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 2.5 

Question 1 :

Solve the linear equation 

\( \dfrac{x}{2}-\dfrac{1}{5}=\dfrac{x}{3}+\dfrac{1}{4} \)

Sol :

\( \dfrac{x}{2}-\dfrac{1}{5}=\dfrac{x}{3}+\dfrac{1}{4} \)

L.C.M of the denominators 2, 3, 4, and 5 is 60

Multiplying both sides by 60 , we obtain

\( \begin{align*}60\bigg(\dfrac{x}{2}-\dfrac{1}{5}\bigg)&=60\bigg(\dfrac{x}{3}+\dfrac{1}{4}\bigg)\\30x-12&=20x+15\\30x-20x&=15+12\\10x&=27\\x&=\dfrac{27}{10}\end{align*} \)

 

Question 2 :

Solve the linear equation 

\( \dfrac{n}{2}-\dfrac{3n}{4}+\dfrac{5n}{6}=21 \)

Sol :

\( \dfrac{n}{2}-\dfrac{3n}{4}+\dfrac{5n}{6}=21 \)

L.C.M of the denominators , 2, 4, and 6, is 12

Multiplying both sides by 12 , we obtain

6n – 9n + 10n = 252

\( \begin{align*}7n&=252\\n&=\dfrac{252}{7}\\n&=36\end{align*} \)

 

Question 3 :

Solve the linear equation 

\( x+7-\dfrac{8x}{3}=\dfrac{17}{6}-\dfrac{5x}{2} \)

Sol :

\( x+7-\dfrac{8x}{3}=\dfrac{17}{6}-\dfrac{5x}{2} \)

L.C.M of the denominators , 2, 3, and 6, is 6

Multiplying both sides by 6 , we obtain

6x + 42 – 16x = 17 – 15x

6x – 16x + 15x = 17 – 42

\( \begin{align*}5x&=-25\\x&=\dfrac{-25}{~~5}\\x&=-5\end{align*} \)

 

Question 4 :

Solve the linear equation 

\( \dfrac{x-5}{3}=\dfrac{x-3}{5} \)

Sol :

\( \dfrac{x-5}{3}=\dfrac{x-3}{5} \)

L.C.M of the denominators 3,5, and is 15

Multiplying both sides by 15 , we obtain

5(x – 5) = 3(x – 3)

\( \begin{align*}5x-25&=3x-9\\5x-3x&=25-9\\2x&=16\\x&=\dfrac{16}{2}\\x&=8\end{align*} \)

 

Question 5 :

Solve the linear equation 

\( \dfrac{3t-2}{4}-\dfrac{2t+3}{3}=\dfrac{2}{3}-t \)

Sol :

\( \dfrac{3t-2}{4}-\dfrac{2t+3}{3}=\dfrac{2}{3}-t \)

L.C.M of the denominators 3,4, and is 12

Multiplying both sides by 12 , we obtain

3(3t – 2) – 4(2t + 3) = 8 – 12t

\( \begin{align*}9t-6-8t-12&=8-12t\\9t-6-8t-12&=8-12t\\3t&=26\\t&=\dfrac{26}{13}\\x&=2\end{align*} \)

 

Question 6 :

Solve the linear equation 

\( m-\dfrac{m-1}{2}=1-\dfrac{m-2}{3} \)

Sol :

\( m-\dfrac{m-1}{2}=1-\dfrac{m-2}{3} \)

L.C.M of the denominators 2, 3 and is 6

Multiplying both sides by 6 , we obtain

6m – 3(m – 1) = 6 – 2(m – 2)

\( \begin{align*}6m-3m+3&=6-2m+4\\6m-3m+3&=6-2m+4\\6m-3m+2m&=6+4-3\\5m&=7\\m&=\dfrac{7}{5}\end{align*} \)

 

Question 7 :

Solve the linear equation

3(t – 3) = 5(2t + 1)

Sol :

3(t – 3) = 5(2t + 1)

3t – 9 = 10t + 5

\( \begin{align*}-9-5&=10t-3t\\-14&=7t\\t&=\dfrac{-14}{7}\\t&=-2\end{align*} \)

 

Question 8 :

Solve the linear equation

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Sol :

15(y – 4) – 2(y – 9) + 5(y + 6) = 0

\( \begin{align*}&15y-60-2y+18+5y+30=0\\&18y-12=0\\&18y=12\\&y=\dfrac{12}{18}=\dfrac{2}{3}\end{align*} \)

 

Question 9 :

Solve the linear equation

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Sol :

3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

\( \begin{align*}&15z-21-18z+22=32z-52-17\\&-3z+1=32z-69\\&-3z-32z=-69-1\\&-35z=-70\\&z=\dfrac{70}{35}=2\end{align*} \)

 

Question 10 :

Simplify and solve the linear equation :

0.25(4f – 3) = 0.05(10f – 9)

Sol :

0.25(4f – 3) = 0.05(10f – 9)

\( \dfrac{1}{4}(4f-3)=\dfrac{1}{20}(10f-9) \)

Multiplying both sides by 20 , we obtain

\( \begin{align*}&5(4f-3)=10f-9\\&20f-15=10f-9\\&20f-10f=-9+15\\&10f=6\\&f=\dfrac{3}{5}=0.6\end{align*} \)

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