RBSE Class 8 Maths Important Questions Chapter 2 Linear Equations in One Variable

Rajasthan Board RBSE Class 8 Maths Important Questions Chapter 2 Linear Equations in One Variable Important Questions and Answers.

RBSE Class 8 Maths Chapter 2 Important Questions Linear Equations in One Variable

Multiple Choice Questions:

Question 1.
The necessary condition for a linear equation is:
(a) Maximum power of variable is one
(b) Maximum power of variable is two
(c) Maximum power of variable is three
(d) none of these
Answer:
(a) Maximum power of variable is one

Question 2.
The solution of 3 + \(\frac{x}{4}\) = 5 is:
(a) x = - 17
(b) x = 8
(c) x = 2
(d) x = 4
Answer:
(b) x = 8

Question 3.
The solution of 3x = 21 is:
(a) 7
(b) 24
(c) 18
(d) 63
Answer:
(a) 7

Question 4.
The solution of a linear equation can be:
(a) any natural number
(b) any rational number
(c) any whole number
(d) none of these
Answer:
(b) any rational number

Question 5.
The solution of x - 4 = 5 is:
(a) 4 × 5
(b) \(\frac{5}{4}\)
(c) 5 - 4
(d) 5 + 4
Answer:
(d) 5 + 4

Fill in the blanks:

Question 1.
Sign of ____________  is used in equations.
Answer:
"="

Question 2.
Just as numbers, variables can also be ____________ from one side of the equation to the other.
Answer:
transformed

Question 3.
We perform the same mathematical operations on both sides of the equation, so that the is ____________ not disturbed.
Answer:
balance

Question 4.
The utility of lenear equation is in their diverse ____________
Answer:
applications

Very Short Answer Type Questions:

Question 1.
What do you understand by LHS in a algebraic equation?
Answer:
By LHS, we mean that left hand side expression in a algebraic equation.

Question 2.
Solve the equation 2x - 3 = 7.
Answer:
2x - 3 = 7
or 2x = 7 + 3
or 2x = 10
or x = \(\frac{10}{2}\) = 5

Question 3.
Solve the equation 2x - 3 = x + 2
Answer:
2x - 3 = x + 2
2x - x = 3 + 2 x = 5

Question 4.
Verily the equation \(\frac{15}{4}\) - 7x = 9 when x = \(-\frac{3}{4}\)
Answer:
LHS = \(\frac{15}{4}-7 \times\left(-\frac{3}{4}\right)\)
= \(\frac{15}{4}+\frac{21}{4}=\frac{36}{4}\) = 9 = R.H.S
 
Question 5.
Solve : \(\frac{x}{3}+\frac{5}{2}=-\frac{3}{2}\)
Answer:
\(\frac{x}{3}+\frac{5}{2}=-\frac{3}{2}\)
or \(\frac{x}{3}=-\frac{3}{2}-\frac{5}{2}=\frac{-3-5}{2}\)
or \(\frac{x}{3}=\frac{-8}{2}\) = -4
or x = - 4 × 3 = - 12