NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.5

1. Find the solution of the linear equation\[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

Ans: We have an equation \[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

The L.C.M of denominators \[\text{2,3,4}\] and \[\text{5}\] is \[\text{60}\]

Multiplying both the sides by \[\text{60}\],

\[\Rightarrow \text{60}\left( \frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}} \right)\text{=60}\left( \frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}} \right)\]

\[\Rightarrow \text{60}\left( \frac{\text{5x-2}}{\text{10}} \right)\text{=60}\left( \frac{\text{4x+3}}{\text{12}} \right)\]

\[\Rightarrow \text{6}\left( \text{5x-2} \right)\text{=5}\left( \text{4x+3} \right)\]

\[\Rightarrow \text{30x-12=20x+15}\]

Shifting \[\text{20x}\] to left hand side and \[\text{12}\] on right hand side

\[\Rightarrow \text{30x-20x=15+12}\]

\[\Rightarrow \text{10x=27}\]

\[\Rightarrow \text{x=}\frac{\text{27}}{\text{10}}\]

2. Find the solution of the linear equation\[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

Ans: We have an equation \[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

The L.C.M of denominators \[\text{2,4}\] and \[\text{6}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{6n-9n+10n}}{\text{12}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{7n=252}\]

Dividing the equation by \[\text{7}\]

\[\Rightarrow \text{n=36}\]

3. Find the solution of the linear equation\[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

Ans: We have an equation \[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

The L.C.M of denominators \[\text{2,3}\] and \[\text{6}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \text{x+7-}\frac{\text{8x}}{\text{3}} \right)\text{=6}\left( \frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}} \right)\]

\[\Rightarrow \text{6}\left( \frac{\text{6x+42-16x}}{\text{6}} \right)\text{=6}\left( \frac{\text{17-15x}}{\text{6}} \right)\]

\[\Rightarrow \text{6x+42-16x=17-15x}\]

Shifting \[\text{15x}\] to left hand side and \[\text{42}\] to right hand side

\[\Rightarrow \text{-10x+15x=17-42}\]

\[\Rightarrow \text{5x=-25}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{x=-5}\]

4. Find the solution of the linear equation\[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

Ans: We have an equation \[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{5}\] is \[\text{15}\]

Multiplying both the sides by \[\text{15}\],

\[\Rightarrow \text{15}\left( \frac{\text{x-5}}{\text{3}} \right)\text{=15}\left( \frac{\text{x-3}}{\text{5}} \right)\]

\[\Rightarrow \text{5}\left( \text{x-5} \right)\text{=3}\left( \text{x-3} \right)\]

\[\Rightarrow \text{5x-25=3x-9}\]

Shifting \[\text{3x}\] to left hand side and \[\text{25}\] to right hand side

\[\Rightarrow \text{5x-3x=-9+25}\]

\[\Rightarrow \text{2x=16}\]

Dividing both the sides by \[\text{2}\]

\[\Rightarrow \text{x=8}\]

5. Find the solution of the linear equation\[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

Ans: We have an equation \[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

The L.C.M of denominators \[\text{3}\] and \[\text{4}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}} \right)\text{=12}\left( \frac{\text{2}}{\text{3}}\text{-t} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{9t-6-8t-12}}{\text{12}} \right)\text{=12}\left( \frac{\text{8-12t}}{\text{12}} \right)\]

\[\Rightarrow \text{t-18=8-12t}\]

Shifting \[\text{12t}\] to left hand side and \[\text{18}\] to right hand side

\[\Rightarrow \text{t+12t=8+18}\]

\[\Rightarrow \text{13t=26}\]

Dividing both the sides by \[\text{13}\]

\[\Rightarrow \text{t=2}\]

6. Find the solution of the linear equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

Ans: We have an equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{2}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \frac{\text{2m-m+1}}{\text{2}} \right)\text{=6}\left( \frac{\text{3-m+2}}{\text{3}} \right)\]

\[\Rightarrow \text{3}\left( \text{m+1} \right)\text{=3}\left( \text{-m+5} \right)\]

\[\Rightarrow \text{3m+3=10-2m}\]

Shifting \[\text{2m}\] to left hand side and \[\text{3}\] to right hand side

\[\Rightarrow \text{3m+2m=10-3}\]

\[\Rightarrow \text{5m=7}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{m=}\frac{\text{7}}{\text{5}}\]

7. Find the solution of the linear equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

Ans: We have an equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

\[\Rightarrow \text{3t-9=10t+5}\]

