RBSE Solutions Class 8 Maths Chapter 11 Linear Equations with One Variable Exercise 11.1

Question 1. \( 6x + 3 = 4x + 11 \)
Answer: We need to solve the equation \( 6x + 3 = 4x + 11 \) for \( x \).
First, gather all the terms with \( x \) on one side and constant numbers on the other side.
\( 6x + 3 = 4x + 11 \)
Subtract \( 4x \) from both sides and subtract \( 3 \) from both sides to move them. This step is called transposing.
\( 6x - 4x = 11 - 3 \)
Now, simplify both sides of the equation.
\( 2x = 8 \)
To find the value of \( x \), divide both sides of the equation by 2. This isolates \( x \).
\( \frac{2x}{2} = \frac{8}{2} \)
\( \implies x = 4 \)
The variable \( x \) is found to be 4 after these algebraic operations.
In simple words: To solve for \( x \), move all \( x \) terms to one side and numbers to the other. Then, simplify and divide to find \( x \).

🎯 Exam Tip: Always check your answer by putting the value of \( x \) back into the original equation to ensure both sides are equal. For this question, \( 6(4) + 3 = 24 + 3 = 27 \) and \( 4(4) + 11 = 16 + 11 = 27 \), so the answer is correct.

Question 2. \( 3(x + 5) = 4x + 9 \)
Answer: We need to find the value of \( x \) from the equation \( 3(x + 5) = 4x + 9 \).
First, distribute the 3 on the left side by multiplying it with both terms inside the bracket.
\( 3x + 15 = 4x + 9 \)
Next, move all the \( x \) terms to one side of the equation and all the constant numbers to the other side.
\( 3x - 4x = 9 - 15 \)
Now, combine the like terms on both sides of the equation.
\( -x = -6 \)
To make \( x \) positive, divide both sides of the equation by -1. This changes the sign of both sides.
\( \frac{-x}{-1} = \frac{-6}{-1} \)
\( \implies x = 6 \)
The value of \( x \) that satisfies this equation is 6.
In simple words: First, multiply the number outside the bracket by everything inside it. Then, move all \( x \) terms to one side and numbers to the other, then solve for \( x \).

🎯 Exam Tip: Remember to apply the distributive property correctly to all terms inside the parentheses. A common mistake is to multiply only the first term.

Question 3. \( 5x + 6 = 21 \)
Answer: We need to solve the given linear equation \( 5x + 6 = 21 \) for \( x \).
The first step is to isolate the term with \( x \) on one side of the equation. We do this by moving the constant term to the other side.
\( 5x = 21 - 6 \)
Now, perform the subtraction on the right side of the equation.
\( 5x = 15 \)
To find \( x \), divide both sides of the equation by the coefficient of \( x \), which is 5.
\( \frac{5x}{5} = \frac{15}{5} \)
\( \implies x = 3 \)
This means that if you replace \( x \) with 3 in the original equation, both sides will be equal.
In simple words: Move the plain number to the other side, then do the math. After that, divide by the number next to \( x \) to find what \( x \) is.

🎯 Exam Tip: When moving a term from one side of an equation to the other, always remember to change its sign (from plus to minus, or minus to plus).

Question 4. \( \frac {x+1}{2}+\frac {x+2}{3} =\frac {2x-5 }{7} +9 \)
Answer: We need to solve the equation \( \frac {x+1}{2}+\frac {x+2}{3} =\frac {2x-5 }{7} +9 \) for \( x \).
The denominators in the equation are 2, 3, and 7. To clear the fractions, we find the Least Common Multiple (LCM) of these denominators. The LCM of 2, 3, and 7 is \( 2 \times 3 \times 7 = 42 \).
Multiply every term in the entire equation by 42 to remove the denominators.
\( 42 \left( \frac{x+1}{2} \right) + 42 \left( \frac{x+2}{3} \right) = 42 \left( \frac{2x-5}{7} \right) + 42(9) \)
This simplifies to:
\( 21(x + 1) + 14(x + 2) = 6(2x - 5) + 378 \)
Next, distribute the numbers into their respective parentheses on both sides.
\( 21x + 21 + 14x + 28 = 12x - 30 + 378 \)
Combine the like terms on each side of the equation.
\( (21x + 14x) + (21 + 28) = 12x + (-30 + 378) \)
\( 35x + 49 = 12x + 348 \)
Now, gather all the \( x \) terms on the left side and constant terms on the right side by transposing them.
\( 35x - 12x = 348 - 49 \)
Simplify both sides.
\( 23x = 299 \)
Finally, divide both sides by 23 to solve for \( x \).
\( \frac{23x}{23} = \frac{299}{23} \)
\( \implies x = 13 \)
The solution to this multi-step equation is \( x = 13 \).
In simple words: First, find the smallest number that all bottom numbers (denominators) can divide into. Multiply every part of the equation by this number to get rid of fractions. Then, open brackets, put all \( x \) terms on one side and numbers on the other, and solve.

