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Class 7 Maths Chapter 2 Question Answers - Linear Equations in One Variable
Ques 1: Show that x = 4 is a solution of the equation: 
Solution: Substituting x = 4 in 
Since, LHS = RHS
∴ x = 4 is a solution of the given equation.
Ques 2: Solve 
Solution: We have 
LCM of 3, 5, 2 and 4 is 60.
∴ The given equation can be expressed as:
⇒ 
⇒ 
⇒ 
⇒ 
Thus, x = 27/10 is the required solution.
Ques 3: Solve for 
Solution: We have 
⇒ 
⇒ 
⇒ 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 
Solution: We have
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 =
Since it becomes 1/2
When 1 is added to its denominator.
i.e 
By cross multiplication, we have
2(x – 2) = x + 1
⇒ 2x – 4 = x + 1
⇒ 2x – x = 1 + 4
⇒ x = 5
⇒ Fraction =
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
[∵ 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 Class 7 Maths Chapter 2 Question Answers - Linear Equations in One Variable is a part of the Class 8 Course Mathematics (Maths) Class 8.
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FAQs on Class 7 Maths Chapter 2 Question Answers - Linear Equations in One Variable
| 1. What are linear equations in one variable? |  |
| 2. How can I solve a linear equation in one variable? | |
Ans. To solve a linear equation in one variable, you need to isolate the variable on one side of the equation. This can be done by performing various operations such as addition, subtraction, multiplication, and division on both sides of the equation until the variable is alone.
| 3. What is the importance of solving linear equations in one variable? | |
Ans. Solving linear equations in one variable is important in various fields such as mathematics, physics, economics, and engineering. It helps in finding unknown quantities, determining relationships between variables, and solving real-world problems involving linear relationships.
| 4. Can linear equations have more than one solution? | |
Ans. No, linear equations in one variable can have only one solution. This is because a linear equation represents a straight line, which intersects the x-axis at only one point. Therefore, there can be at most one value of the variable that satisfies the equation.
| 5. What is the significance of the slope-intercept form of a linear equation? | |
Ans. The slope-intercept form of a linear equation, y = mx + b, is significant as it provides useful information about the equation. 'm' represents the slope of the line, indicating the rate of change, while 'b' represents the y-intercept, which is the value of y when x = 0. This form allows for easy interpretation and graphing of the equation.
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