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Ex.2.3 Q1 Linear Equations in One Variable Solution - NCERT Maths Class 8

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Question

Solve the following equations and check your results:

(i) \(3x = 2x + 18\)

(ii) \(5t - 3 = 3t - 5 \)

(iii) \(5x + 9 = 5 + 3x\)

(iv) \(4z + 3 = 6 + 2z \)

(v) \(2x-1 = 14 - {\rm{ }}x\)

(vi) \(8x + 4 = 3\left( {x{\rm{ }} - {\rm{ }}1} \right) + 7\)

(vii) \(\begin{align}x\,\, = \,\,\frac{4}{5}\,(x\, + \,10) \end{align}\)

(viii) \(\begin{align}\frac{{2x}}{3}\, + \,1\, = \,\frac{{7x}}{{15}}\, + \,3\end{align}\)

(ix) \(\begin{align}2y\, + \,\frac{5}{3}\,\, = \,\frac{{26}}{3}\, - \,y\end{align}\)

(x)  \(\begin{align}3m\, = \,5m\, - \,\frac{8}{5}\end{align}\)


What is known?

 Video Solution
Linear Equations
Ex 2.3 | Question 1

Text Solution

Equations

What is unknown?

Value of the variable

Reasoning:

In algebraic equations, arithmetic operations can be executed on variables treating them as numbers.

(i) \(3x = 2x + 18\)

Steps:

Subtracting \(2x\) on both sides

\[\begin{align}3x-2x &= 2x-2x + 18\\x&= 18\end{align}\]

(ii) \(~5t-3=3t-5\)

Steps:

 Subtracting \(3t\) on both sides

\[\begin{align}5t-3-3t&= 3t-3t - 5\\2t-3 &= - 5 \end{align} \]

Transposing \(- 3\) to RHS

\[ 2t = - 5 + 3 \]

Dividing both sides by \(2\)

\[ \begin{align} t\, &= \,\frac{{ - 2}}{2}  \\t &= - 1{\rm{ }}\\\end{align}\]

(iii) \(5x + 9 = 5 + 3x\)

Steps:

Subtracting \(3x\) from both sides,

\[\begin{align} 5x - 3x + 9 &= 5 + 3x - 3x  \\2x + 9 &= 5\\2x &= 5 - 9 \\2x &= - 4 \end{align} \]

Dividing by \(2\) on both sides

\[\begin{align} x &= \frac{{ - 4}}{2}\\x &= - 2\end{align}\]

(iv) \(4z + 3 = 6 + 2z \)

Steps:

\[4z + 3 = 6 + 2z\]

subtracting \(2z\) from both sides

\[ \begin{align} 4z-2z + 3 &= 6 + 2z-2z \\ 2z + 3 &= 6\\2z &= 6-3\\2z& = 3 \end{align} \]

Dividing by \(2\) on both sides

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

(v) \(2x-1 = 14 - {\text{ }}x\)

Steps:

\[2x-1 = 14 - x\]

Adding \(x\) on both sides

\[\begin{align} 2x + x-1&= 14-x + x \\3x-1 &= 14\\3x &= 15 \end{align} \]

Dividing \(3\) on both sides

\[ \begin{align} x &=\frac{{15}}{3} \\x &= 5\\\end{align}\]

(vi) \(\begin{align}8x + 4 = 3\left( {x - 1} \right) + 7\end{align}\)

Steps:

\[8x + 4 = 3 \times x + 3 \times \left( - 1 \right) + 7 \]

Applying BODMAS to RHS

\[ \begin{align} 8x + 4 &= 3x - 3 + 7\\8x + 4{\rm{ }} &= 3x + 4\\8x &= 3x + 4-4\\8x &= {\rm{ }}3x{\rm{ }} \end{align} \]

Subtracting \(3x\) both sides

\[ \begin{align} 8x-3x &= 3x-3x \\5x &= 0\\x\, &= \,0\end{align}\]

(vii) \(\begin{align}x\,\, = \,\,\frac{4}{5}\,(x\, + \,10)\end{align}\)

Steps:

Multiplying both sides by \(5\)

\[5\, \times \,x= 5 \times \frac{4}{5}\left( {x + 10} \right)\]
Applying BODMAS on RHS

\[\begin{align} 5x &= 4{\rm{ }}\left( {x + 10} \right) \\5x&= 4x + 40\end{align}\]

Subtracting \(4x\) on both sides

\[\begin{align} 5x-4x&= {\rm{ }}4x-4x + 40\\x &= 40\end{align}\]

(viii) \(\begin{align}\frac{{2x}}{3}\, + \,1\, = \,\frac{{7x}}{{15}}\, + \,3\end{align}\)

Steps:

Transposing variables on one side and constant on other side

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

LCM on LHS

\[\frac{{5(2x)\, - \,1(7x)}}{{15}} = \,\,2\]

We get,

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

Multiplying \(15\) on both sides

\[\begin{align} 15\,\, \times \,\,\frac{{(10x\,\, - \,\,7x)}}{{15}}\,\, &= \,15\, \times \,2 \\3x& = 30\\{\rm{Ans: }}x &= 10\end{align}\]

(ix) \(\begin{align}2y\, + \,\frac{5}{3}\,\, = \,\frac{{26}}{3}\, - \,y\end{align}\)

Steps:

Transposing variables on one side and constant on other side

\[\begin{align}2y\, + \,y&= \,\frac{{26}}{3}\, - \,\frac{5}{3} \\3y\,\,\, &= \,\frac{{26\, - \,5}}{3}\,\\3y\,\, &= \,\,\frac{{21}}{3}\end{align}\]

We get \(3y = 7\) (Dividing with \(3\) on both sides)

\({\text{Ans:}}\,y\,\, = \,\,\frac{7}{3}\)

(x) \(3m\, = \,5m\, - \,\frac{8}{5}\)

Steps:

Transporting variable to R.H.S and constants to L.H.S

\[\begin{align}\frac{8}{5}&= \,5m\, - \,3m\ \\ \frac{8}{5}\,\,& = \,\,2m \end{align} \]

Dividing both sides by \(2\)

\[\begin{align}m\,\, &= \,\,\frac{4}{5}\\{\rm{Ans:}}\,\,m\,\, &= \,\,\frac{4}{5}\end{align}\]

  
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