Shifting \[\text{5}\] to left hand side and \[\text{3t}\] to right hand side

\[\Rightarrow \text{-9-5=10t-3t}\]

\[\Rightarrow \text{-14=7t}\]

Dividing both the sides by \[\text{7}\]

\[\Rightarrow \text{-2=t}\]

8. Find the solution of the linear equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

Ans: We have an equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

\[\Rightarrow \text{15y-60-2y+18+5y+30=0}\]

\[\Rightarrow \text{18y-12=0}\]

Shifting \[\text{12}\] to right hand side

\[\Rightarrow \text{18y=12}\]

Dividing both the sides by \[\text{18}\]

\[\Rightarrow \text{y=}\frac{\text{12}}{\text{18}}\]

\[\Rightarrow \text{y=}\frac{\text{2}}{\text{3}}\]

9. Find the solution of the linear equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

Ans: We have an equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

\[\Rightarrow \text{15z-21-18z+22=32z-52-17}\]

\[\Rightarrow \text{-3z+1=32z-69}\]

Shifting \[\text{69}\] to left hand side and \[\text{3z}\] to right hand side

\[\Rightarrow \text{1+69=32z+3z}\]

\[\Rightarrow \text{70=35z}\]

Dividing both the sides by \[\text{35}\]

\[\Rightarrow \text{z=2}\]

10. Find the solution of the linear equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

Ans: We have an equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

\[\Rightarrow \text{f-0}\text{.75=0}\text{.5f-0}\text{.45}\]

Shifting \[\text{0}\text{.5f}\] to left hand side and \[\text{0}\text{.75}\] to right hand side

\[\Rightarrow \text{0}\text{.5f=-0}\text{.45+0}\text{.75}\]

\[\Rightarrow \text{0}\text{.5f=0}\text{.3}\]

Dividing both the sides by \[\text{0}\text{.5}\]

\[\Rightarrow \text{f=0}\text{.6}\]

Why is the NCERT Class 8 Maths Chapter 2 on Linear Equation Important?

Students start learning the basic concepts of higher Mathematics 8th standard onwards. The chapter of Linear Equation is very crucial for the understanding of almost everything in Mathematics in the later classes. Linear Equation is the basis of solving almost all problems in Maths. From Algebra, Calculus, to Geometry and Arithmetic, the knowledge about Linear Equation plays a paramount role. You can study NCERT Solution for Class 8 Maths Chapter 2 Exercise 2.5 from Vedantu’s notes which are available as free PDF.

If you know how to solve a linear Equation, then only you can solve more complicated problems in the future. The Linear Equations in One Variable Class 8 chapter is immensely interesting but requires practice, thanks to the nature of the subject, Maths. Scoring high in Maths require daily practice and indulgence. Vedantu’s notes make studying Maths interesting and easy with its notes.

The notes contain solved problems and their explanations, question paper sets and everything that you need to know about the topic. You can now make your Maths preparation fun by downloading the NCERT Solutions for Class 8 Maths Chapter 2 PDF.

What Will You Learn in Cbse Class 8 Maths Chapter 2 Linear Equation 2.5?

The Chapter, ‘Linear Equations’ teaches you how to find out an unknown variable by using one or more real numbers. In this chapter, you will learn to solve a few complicated problems of Linear Equation. If you study this chapter with the help of Vedantu’s notes, it will become quite easy to get hold of the later chapters of Maths. Solving any problem to get the value of an unknown variable will be the easiest task for you. Don’t miss the PDF that’s available to you for free. Download the PDF on Linear Equation now.

Why Vedantu and How Can its Ncert Solutions for Class 8 Maths Chapter 2 Exercise 2.5 Note Help You?

The Chapter on Linear Equation in One Variable contains all the important questions and solutions that the chapter deals with. Don’t miss out on anything related to Linear Equation. You need to be well versed with all the concepts before your exams, including the marks allocation and pattern of the questions.

When it comes to your search for the notes that you actually require, Vedantu’s notes are the best. The notes simplify the most difficult concepts for a better understanding. The notes have been crafted for your benefit. The solutions that are provided are very easy to understand. You don’t need to go through the solution, again and again, to understand how the problem must be smartly solved. Vedantu knows what students require and thus provides you with the same. The PDF for Class 8 Maths Chapter 2 is available for free. Anybody can download it and study these for their examination. Refer to Vedantu and beat the exam-stress.