EXERCISE 2.3
Q 1 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 2 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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 =5t-3
=5(-1)-3
=-5-3=-8
R.H.S = 3t - 5
R.H.S = 3t - 5
=3(-1)-5
=-3-5=-8
L.H.S = R.H.S
Hence, the result obtained above is correct
Q 3 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 4 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 5 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 6 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 7 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 8 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 9 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 10 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Hence, the result obtained above is correct
Q 3 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 4 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 5 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 6 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 7 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 8 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 9 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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
Q 10 - Ex 2.3 - Linear Equations in One Variable - NCERT Maths Class 8th - Chapter 2
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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