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

Text Solution

What is known?

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

\[\begin{align}5t-3-3t&= 3t-3t - 5\\2t-3 &= - 5\\2t &= - 5 + 3 \qquad\left( {{\text{Transposing - 3 to RHS}}} \right)\\t\, &= \,\frac{{ - 2}}{2} \qquad \quad \left( {{\text{Dividing both sides by 2}}} \right)\\t &= - 1{\rm{ }}\\\end{align}\]

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

Steps:

\[\begin{align}5x - 3x + 9 &= 5 + 3x - 3x \quad ({\text{Subtracting }}3x{\text{ from both sides}})\\2x + 9 &= 5\\2x &= 5 - 9\\2x &= - 4 \qquad  \qquad \quad    ({\text{Dividing by }}2{\text{ on both sides}})\\x &= \frac{{ - 4}}{2}\\x &= - 2\end{align}\]

(vi) \(4z + 3 = 6 + 2z \)

Steps:

\[\begin{align}4z + 3{\rm{ }} &= {\rm{ }}6 + 2z\\4z-2z + 3 &= 6 + 2z-2z\qquad\left( {{\rm{subtracting \;}}2z\;{\rm{ from\; both \;sides}}} \right)\\2z + 3 &= 6\\2z &= 6-3\\2z& = 3{\rm{ }} \qquad\qquad\qquad\left( {{\text{Dividing by }}2{\text{ on both sides}}} \right)\\z &= \,\frac{3}{2}\end{align}\]

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

Steps:

\[\begin{align}2x-1& = 14 - x\\2x + x-1&= 14-x{\rm{ }} + {\rm{ }}x{\rm{ }} \quad\left( {{\text{Adding }}x{\text{ on both sides}}} \right)\\3x-1 &= {\rm{ }}14\\3x{\rm{ }} &= {\rm{ }}15{\rm{ }}\\x\,\, &= \,\,\frac{{15}}{3}\quad\qquad\left( {{\text{Dividing }}3{\text{ on both sides}}} \right)\\x &= 5\\\end{align}\]

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

Steps:

\[\begin{align}8x + 4 &= 3 \times x + 3 \times \left( { - 1} \right) + 7 \quad \left( {{\text{Applying BODMAS to RHS}}} \right)\\8x + 4 &= 3x - 3 + 7\\8x + 4{\rm{ }} &= 3x + 4\\8x &= 3x + 4-4\\8x &= {\rm{ }}3x{\rm{ }} & \\8x-3x &= 3x-3x\quad \qquad \qquad \quad\left( {{\text{Subtracting 3}}x{\text{ both sides}}} \right)\\5x &= 0\\x\, &= \,0\end{align}\]

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

Steps:

\[\begin{align}5\, \times \,x&= 5 \times \frac{4}{5}\left( {x + 10} \right)\quad\left( {{\text{Multiplying both sides by 5}}} \right)\\5x &= 4{\rm{ }}\left( {x + 10} \right){\rm{ }}\qquad \quad \left( {{\text{Applying BODMAS on RHS}}} \right)\\5x&= 4x + 40 \qquad \quad \quad  \left( {{\text{Subtracting 4}}x{\text{ on both sides}}} \right)\\5x-4x&= {\rm{ }}4x-4x + 40\\x &= 40\end{align}\]

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

Steps:

\(\begin{align}\frac{{2x}}{3}\, - \,\frac{{7x}}{{15}}\, &= \,\,\,3\, - \,1 \qquad  \left( {{\text{Transposing variables on one side and constant on other side}}} \right) \hfill \\\frac{{5(2x)\, - \,1(7x)}}{{15}}\,\,& = \,\,2 \qquad \left( {{\text{LCM on LHS}}} \right) \hfill \\\end{align} \)

We get,

\[\begin{align}\frac{{10x\,\, - \,\,7x}}{{15}}\,\, &= \,2\\15\,\, \times \,\,\frac{{(10x\,\, - \,\,7x)}}{{15}}\,\, &= \,15\, \times \,2 \left( {{\text{Multiplying 15 on both sides}}} \right)\\3x& = 30\\{\rm{Ans: }}x &= 10\end{align}\]

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

Steps:

\[\begin{align}2y\, + \,y&= \,\frac{{26}}{3}\, - \,\frac{5}{3}  \quad\left( {{\text{Transposing variables on one side and constant on other side}}} \right)\\3y\,\,\, &= \,\frac{{26\, - \,5}}{3}\,\\3y\,\, &= \,\,\frac{{21}}{3}\end{align}\]

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

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

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

Steps:

\[\begin{align}\,\frac{8}{5}\,&= \,5m\, - \,3m  \qquad \left( {{\text{Transporting variable to R}}{\text{.H}}{\text{.S and constants to L}}{\rm{.H}}{\rm{.S}}} \right)\\\frac{8}{5}\,\,& = \,\,2m\qquad\qquad \left( {{\text{Dividing both sides by 2}}} \right)\\m\,\, &= \,\,\frac{4}{5}\\{\rm{Ans:}}\,\,m\,\, &= \,\,\frac{4}{5}\end{align}\]

  
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