Get the most accurate RBSE Solutions for Class 8 Mathematics Chapter 11 Linear Equations with One Variable here. Updated for the 2026-27 academic session, these solutions are based on the latest RBSE textbooks for Class 8 Mathematics. Our expert-created answers for Class 8 Mathematics are available for free download in PDF format.
Detailed Chapter 11 Linear Equations with One Variable RBSE Solutions for Class 8 Mathematics
For Class 8 students, solving RBSE textbook questions is the most effective way to build a strong conceptual foundation. Our Class 8 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 11 Linear Equations with One Variable solutions will improve your exam performance.
Class 8 Mathematics Chapter 11 Linear Equations with One Variable RBSE Solutions PDF
Rajasthan Board RBSE Class 8 Maths Chapter 11 Linear Equations with One Variable In Text Exercise
Page No: 125
Question. Let us practice to solve an equation: Equation \( 3x + 9 = 15 \) with solution \( x = 2 \).
| On both sides | New equation | Solution |
|---|---|---|
| 1. Adding 2 | \( 3x + 11 = 17 \) | \( x = ... \) |
| 2. Subtracting 3 | \( 3x + 6 = 12 \) | \( x = ... \) |
| 3. Multiplying by 2 | \( 6x + 18 = ... \) | \( x = ... \) |
| 4. Dividing by 3 | \( ... = 5 \) | \( x = ... \) |
Answer: This table shows how different operations affect an equation, assuming the starting solution is \( x = 2 \). The goal is to see how the equation changes while keeping the solution the same.
| On both sides | New equation | Solution |
|---|---|---|
| 1. Adding 2 | \( 3x + 11 = 17 \) | \( x = 2 \) |
| 2. Subtracting 3 | \( 3x + 6 = 12 \) | \( x = 2 \) |
| 3. Multiplying by 2 | \( 6x + 18 = 30 \) | \( x = 2 \) |
| 4. Dividing by 3 | \( x + 3 = 5 \) | \( x = 2 \) |
🎯 Exam Tip: Remember that whatever operation you perform on one side of an equation, you must perform the exact same operation on the other side to keep the equation balanced and the solution valid.
Solve the Following Equations
Question 1. Solve the equation: \( \frac {2x}{x+6 }=1 \)
Answer: We have the equation: \( \frac {2x}{x+6 } = 1 \)
Now, we cross-multiply to simplify.
\( 2x = 1 \cdot (x+6) \)
\( 2x = x + 6 \)
To solve for \( x \), transpose \( x \) to the left side.
\( 2x - x = 6 \)
\( x = 6 \)
In simple words: To find \( x \), we first multiply both sides by \( (x+6) \). Then, we move all \( x \) terms to one side and numbers to the other. Finally, we get the value of \( x \).
🎯 Exam Tip: Remember to check your solution by plugging the value of \( x \) back into the original equation to ensure it balances. Also, be careful when \( x \) appears in the denominator, as \( x+6 \) cannot be zero.
Question 2. Solve the equation: \( 10 = x + 3 \)
Answer: We start with the equation: \( 10 = x + 3 \)
To find \( x \), we need to isolate it on one side. So, we transpose 3 to the left side.
\( 10 - 3 = x \)
\( 7 = x \)
This means: \( x = 7 \)
In simple words: To find \( x \), simply subtract 3 from 10. This gives us the value of \( x \).
🎯 Exam Tip: When transposing a term from one side of an equation to the other, always change its sign (plus becomes minus, minus becomes plus).
Question 3. Solve the equation: \( 16 = 7x - 9 \)
Answer: We have the equation: \( 16 = 7x - 9 \)
First, transpose -9 to the left side by adding it.
\( 16 + 9 = 7x \)
\( 25 = 7x \)
Now, to find \( x \), divide both sides by 7.
\( \frac {7x}{7} = \frac {25}{7} \)
\( x = \frac {25}{7} \)
In simple words: First, move the number 9 to the other side by changing its sign. Then, divide both sides by 7 to get what \( x \) is equal to.
🎯 Exam Tip: Remember that division is the opposite of multiplication. If a number multiplies \( x \), divide by that number to isolate \( x \).
Question 4. Solve the equation: \( \frac {x+5}{ x } = 2\frac {2}{3} \)
Answer: We have the equation: \( \frac {x+5}{ x } = 2\frac {2}{3} \)
First, convert the mixed fraction to an improper fraction.
\( 2\frac {2}{3} = \frac {(2 \cdot 3) + 2}{3} = \frac {6+2}{3} = \frac {8}{3} \)
So, the equation becomes: \( \frac {x+5}{ x } = \frac {8}{3} \)
Next, we cross-multiply.
\( 3(x+5) = 8x \)
Now, distribute the 3 on the left side.
\( 3x + 15 = 8x \)
To solve for \( x \), transpose \( 3x \) to the right side.
\( 15 = 8x - 3x \)
\( 15 = 5x \)
Now, divide both sides by 5.
\( \frac {15}{5} = \frac {5x}{5} \)
\( 3 = x \)
Thus, \( x = 3 \).
In simple words: First, change the mixed number to a simple fraction. Then, multiply across the equal sign. After that, gather all the \( x \) terms on one side and the numbers on the other to find what \( x \) is.
🎯 Exam Tip: Always convert mixed fractions to improper fractions before starting algebraic operations to avoid mistakes. Also, double-check your distribution when multiplying into a bracket.
Question 3. Let's revise to make mathematical sentences. If there is any number \( x \), then fill in the blanks.
Answer: We will write the mathematical expression for each statement.
(i) 5 more than the number = \( x + 5 \)
(ii) 3 less than the number = \( x - 3 \)
(iii) Half of the number = \( \frac {1}{2}x \)
(iv) 7 less half of the number = \( \frac {1}{2}x - 7 \)
(v) 4 more than one third of the number = \( \frac {1}{3}x + 4 \)
(vi) 6 more than triple of the number = \( 3x + 6 \)
(vii) 3 less than 5 times of the number = \( 5x - 3 \)
In simple words: When we are given a word phrase about a number, we can write it using math symbols like \( + \), \( - \), or fractions with \( x \). "More than" means add, "less than" means subtract, "half of" means multiply by \( \frac{1}{2} \), and "triple" means multiply by 3.
🎯 Exam Tip: Pay close attention to keywords like 'more than', 'less than', 'times', and 'of' as they indicate specific mathematical operations. The phrase "less than" often means the number comes after the operation, like "7 less than half of x" is \( \frac{1}{2}x - 7 \).
Free study material for Mathematics
RBSE Solutions Class 8 Mathematics Chapter 11 Linear Equations with One Variable
Students can now access the RBSE Solutions for Chapter 11 Linear Equations with One Variable prepared by teachers on our website. These solutions cover all questions in exercise in your Class 8 Mathematics textbook. Each answer is updated based on the current academic session as per the latest RBSE syllabus.
Detailed Explanations for Chapter 11 Linear Equations with One Variable
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 8 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 8 students who want to understand both theoretical and practical questions. By studying these RBSE Questions and Answers your basic concepts will improve a lot.
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