Linear Equations in One Variable

Linear Equation in One Variable Class 8

Linear Equations in One Variable Class 8 Practice Questions Linear Equations in One Variable

Linear Equation in One Variable Class 8

Linear Equations in One Variable Class 8 Practice Questions
A linear equation with single variable, where degree of equation is one is known as linear equation in one variable. The general form of linear equation in one variable is ax + b = 0, where a, b are real numbers and the value of a not equal to 0. x = -b/a is the unique solution for the equation. Solving linear equations means finding the solution of the equation by isolating the variable. The exemplar practice questions for class 8 Mathematics linear equations are very important for students in order to develop logical and mental ability. All questions are designed as per latest CBSE Syllabus.

Below you find a list of important questions with solutions for grade 8 math have been prepared by subject experts based on the latest CBSE guidelines. Students get step-by-step explanation to every question based on linear equations in one variable. These questions can also prove to be of valuable help to students in their assignments and preparation of boards and competitive exams. Let us solve linear equations questions for class 8:

Step 1:

4(2x + 2) - 3 = 5(x + 3)

4(2x) + 4(2) - 3 = 5(x) + 5(3)

8x + 8 - 3 = 5x + 15

8x + 8 - 3 = 5x + 15

8x + 5 = 5x + 15

8x - 5x = 15 - 5

3x = 10

Step 3: Isolate x by dividing each side by 3

3x/3 = 10/3

or x = 10/3

Therefore the value of linear equation is 10/3.

4(x + 3) - 4 = 8($\frac{x}{2}$ + 1)

4x + 12 = 4x + 8

Step 2: Combining the like terms such that keeping the terms containing variable on one side of the equation and the constants or numbers on the other side of the equation

4x - 4x = 8 - 12

Step 3: Solving and simplifying. We see that it is $4x$ minus $4x$ on the left side of the equation which cancels out. On cancellation we get $0$ the left side. On the right side it is 8 -12 = -4, resulting in no terms having variable $x$ that could be solved.

0 = -4 (Which is not true)

Step 4: As there is no solution obtained in the given pair of equation, we can conclude that the equations are inconsistent in form.

x + 5x + 4 = 3 (2x - 1)

Solution:

Step 1: Write down the equation given

x + 5x + 4 = 3 (2x - 1)

We see that there is existence of variables on both sides of the equal to. Same goes with the numbers as they are present on either side of the equal to.

Step 2: Using distributive property on right side of the equation.

x + 5x + 4 = 3 (2x) - 3(1)

x + 5x + 4 = 6x - 3

Step 3: We need to separate the terms containing variables on one side and the constants on the other side of the equal to. Combining the like terms shall give us

x + 5x - 6x = - 3 - 4

Step 4: Solving either side of the equation. So, it becomes

x - x = -7

0 = -7

Here the terms in $x$ cancels each other. As a result there are no terms in $x$ left in the equation to solve. Hence, no solution

Step 4: As there is no solution obtained in the given linear equation, we can conclude that the equations are inconsistent in form.

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