CBSE Class 8  >  Class 8 Notes  >  Mathematics (Maths)   >  Solved Questions: Linear Equations in One Variable

Solved Questions: Linear Equations in One Variable

Ques 1: Show that x = 4 is a solution of the equation: Solved Questions: Linear Equations in One Variable

Solution: Substituting x = 4 in  Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable

Since, LHS = RHS

∴ x = 4 is a solution of the given equation.


Ques 2: Solve  Solved Questions: Linear Equations in One Variable

Solution: We have Solved Questions: Linear Equations in One Variable 
LCM of 3, 5, 2 and 4 is 60.

∴ The given equation can be expressed as:

Solved Questions: Linear Equations in One Variable

Solved Questions: Linear Equations in One Variable

Solved Questions: Linear Equations in One Variable

Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable

Thus, x = 27/10 is the required solution.


Ques 3: Solve for Solved Questions: Linear Equations in One Variable

Solution: We have Solved Questions: Linear Equations in One Variable

Solved Questions: Linear Equations in One Variable
Solved Questions: Linear Equations in One Variable

⇒ 3x - 4 + 44 - 4x - 3 = 2x + 4

⇒ 3x - 4x - 2x = 4 + 3 - 44 + 4

⇒ 3x - 6x = 11 - 44

⇒ -3x = -33 ⇒ x = 11


Ques 4: Solve for  Solved Questions: Linear Equations in One Variable

Solution: We have   Solved Questions: Linear Equations in One Variable 

By cross multiplication, we get:

(2 + x)(7 - x) = (5 - x)(4 + x)

⇒ 2(7 - x) + x(7 - x) = 5(4 + x) - x(4 + x) 

⇒ 14 - 2x + 7x - x2 = 20 + 5x - 4x - x2 

⇒ -x2 + x2 - 2x + 7x - 5x + 4x = 20 - 14 

⇒ -7x + 7x + 4x = 6

⇒  4x = 6 ⇒ x = 6/4 or 3/2

Thus, the solution of the given equation is x = 3/2


Ques 5: A number is such that it is as much greater than 65 as it is less than 91. Find the number.
Solution: Let the number be x.
Since, we have [The number] - 65 = 91 - [The number]

⇒ x - 65 = 91 - x
⇒ x + x = 91 + 65
⇒ 2x = 156

⇒ x = 156/2 = 78

Thus, the required number is 78.


Ques 6: The numerator of a fraction is 2 less than the denominator. If 1 is added to its denominator, it becomes 1/2. Find the fraction.
Solution: Let the denominator of the fraction be x.

∴ Numerator = x - 2

The fraction =Solved Questions: Linear Equations in One Variable

Since it becomes 1/2

When 1 is added to its denominator.

i.e   Solved Questions: Linear Equations in One Variable

By cross multiplication, we have
2(x - 2) = x + 1

⇒ 2x - 4 = x + 1
⇒ 2x - x = 1 + 4
⇒ x = 5

⇒ Fraction =Solved Questions: Linear Equations in One Variable


Ques 7: After 24 years I shall be 3 times as old as I was 4 years ago. Find my present age.
Solution: Let my present age be x years.

∴ After 24 years, my age will be (x + 24) years.
4 years ago, my age was (x - 4) years.

According to the given condition, we have

(x + 24) = 3(x - 4)

⇒ x + 24 = 3x - 12
⇒ x - 3x = -12 - 24
⇒ -2x = -36

⇒ x = -36/-2 = 18

Thus, my present age is 18 years.


Ques 8: If the sum of two numbers is 30 and their ratio is 2 : 3, then find the numbers.
Solution: Let one of the numbers be x.

∴ The other number = (30 - x)

According to the condition, we have
Solved Questions: Linear Equations in One Variable  [∵ The ratio number is 2 : 3]

⇒ 3x = 2(30 - x) [By cross multiplication] 

⇒ 3x = 60 - 2x 

⇒ 3x + 2x = 60 

⇒ 5x = 60

⇒ x = 60/5 = 12

∴ 30 - x = 30 - 12 = 18 

Thus, the required numbers are 12 and 18.

The document Solved Questions: Linear Equations in One Variable is a part of the Class 8 Course Mathematics (Maths) Class 8.
All you need of Class 8 at this link: Class 8

FAQs on Solved Questions: Linear Equations in One Variable

1. How do I solve linear equations in one variable with fractions?
Ans. To solve linear equations with fractions, multiply all terms by the least common multiple (LCM) of denominators to eliminate fractions, then isolate the variable using addition, subtraction, multiplication, or division. For example, in (x/2) + 3 = 5, multiply by 2 to get x + 6 = 10, so x = 4. This method simplifies CBSE Class 8 equations significantly and reduces calculation errors.
2. Why do I get different answers when I solve the same linear equation two different ways?
Ans. Different methods yield identical solutions when executed correctly; discrepancies indicate computational errors, not flawed approaches. Whether using transposition, balancing both sides, or systematic substitution, the answer remains constant. Check your arithmetic carefully-arithmetic mistakes are common when moving terms across the equals sign or simplifying expressions in one-variable equations.
3. What's the difference between linear equations and linear inequalities in Class 8?
Ans. Linear equations use the equals sign (=) and have exact solutions, while linear inequalities use symbols like <, >, ≤, or ≥ and have solution sets or ranges. For instance, x + 2 = 5 gives x = 3 (one answer), but x + 2 > 5 gives x > 3 (all numbers greater than 3). Both are foundational CBSE concepts but differ fundamentally.
4. How do I check if my answer is correct after solving a linear equation?
Ans. Substitute your solution back into the original equation and verify both sides equal the same value. If x = 3 solves 2x + 1 = 7, then 2(3) + 1 should equal 7, which it does. This verification step catches mistakes immediately and builds confidence before submitting exam answers on linear equations in one variable.
5. What mistakes do students make most often when solving linear equations with variables on both sides?
Ans. Common errors include forgetting to apply operations to all terms, moving terms incorrectly across the equals sign, and miscombining like terms. For example, in 3x + 5 = x + 13, students often forget to subtract x from both sides properly. Careful step-by-step transposition and double-checking variable collection prevents these frequent mistakes in CBSE mathematics assessments.
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