul's present age is $2r + 5 = 2(12) + 5 = 24 + 5 = 29$ years.
Rina's present age is 12 years and Rahul's present age is 29 years.
Step 6: Check the solution.
- Rina's age = 12 years, Rahul's age = 29 years.
- Is Rahul's age 5 years more than twice Rina's age? Twice Rina's age is $2 \times 12 = 24$. 5 more than 24 is $24 + 5 = 29$. Yes.
- Is the sum of their ages 41 years? $12 + 29 = 41$. Yes.
All conditions are satisfied. The solution is correct.
Example 4. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator is increased by 2, the new number is $\frac{1}{2}$. Find the original rational number.
Answer:
Step 1: Understand the problem. We are dealing with a rational number. We are given a relationship between its numerator and denominator. We are also told what happens to the fraction when the numerator and denominator are changed in specific ways, resulting in a new fraction $\frac{1}{2}$. We need to find the original rational number.
Step 2: Assign a variable. Let the denominator of the original rational number be $d$.
Step 3: Formulate the equation.
- Denominator = $d$.
- The numerator is 3 less than the denominator, so the numerator is $d - 3$.
- The original rational number is $\frac{d-3}{d}$. (Note that for this to be a rational number, $d$ must be a non-zero integer. Also, from the problem, if $d-3=0$, i.e., $d=3$, the numerator is 0, which is a valid case. If $d=0$, the original number is undefined, but the problem structure implies $d \neq 0$).
- The denominator is increased by 5, so the new denominator is $d + 5$. (Note that $d+5$ must also be non-zero).
- The numerator is increased by 2, so the new numerator is $(d - 3) + 2 = d - 1$.
- The new rational number is $\frac{d-1}{d+5}$.
- We are given that the new rational number is equal to $\frac{1}{2}$.
So, we write the equation:
$\frac{d-1}{d+5} = \frac{1}{2}$
... (i)
Step 4: Solve the equation.
We have an equation with fractions on both sides. The easiest way to solve this is by cross-multiplication.
$2 \times (d - 1) = 1 \times (d + 5)$
(Cross-multiplying)
Apply the distributive property on both sides:
$2d - 2 = d + 5$
... (ii)
Now, solve the linear equation $2d - 2 = d + 5$. Collect variable terms on one side (transpose $d$ to LHS) and constant terms on the other side (transpose -2 to RHS):
$2d - d = 5 + 2$
(Transposing $d$ to LHS and -2 to RHS)
$d = 7$
Step 5: Interpret the solution.
The denominator of the original rational number, $d$, is 7. Since $d=7 \neq 0$, this is a valid denominator.
The numerator of the original rational number is $d - 3 = 7 - 3 = 4$.
The original rational number is $\frac{\text{Numerator}}{\text{Denominator}} = \frac{4}{7}$.
Step 6: Check the solution.
- Original number is $\frac{4}{7}$. Is the numerator (4) 3 less than the denominator (7)? $7 - 4 = 3$. Yes.
- If the denominator is increased by 5, the new denominator is $7 + 5 = 12$.
- If the numerator is increased by 2, the new numerator is $4 + 2 = 6$.
- The new number is $\frac{6}{12}$. Is this equal to $\frac{1}{2}$? Yes, $\frac{6}{12} = \frac{6 \div 6}{12 \div 6} = \frac{1}{2}$.
All conditions are satisfied. The solution is correct.
The original rational number is $\frac{4}{7}$.