Exercise 2.5
Q 1 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}60\bigg(\dfrac{x}{2}-\dfrac{1}{5}\bigg)&=60\bigg(\dfrac{x}{3}+\dfrac{1}{4}\bigg)\\30x-12&=20x+15\\30x-20x&=15+12\\10x&=27\\x&=\dfrac{27}{10}\end{align*} \)
Q 2 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}7n&=252\\n&=\dfrac{252}{7}\\n&=36\end{align*} \)
Q 3 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}5x&=-25\\x&=\dfrac{-25}{~~5}\\x&=-5\end{align*} \)
Q 4 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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)
\( \begin{align*}5x-25&=3x-9\\5x-3x&=25-9\\2x&=16\\x&=\dfrac{16}{2}\\x&=8\end{align*} \)
Q 5 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}9t-6-8t-12&=8-12t\\9t-6-8t-12&=8-12t\\3t&=26\\t&=\dfrac{26}{13}\\x&=2\end{align*} \)
Q 6 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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)
\( \begin{align*}6m-3m+3&=6-2m+4\\6m-3m+3&=6-2m+4\\6m-3m+2m&=6+4-3\\5m&=7\\m&=\dfrac{7}{5}\end{align*} \)
Q 7 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
Question 7 :
Solve the linear equation
3(t - 3) = 5(2t + 1)
Sol :
3(t - 3) = 5(2t + 1)
3t - 9 = 10t + 5
\( \begin{align*}-9-5&=10t-3t\\-14&=7t\\t&=\dfrac{-14}{7}\\t&=-2\end{align*} \)
Q 8 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}&15y-60-2y+18+5y+30=0\\&18y-12=0\\&18y=12\\&y=\dfrac{12}{18}=\dfrac{2}{3}\end{align*} \)
Q 9 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
\( \begin{align*}&15z-21-18z+22=32z-52-17\\&-3z+1=32z-69\\&-3z-32z=-69-1\\&-35z=-70\\&z=\dfrac{70}{35}=2\end{align*} \)
Q 10 - Ex 2.5 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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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