🎯 Exam Tip: Always find the LCM of all denominators and multiply the *entire* equation by it to simplify. Remember to multiply constant terms too, not just the fractions.

Question 5. \( \frac {3x-2}{5} =4-\left(\frac {x+2 }{ 3 } \right) \)
Answer: We need to solve the equation \( \frac {3x-2}{5} =4-\left(\frac {x+2 }{ 3 } \right) \) for \( x \).
The denominators present in the equation are 5 and 3. The Least Common Multiple (LCM) of 5 and 3 is 15. We will multiply every term in the equation by 15 to clear the fractions.
\( 15 \left( \frac{3x-2}{5} \right) = 15(4) - 15 \left( \frac{x+2}{3} \right) \)
This simplifies to:
\( 3(3x - 2) = 60 - 5(x + 2) \)
Now, distribute the numbers into the parentheses on both sides of the equation.
\( 9x - 6 = 60 - 5x - 10 \)
Combine the constant terms on the right side.
\( 9x - 6 = 50 - 5x \)
Next, move all the terms containing \( x \) to the left side and all the constant terms to the right side by transposing them.
\( 9x + 5x = 50 + 6 \)
Simplify both sides of the equation.
\( 14x = 56 \)
Finally, divide both sides by 14 to solve for \( x \).
\( \frac{14x}{14} = \frac{56}{14} \)
\( \implies x = 4 \)
The value of \( x \) that satisfies this equation is 4.
In simple words: Multiply all parts of the equation by the smallest common multiple of the bottom numbers to remove fractions. Then, open brackets, move \( x \) terms to one side, numbers to the other, and divide to find \( x \).

🎯 Exam Tip: Be careful with the minus sign before a fraction, like \( -\left(\frac{x+2}{3}\right) \). When you multiply by the LCM, the minus sign applies to the *entire* numerator of that fraction, so \( -5(x+2) \) becomes \( -5x - 10 \), not \( -5x + 10 \).

Question 6. \( \frac {x+2}{2}+\frac {x+4 }{ 3 } =\frac {x+6 }{4}+\frac {x+8 }{5} \)
Answer: We need to solve the given linear equation \( \frac {x+2}{2}+\frac {x+4 }{ 3 } =\frac {x+6 }{4}+\frac {x+8 }{5} \) for \( x \).
First, find the Least Common Multiple (LCM) of the denominators 2, 3, 4, and 5.
The LCM of 2, 3, 4, 5 is 60.
Multiply every term in the equation by 60 to eliminate the fractions.
\( 60 \left( \frac{x+2}{2} \right) + 60 \left( \frac{x+4}{3} \right) = 60 \left( \frac{x+6}{4} \right) + 60 \left( \frac{x+8}{5} \right) \)
This simplifies to:
\( 30(x + 2) + 20(x + 4) = 15(x + 6) + 12(x + 8) \)
Next, distribute the numbers into the parentheses on both sides of the equation.
\( 30x + 60 + 20x + 80 = 15x + 90 + 12x + 96 \)
Combine the like terms (terms with \( x \) and constant terms) on each side.
\( (30x + 20x) + (60 + 80) = (15x + 12x) + (90 + 96) \)
\( 50x + 140 = 27x + 186 \)
Now, move all the \( x \) terms to the left side and all the constant numbers to the right side by transposing.
\( 50x - 27x = 186 - 140 \)
Simplify both sides of the equation.
\( 23x = 46 \)
Finally, divide both sides by 23 to find the value of \( x \).
\( \frac{23x}{23} = \frac{46}{23} \)
\( \implies x = 2 \)
Thus, the solution to this equation is \( x = 2 \).
In simple words: Find the smallest common number for all bottom parts of the fractions. Multiply every single piece of the equation by this number to remove fractions. Then, tidy up by opening brackets, collecting \( x \) terms on one side, numbers on the other, and finally divide to get \( x \).

