# Excercise 2.3 Linear Equations in One Variable- NCERT Solutions Class 8

Go back to  'Linear Equations in One Variable'

## Chapter 2 Ex.2.3 Question 1

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?

### 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}