Class 7 RS Aggarwal Chapter-7 Linear Equations in One Variable Solutions - Free PDF Download

Linear Equations

A linear equation is one in which the variable's highest power is always 1. It is also referred to as a one-degree equation. A linear equation in one variable has the standard form Ax + B = 0. In this case, x is a variable, while A and B are constants.

What Exactly is a Linear Equation?

A linear equation is an equation with the highest degree of one. This means that in a linear equation, no variable has an exponent greater than 1. 

The Formula for Linear Equation

A linear equation is expressed using the Linear Equation Formula. A linear equation, for example, can be expressed in the standard form, the slope-intercept form, or the point-slope form. Now, let us look at the standard form of a linear equation and see how it is expressed. We can see that it varies depending on the number of variables, and it is important to remember that the highest (and only) degree of all variables in the equation should be 1.

Linear Equations: One Variable

A linear equation in one variable is an equation with a single variable. It has the form Ax + B = 0, where A and B are any two integers and x is an unknown variable with a single solution. It is the most straightforward way to represent a mathematical statement. The degree of this equation is always equal to one. A linear equation with one variable is very simple to solve. To obtain the value of the unknown variable, the variables are separated and brought to one side of the equation, while the constants are combined and brought to the other side of the equation.

Solve the Following Linear Equation in One Variable: 4x + 8 = 20

To solve the given equation, we move the numbers to the right side of the equation while leaving the variable on the left. That is, 4x = 20 - 8. So, when we solve for the value of x, we get 4x = 20 - 8. Finally, the value of x = 12/4 = 3 is found.

Linear Equations: How Do You Solve Them?

A weighing scale with equal weights on both sides is equivalent to an equation. It still holds true if we add or subtract the same number from both sides of an equation. Similarly, it is correct to multiply or divide the same number into both sides of an equation. We move the variables to one side of the equation and the constant to the other, and then we calculate the value of the unknown variable. This is how you solve a linear equation with a single variable. 

Solve the equation 5x - 4 = 11 as an example.

To maintain the equilibrium, we do mathematical operations on the Left-hand Side (LHS) and Right-hand Side (RHS). So, let us add four on both sides to reduce the LHS to three times. 5x - 4 + 4 = 5x and 11 + 4 = 15 are the new LHS and RHS, respectively. Let's divide both sides by 5 to get the LHS down to x. As a result, x = 3. This is one method for solving linear equations with a single variable.

RS Aggarwal Solutions for Class 7 Math Chapter 7

Linear Equations in One Variable Class 7

Any equation is written in the form ax + b = 0 here, ‘a’ and ‘b’ are real numbers, and ‘x’ is a variable, called Linear Equation in one Variable. These equations have the variable of order one only.

For example, 5x + 13 = 16 is a linear equation with a single variable. Such an equation possesses only one solution.

For the above example, the solution is x = 3/5.

Linear Equations in One Variable: RS Aggarwal Solutions for Class 7, Chapter 7

In standard form Linear Equation in One Variable is:

ax + b = 0

Here, ‘a' and ‘b' both are real numbers that do not equal zero.

Steps to follow while solving Linear Equations in One Variable

Follow the below-given steps while solving such equations:

• Step 1: If the given equation is in fraction form, first convert it into simple form.

• Step 2: Try to convert the equation in the most simple form.

• Step 3: Now, bring the variable on any side of the equation.

• Step 4: Solve the equation to get the answer.

Solved Examples of Linear Equations in One Variable Class 7 RS Aggarwal

Let us take the help of a few examples to understand the concept in a more abstract way. We will go in the same stepwise way as mentioned above.

Example 1: Solve the equation 6x - 8 = 5x + 16.

Step 1: As the equation is already in simplified form with no fraction, we can directly jump to step number third.

Step 2: Now bring all the variable containing terms on one side of the equation, i.e.,

6x - 5x - 8 = 16

Step 3: Bring the constants on the other side of the equation, i.e.,

6x - 5x = 16 + 8

Step 4: Now solve both the sides to get the required answer.

x = 24

We have a trick to verify the answer by ourselves. For this, put the value x=24 in the given equation and if you get some value on both sides then your answer is absolutely correct, i.e.,

6(24) - 8 = 5(24) + 16

144 - 8 = 120 + 16

136 = 136

So, our answer is right.

Example 2: Solve the equation \[\frac{6n}{2}+\frac{8n}{4}=n+3\]

Solution: Refer to the steps:

Step 1: The equation is in fraction form. First, convert it into a simple form

\[\frac{6n}{2}+\frac{8n}{4}=n+3\]

3n + 2n = n + 3

Step 2: Now gather the variables on one side and constants on another side of the equation, i.e.,

3n + 2n - n = 3

Step 3: Now solve the equation

4n = 3

n = \[\frac{3}{4}\]

You do not have to elaborate each step once you are familiar with the drill.

Class 7 Math RS Aggarwal Solutions Chapter 7: Exercises

Let us now go through the contents of RS Aggarwal for class 7, chapter 7 solutions. It consists of the following three exercises in all:

• Exercise  (Ex 7A) 7.1

• Exercise  (Ex 7B) 7.2

• Exercise (Ex 7C) 7.3

The first exercise, 7A, explains the basic concepts of the linear Equations in one variable. In this exercise, the idea is given with simple and basic examples. Through these examples, students will learn to distinguish these types of equations from other high-ordered equations. It consists of 32 questions in all.

Second exercise 7B, it has questions a bit more tedious as compared to exercise 7A. This exercise introduces you to a fraction containing problems with easy solutions. It has 35 questions in total.

Third exercise 7C, it has questions in more complex form when compared with earlier exercise. There are 18 questions in this exercise. You can also refer to test papers on page number 118 when done with all three exercises.

Linear Equations in One Variable can quickly be learned with RS Aggarwal solutions for class 7 math, chapter 7 given by the author. Exercises in this book are compiled in such a way that it prepares the child for tedious and complex questions. The complete chapter is available in a stepwise way, where the difficulty of questions grows with each exercise.

First it introduces the child to the Linear Equations with simple definitions and problems. Then it slowly introduces them to difficult problems that will boost their problem-solving techniques. Test Papers help the students to increase their accuracy in problem solving.

RS Aggarwal for Class 7, Solutions for Chapter 7: Preparation Tips

Students can remove their exam phobia with the solved questions and examples given in the R S Aggarwal for Class 7, Solutions for Chapter 7. You should attempt these questions with time restrictions that will train your mind for the exam. R S Aggarwal for Class 7, Solutions for Chapter 7 are easy to understand without anybody’s aid that saves your time and money. Many questions come from this topic in various national-level exams and aptitude tests, which can easily be understood with the help of this study material.

Conclusion 

Vedantu provides Solutions to all the questions based on this chapter from RS Aggarwal Mathematics. You can find the solutions of all the chapters here. In this attached PDF which is also downloadable for free, students are advised to go through all the questions carefully and understand the concepts used to solve these questions. This will help the students immensely in their examinations.