## RD Sharma Class 8 Solutions Chapter 9 - Linear Equation In One Variable (Ex 9.4) Exercise 9.4 - Free PDF

## FAQs on RD Sharma Class 8 Solutions Chapter 9 - Linear Equation In One Variable (Ex 9.4) Exercise 9.4

**1.What are some of the topics discussed in this chapter?**

**The following are some of the major issues or concepts covered in this chapter:**

An algebraic equation is a variable-based equivalence. The value of the expression on one side of the equality sign is equal to the value of the expression on the other side of the equality sign.

Every rational integer can be used as the answer to a linear equation.

Both sides of an equation can have linear expressions.

Variables, like numbers, can be moved from one side of an equation to the other.

Occasionally, the expressions that makeup equations must be simplified before they can be solved using standard procedures. Certain equations may not be linear to start with, however, by multiplying both sides of the equation by an appropriate expression, they can be made linear.

The versatility of linear equations can be seen in their many applications, which include problems involving numbers, ages, and perimeters. To learn more, click here.

**2. What will the students get to learn through this chapter?**

Solutions for Class 8 Maths by RD Sharma six problems in Chapter 9 deal with linear equations in one variable.

**The following are some of the important themes presented in this chapter:**

9.1 The Beginning

9.2 Solving Equations with Linear Expressions on One Side and Numerical Expressions on the Other 9.3 Some Examples

9.4 Solving Equations With Both Sides of the Variable

9.5 Additional Applications

9.6 Taking Equations and Reducing Them to a Simpler Form

9.7 Equations That Can Be Reduced to Linear Form You will be able to answer all of the questions about linear equations in exams if you practise all of these activities.

3. Solve this: In Mahuli village, there is a short rectangular land set aside for a school. The plot's length and width are 11:4 in proportion. The village panchayat will spend ﹩75,000 to fence the area at a rate of 100 ﹩ per metre. What are the plot's dimensions?

Let the rectangle plot's length be 11x and its width is 4x.

Fencing per metre rate = 100

The total cost of the fencing is $75,000.

2(l+b) = 2(11x + 4x) = 215x = 30x Perimeter of the plot = 2(l+b) = 2(11x + 4x) = 215x = 30x

30 x 100 = total amount of fencing,

In response to the question,

75000 = (30 x 100)

75000 = 3000x

x = 75000/3000 x = 75000/3000 x = 75000/3000 x = 7

x = 25 x = 25 x = 25 x = 25 x =

The plot's length is 11x = 11 25 = 275m.

The plot's width is 4 25 = 100 metres.