🎯 Exam Tip: When dealing with multiple fractions, finding the LCM correctly is the crucial first step. If the LCM is incorrect, the entire solution will be wrong. Double-check your LCM calculation.

Question 7. \( 0.6x + 0.25x = 0.45x + 1.2 \)
Answer: We need to solve the decimal equation \( 0.6x + 0.25x = 0.45x + 1.2 \) for \( x \).
First, it's often easier to work with whole numbers. Convert the decimal fractions into simple fractions to clearly see the denominators.
\( \frac{6}{10}x + \frac{25}{100}x = \frac{45}{100}x + \frac{12}{10} \)
The denominators are 10 and 100. The Least Common Multiple (LCM) of 10 and 100 is 100.
Multiply every term in the entire equation by 100 to remove the denominators.
\( 100 \left( \frac{6}{10}x \right) + 100 \left( \frac{25}{100}x \right) = 100 \left( \frac{45}{100}x \right) + 100 \left( \frac{12}{10} \right) \)
This simplifies to:
\( 60x + 25x = 45x + 120 \)
Next, combine the like terms on the left side of the equation.
\( 85x = 45x + 120 \)
Now, move the term with \( x \) from the right side to the left side by subtracting \( 45x \) from both sides.
\( 85x - 45x = 120 \)
Simplify the left side.
\( 40x = 120 \)
Finally, divide both sides by 40 to solve for \( x \).
\( \frac{40x}{40} = \frac{120}{40} \)
\( \implies x = 3 \)
The solution to the given equation is \( x = 3 \).
In simple words: Change the decimal numbers to fractions. Find the smallest number that all bottom parts can divide into, and multiply every part of the equation by it. This clears the fractions. Then, combine like terms, move \( x \) terms to one side, and solve.

🎯 Exam Tip: When converting decimals to fractions, remember that the number of digits after the decimal point tells you the denominator (e.g., one digit for 10, two digits for 100). This helps in finding the correct LCM to clear decimals.

Question 8. \( 2.5x - 7 = 0.5x + 13 \)
Answer: We need to solve the equation \( 2.5x - 7 = 0.5x + 13 \) for \( x \).
First, gather all terms containing \( x \) on one side of the equation and all constant numbers on the other side.
Move \( 0.5x \) from the right side to the left side by subtracting it, and move \( -7 \) from the left side to the right side by adding it.
\( 2.5x - 0.5x = 13 + 7 \)
Now, combine the like terms on both sides of the equation.
\( 2x = 20 \)
To find the value of \( x \), divide both sides of the equation by 2, which is the coefficient of \( x \).
\( \frac{2x}{2} = \frac{20}{2} \)
\( \implies x = 10 \)
The solution to the equation, where \( x \) is the unknown, is 10.
In simple words: Bring all the \( x \) parts to one side and all the plain numbers to the other. Then, do the addition or subtraction on both sides. Finally, divide to find what \( x \) is.

🎯 Exam Tip: When moving terms across the equals sign, always change their operation (addition becomes subtraction, multiplication becomes division, and vice versa). This is fundamental for solving equations.

Question 9. \( \frac {7x+4}{x+2} = -\frac{4}{3} \)
Answer: We need to solve the equation \( \frac {7x+4}{x+2} = -\frac{4}{3} \) for \( x \).
This equation involves fractions on both sides. To solve it, we can use cross-multiplication. This means multiplying the numerator of one fraction by the denominator of the other.
\( 3(7x + 4) = -4(x + 2) \)
Next, distribute the numbers into the parentheses on both sides of the equation.
\( 21x + 12 = -4x - 8 \)
Now, move all the terms containing \( x \) to the left side and all the constant numbers to the right side by transposing them.
\( 21x + 4x = -8 - 12 \)
Combine the like terms on both sides.
\( 25x = -20 \)
Finally, divide both sides by 25 to solve for \( x \).
\( \frac{25x}{25} = \frac{-20}{25} \)
\( \implies x = -\frac{4}{5} \)
The solution to this rational equation is \( x = -\frac{4}{5} \).
In simple words: When you have fractions equal to each other, multiply the top of one by the bottom of the other (cross-multiply). Then, open brackets, move \( x \) terms to one side, numbers to the other, and divide to get your answer.

🎯 Exam Tip: Always remember to distribute the negative sign correctly when cross-multiplying. In this case, \( -4(x+2) \) becomes \( -4x - 8 \), not \( -4x + 8 \).