NCERT Solutions for class 8 Maths chapter 2 Linear Equation In One Variable

Exercise 2.1  Exercise 2.2  Exercise 2.3  Exercise 2.4  Exercise 2.5  Exercise 2.6

EXERCISE 2.3

Question 1 :

Solve and check result : 3x = 2x + 18

Sol :

3x = 2x + 18

On transposing 2x to L.H.S , we obtain

3x – 2x = 18

x = 18

L.H.S ​\( \begin{align*}&=3x\\&=3\times{}18\\&=54\end{align*} \)

R.H.S ​\( \begin{align*}&=2x+18\\&=2\times{18}+18\\&=36+18\\&=54\end{align*} \)

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 2 :

Solve and check result : 

5t – 3 = 3t – 5

Sol :

5t – 3 = 3t – 5

On transposing 3t to L.H.S and -3 to R.H.S , we obtain

5t – 3t = -5 – (-3)

2t = -2

On dividing both sides by 2 , we obtain

t = -1

L.H.S ​\( \begin{align*}\end{align*} \)

R.H.S ​\( \begin{align*}\end{align*} \)

L.H.S = R.H.S

Question 3 :

Solve and check result :

5x +9 = 5 + 3x

Sol :

On transposing 3x to L.H.S and 9 to R.H.S , we obtain 

5x – 3x = 5 – 9

2x = -4

On dividing both sides by 2 , we obtain

x = -2

L.H.S  ​\( \begin{align*}&=5x+9\\&=5\times(-2)+9\\&=-10+9\\&=-1\end{align*} \)

R.H.S  ​\( \begin{align*}&=5+3\times(-2)\\&=5-6\\&=-1\end{align*} \)

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 4 :

Solve and check result :

4z + 3 = 6 +2z

Sol :

4z + 3 = 6 +2z

On transposing 2z to L.H.S and 3 to R.H.S , we obtain 

4z – 2z = 6 – 3

2z = 3

Dividing both sides by 2 , we obtain

\( z=\dfrac{3}{2} \)

 

L.H.S  ​\( \begin{align*}&=4z+3\\&=4\times{\dfrac{3}{2}}+3\\&=6+3\\&=9\end{align*} \)

 

R.H.S  ​\( \begin{align*}&=6+2z\\&=6+2\times\dfrac{3}{2}\\&=6+3\\&=9\end{align*} \)

 

L.H.S = R.H.S

Hence, the result obtained above is correct

Question 5 :

Solve and check result :

2x – 1 = 14 – x

Sol :

2x – 1 = 14 – x

Transposing x to L.H.S and 1 to R.H.S , we obtain

2x + x = 14 + 1

3x = 15

Dividing both sides by 3 , we obtain

x = 5

L.H.S  ​\( \begin{align*}&=2x-1\\&=2\times(5)-1\\&=10-1\\&=9\end{align*} \)

 

R.H.S  ​\( \begin{align*}&=14-x\\&=14-5\\&=9\end{align*} \)

 

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 6 :

Solve and check result :

8x + 4 = 3(x – 1)+ 7

Sol :

8x + 4 = 3(x – 1)+ 7

8x + 4 = 3x – 3 + 7

On transposing 3x to L.H.S and 4 to R.H.S , we obtain

8x – 3x = – 3 + 7 – 4

5x = – 7 + 7

x = 0

L.H.S  ​\( \begin{align*}&=8x+4\\&=8\times(0)+4\\&=4\end{align*} \)

R.H.S  ​\( \begin{align*}&=3(x-1)+7\\&=3(0-1)+7\\&=-3+7\\&=4\end{align*} \)

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 7 :

Solve and check result :

\( x=\dfrac{4}{5}(x+10) \)

Sol :

\( x=\dfrac{4}{5}(x+10) \)

Multiply both sides by 5 , we obtain 

5x = 4(x + 10)

5x = 4x + 40

On transposing 4x to L.H.S , we obtain

5x – 4x = 40

x = 40 

L.H.S = x = 40

R.H.S  ​\( \begin{align*}&=\dfrac{4}{5}(x+10)\\&=\dfrac{4}{5}(40+10)\\&=\dfrac{4}{5}\times{50}\\&=40\end{align*} \)

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 8 :

Solve and check result :

\( \dfrac{2x}{3}+1=\dfrac{7x}{15}+3 \)

Sol :

\( \dfrac{2x}{3}+1=\dfrac{7x}{15}+3 \)

On transposing ​\( \dfrac{7x}{15} \)​ to L.H.S and 1 to R.H.S , we obtain

\( \dfrac{2x}{3}-\dfrac{7x}{15}=3-1 \)

\( \dfrac{5\times2x-7x}{15}=2 \)

\( \dfrac{3x}{15}=2 \)

Multiplying both sides by 5 , we obtain

x = 10

L.H.S  ​\( \begin{align*}&=\dfrac{2x}{3}+1\\&=\dfrac{2\times{10}}{3}+1\\&=\dfrac{2\times{10}+1\times3}{3}\\&=\dfrac{23}{3}\end{align*} \)

 

R.H.S  ​\( \begin{align*}&=\dfrac{7x}{15}+3\\&=\dfrac{7\times10}{15}+3\\&=\dfrac{7\times2}{3}+3\\&=\dfrac{14}{3}+3\\&=\dfrac{14+3\times3}{3}\\&=\dfrac{23}{3}\end{align*} \)

 

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 9 :

Solve and check result :

\( 2y+\dfrac{5}{3}=\dfrac{26}{3}-y \)

Sol :

\( 2y+\dfrac{5}{3}=\dfrac{26}{3}-y \)

On transposing y to L.H.S and ​\( \dfrac{5}{3} \)​ to R.H.S , we obtain 

\( 2y+y=\dfrac{26}{3}-\dfrac{5}{3} \)

\( 3y=\dfrac{21}{3}=7 \)

Dividing both sides by 3 , we obtain

\( y=\dfrac{7}{3} \)

 

L.H.S  ​\( \begin{align*}&=2y+\dfrac{5}{3}\\&=2\times\dfrac{7}{3}+\dfrac{5}{3}\\&=\dfrac{14}{3}+\dfrac{5}{3}\\&=\dfrac{19}{3}\end{align*} \)

 

R.H.S  ​\( \begin{align*}&=\dfrac{26}{3}-y\\&=\dfrac{26}{3}-\dfrac{7}{3}\\&=\dfrac{19}{3}\end{align*} \)

L.H.S = R.H.S

Hence, the result obtained above is correct

 

Question 10 :

Solve and check result :

\( 3m=5m-\dfrac{8}{5} \)

Sol :

\( 3m=5m-\dfrac{8}{5} \)

On transposing 5m to L.H.S , we obtain

\( 3m-5m=-\dfrac{8}{5} \)

\( -2m=-\dfrac{8}{5} \)

Dividing both sides by -2 , we obtain

\( m =\dfrac{4}{5} \)

L.H.S  ​\( \begin{align*}&=3m\\&=3\times\dfrac{4}{5}\\&=\dfrac{12}{5}\end{align*} \)

 

R.H.S  ​\( \begin{align*}&=5m-\dfrac{8}{5}\\&=5\times\dfrac{4}{5}-\dfrac{8}{5}\\&=\dfrac{12}{5}\end{align*} \)

 

L.H.S = R.H.S

Hence, the result obtained above is correct

 

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