NCERT Solutions for Class 8 Maths Chapter 2: Linear Equations in One Variable - Exercise 2.3

1. Solve and Verify the Equation \[\text{3x=2x+18}\].

Ans: We have an equation \[\text{3x=2x+18}\]

To solve the equation, we will shift \[\text{2x}\] to left hand side 

\[\text{3x-2x=18}\]

\[\text{x=18}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{3x=3 }\!\!\times\!\!\text{ 18}\]

\[\text{54}\]

R.H.S

\[\text{2x+18=2 }\!\!\times\!\!\text{ 18+18}\]

\[\text{54}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

2. Solve and Verify the Equation \[\text{5t-3=3t-5}\].

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

To solve the equation, we will shift \[\text{3t}\] to left hand side and \[\text{3}\] to the right hand side, now,

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

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

Dividing the equation by \[\text{2}\]

\[\text{t=-1}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{5t-3=}\left( \text{5 }\!\!\times\!\!\text{ -1} \right)\text{-3}\]

\[\text{-5-3=-8}\]

R.H.S

\[\text{3t-5=}\left( \text{3 }\!\!\times\!\!\text{ -1} \right)\text{-5}\]

\[\text{-3-5=-8}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

3. Solve and Verify the Equation \[\text{5x+9=5+3x}\].

Ans: We have an equation \[\text{5x+9=5+3x}\]

To solve the equation, we will shift \[\text{3x}\] to left hand side and \[\text{9}\] to the right hand side, now,

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

\[\text{2x=-4}\]

Dividing the equation by \[\text{2}\]

\[\text{x=-2}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{5x+9=}\left( \text{5 }\!\!\times\!\!\text{ -2} \right)\text{+9}\]

\[\text{-10+9=-1}\]

R.H.S

\[\text{5+3x=5+}\left( \text{3 }\!\!\times\!\!\text{ -2} \right)\]

\[\text{5-6=-1}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

4. Solve and Verify the Equation \[\text{4z+3=6+2z}\].

Ans: We have an equation \[\text{4z+3=6+2z}\]

To solve the equation, we will shift \[\text{2z}\] to left hand side and \[\text{3}\] to right hand side, now,

\[\text{4z-2z=6-3}\]

\[\text{2z=3}\]

Dividing the equation by \[\text{2}\]

\[\text{z=}\frac{\text{3}}{\text{2}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{4z+3=}\left( \text{4 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)\text{+3}\]

\[\text{6+3=9}\]

R.H.S

\[\text{6+2z=6+}\left( \text{2 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)\]

\[\text{6+3=9}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

5. Solve and Verify the Equation \[\text{2x-1=14-x}\].

Ans: We have an equation \[\text{2x-1=14-x}\]

To solve the equation, we will shift \[\text{x}\] to left hand side and \[\text{1}\] to the right hand side, now,

\[\text{2x+x=14+1}\]

\[\text{3x=15}\]

Dividing the equation by \[\text{3}\]

\[\text{x=5}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{2x-1=}\left( \text{2 }\!\!\times\!\!\text{ 5} \right)\text{-1}\]

\[\text{10-1=9}\]

R.H.S

\[\text{14-x=14-5}\]

\[\text{14-5=9}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

6. Solve and Verify the Equation \[\text{8x+4=3}\left( \text{x-1} \right)\text{+7}\].

Ans: We have an equation \[\text{8x+4=3}\left( \text{x-1} \right)\text{+7}\]

\[\text{8x+4=3x-3+7}\]

\[\text{8x+4=3x+4}\]

To solve the equation, we will shift \[\text{3x}\] to left hand side and \[4\] to the right hand side, now,

\[\text{8x-3x=4-4}\]

\[\text{5x=0}\]

\[\text{x=0}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{8x+4=}\left( \text{8 }\!\!\times\!\!\text{ 0} \right)\text{+4}\]

\[\text{0+4=4}\]

R.H.S

\[\text{3}\left( \text{x-1} \right)\text{+7=3}\left( \text{0-1} \right)\text{+7}\]

\[\text{-3+7=4}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

7. Solve and Verify the Equation \[\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\].

Answer: We have an equation \[\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\]

\[\text{x=}\frac{\text{4x+40}}{\text{5}}\]

Multiplying the equation by \[\text{5}\]

\[\text{5x=4x+40}\]

To solve the equation, we will shift \[\text{4x}\] to left hand side, now,

\[\text{5x-4x=40}\]

\[\text{x=40}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{x=40}\]

R.H.S

\[\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\text{=}\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)\]

\[\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)\text{=}\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}\]

\[\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}\text{=40}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

8. Solve and Verify the Equation \[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}\].

Ans: We have an equation \[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}\]

To solve the equation, we will shift \[\frac{\text{7x}}{\text{15}}\] to left hand side and \[\text{1}\] to right hand side, now,

\[\frac{\text{2x}}{\text{3}}\text{-}\frac{\text{7x}}{\text{15}}\text{=3-1}\]

\[\frac{\text{10x-7x}}{\text{15}}\text{=2}\]

Multiplying \[15\] to the equation, we get

\[\text{3x=30}\]

\[\text{x=10}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}\]

\[\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}\text{=}\frac{\text{23}}{\text{3}}\]

R.H.S

\[\frac{\text{7x}}{\text{15}}\text{+3=}\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}\]

\[\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}\text{=}\frac{\text{115}}{\text{15}}\]

\[\frac{\text{115}}{\text{15}}\text{=}\frac{\text{23}}{\text{3}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

9. Solve and Verify the Equation \[\text{2y+}\frac{\text{5}}{\text{3}}\text{+}\frac{\text{26}}{\text{3}}\text{-y}\].

Ans: We have an equation \[\text{2y+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{26}}{\text{3}}\text{-y}\]

To solve the equation, we will shift \[y\] to left hand side and \[\frac{\text{5}}{\text{3}}\] to right hand side, now,

\[\text{2y+y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{5}}{\text{3}}\]

\[\text{3y=}\frac{\text{21}}{\text{3}}\]

\[\text{3y=7}\]

\[\text{y=}\frac{\text{7}}{\text{3}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{2y+}\frac{\text{5}}{\text{3}}\text{=2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}\]

\[\text{2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{14+5}}{\text{3}}\]

\[\frac{\text{14+5}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}\]

R.H.S

\[\frac{\text{26}}{\text{3}}\text{-y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}\]

\[\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

10. Solve and Verify the Equation \[\text{3m=5m-}\frac{\text{8}}{\text{5}}\].

Ans: We have an equation \[\text{3m=5m-}\frac{\text{8}}{\text{5}}\]

To solve the equation, we will shift \[\text{5m}\] to left hand side, now,

\[\text{3m-5m=-}\frac{\text{8}}{\text{5}}\]

\[\text{-2m=-}\frac{\text{8}}{\text{5}}\]

Dividing the equation by \[\text{-2}\], we get,

\[\text{m=}\frac{4}{\text{5}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{3m=3 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{5}}\]

\[\frac{\text{3 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{=}\frac{\text{12}}{\text{5}}\]

R.H.S

\[\text{5m-}\frac{\text{8}}{\text{5}}\text{=}\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}\]

\[\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}=\frac{12}{\text{5}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (Ex 2.3) Exercise 2.3

Opting for the NCERT solutions for Ex 2.3 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 2.3 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 2 Exercise 2.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 2 Exercise 2.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 8 Maths Chapter 2 Exercise 2.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

NCERT Solutions Class 8 Maths Chapter 2 All Exercises