# Excercise 2.5 Linear Equations in One Variable- NCERT Solutions Class 8

Go back to  'Linear Equations in One Variable'

## Chapter 2 Ex.2.5 Question 1

Solve the linear equation \begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

\begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}

LCM of the denominators, $$2, 3, 4,$$ and $$5$$, is $$60.$$

Multiplying both sides by $$60,$$ we obtain

\begin{align}60\left( {\frac{x}{2} - \frac{1}{5}} \right) = 60\left( {\frac{x}{3} + \frac{1}{4}} \right)\end{align}

Opening the  brackets, we get,
\begin{align}30x - 12 &= 20x + 15 \\30x - 20x &= 15 + 12\\10x &= 27\quad \quad \\\,\,\,\,\,x &= \frac{{27}}{{10}}\end{align}

## Chapter 2 Ex.2.5 Question 2

Solve the linear equation \begin{align}\frac{n}{2} - \frac{{3n}}{4} + \frac{{5n}}{6} = 21\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$

LCM of the denominators, $$2, 4,$$ and $$6,$$ is $$12.$$

Multiplying both sides by $$12,$$ we obtain

\begin{align}6n - 9n + 10n &= 2527\\7n &= 252 \\n &= \frac{{252}}{7}\\n &= 36 \\\end{align}

## Chapter 2 Ex.2.5 Question 3

Solve the linear equation \begin{align}x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

\begin{align}x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}\end{align}

LCM of the denominators,$$2, 3,$$ and $$6,$$ is $$6.$$

Multiplying both sides by $$6,$$ we obtain

\begin{align}6x + 42 - 16x &= 17 - 15x \\6x - 16x + 15x &= 17 - 42 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5x &= - 25 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= \frac{{ - 25}}{5} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= - 5 \\\end{align}

## Chapter 2 Ex.2.5 Question 4

Solve the linear equation \begin{align}\frac{{x - 5}}{3} = \frac{x - 3}{5}\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominatorss to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

\begin{align}\frac{{x - 5}}{3} = \frac{{x - 3}}{5}\end{align}

LCM of the denominators, $$3$$ and $$5,$$ is $$15.$$

Multiplying both sides by $$15,$$ we obtain

\begin{align}5\left( {x - 5} \right)&= 3\left( {x - 3} \right) \end{align}

Opening the brackets we get,

\begin{align}5x - 25 &= 3x - 9 \\5x - 3x &= 25 - 9 \\2x &= 16 \\\,\,x& = \frac{{16}}{2} \\\,\,x &= 8 \\\end{align}

## Chapter 2 Ex.2.5 Question 5

Solve the linear equation\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}

LCM of the denominators, $$3$$ and $$4,$$ is $$12.$$

Multiplying both sides by $$12,$$ we obtain

\begin{align}3\left( {3t \! - \! 2} \right) \! - \! 4\left( {2t \! + \! 3} \right) &= \! 8 \! - \! 12t \\9t \! - \! 6 \! - \! 8t \! - \! 12 &= \! 8 \! - \! 12t\, \\ \quad\quad\quad\text{(Opening }& \text{the brackets)}\\9t \! - \! 8t \! + \! 12t &= \! 8 \! + \! 6 \! + \! 12 \\13t &= \! 26 \\t &= \! \frac{{26}}{{13}} \\t &= \! 2 \\\end{align}

## Chapter 2 Ex.2.5 Question 6

Solve the linear equation \begin{align}m - \frac{{m - 1}}{2} = 1 - \frac{{m - 2}}{3}\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.

Steps:

LCM of the denominators, $$2$$ and $$3,$$ is $$6.$$

Multiplying both sides by $$6,$$ we obtain

\begin{align}6m \! - \! 3\left( {m\! -\! 1} \right) & \! = \! 6 \!-\! 2\left( {m \! - \! 2} \right)\\6m \! - \! 3m \! + \! 3 & \! = \! 6 \! - \! 2m \! + \! 4 \\\quad\quad\quad\text{(Opening }& \text{the brackets)}\\6m\! -\! 3m \! + \! 2m & \! = \! 6 \! + \! 4 \! - \! 3\\5m & \! = \! 7\\\,\,m & \! = \! \frac{7}{5}\end{align}

## Chapter 2 Ex.2.5 Question 7

Simplify and solve the linear equation $$3(t - 3) = 5(2t + 1)$$

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

First open the brackets then transpose variable to one side and constant to another side.

Steps:

\begin{align}3\left( {t \! - \! 3} \right) & \! = \! 5\left( {2t \! + \! 1} \right) \\3t \! - \! 9 & \! = \! 10t \! + \! 5 \\\quad\quad\quad\text{(Opening }& \text{the brackets)}\\- 9 \! - \! 5 & \! = \! 10t \! - \! 3t \\ - 14 & \! = \! 7t \\t & \! = \! \frac{{ - 14}}{7} \\t & \! = \! - 2 \\\end{align}

## Chapter 2 Ex.2.5 Question 8

Simplify and solve the linear equation \begin{align}15(y \! - \! 4) \! - \! 2(y \! - \! 9) \! + \! 5(y \! + \! 6) \!= \! 0\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

First open the brackets then transpose variable to one side and constant to another side.

Steps:

\begin{align}15\left( {y \! - \! 4} \right) \! - \! 2\left( {y \! - \! 9} \right) \! + \! 5\left( {y \! + \! 6} \right)& \! = \! 0 \\15y \! - \! 60 \! - \! 2y \! + \! 18 \! + \! 5y \! + \! 30 & \! = \! 0 \\\qquad\text{(Opening } \text{the brackets)}\\ \\18y \! - \! 12 & \! = \! 0 \\18y & \! = \! 12 \\y & \! = \! \frac{{12}}{{18}} \\ & \! = \! \frac{2}{3} \end{align}

## Chapter 2 Ex.2.5 Question 9

Simplify and solve the linear equation \begin{align}3(5z \!-\! 7) \!-\! 2(9z - 11) \!=\! 4(8z \!-\! 13) \!-\! 17\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

First open the brackets then transpose variable to one side and constant to another side.

Steps:

\begin{align}3(5z\! -\!7)\!-\!2(9z\!-\!11)&\!=\!4(8z\!-\!13)\!-\!17\\15z - 21 - 18z + 22& \!= \!32z - 52 - 17 \\\qquad\qquad\qquad\text{(Opening }& \text{the brackets)}\\\\- 3z + 1 &= 32z - 69\\ - 3z - 32z& = - 69 - 1\\ - 35z &= - 70\\z& = \frac{{70}}{{35}}\\z\, &= 2\\\end{align}

## Chapter 2 Ex.2.5 Question 10

Simplify and solve the linear equation \begin{align}0.25(4f - 3) = 0.05(10f - 9)\end{align}

### Solution

What is known?

Equation

What is unknown?

Value of the variable

Reasoning:

First open the brackets then transpose variable to one side and constant to another side.

Steps:

\begin{align}0.25\left( {4f - 3} \right)&= 0.05\left( {10f - 9} \right)\\\frac{1}{4}\left( {4f - 3} \right) &= \frac{1}{{20}}\left( {10f - 9} \right)\end{align}

Multiplying both sides by $$20$$ we obtain

\begin{align}5\left( {4f - 3} \right) &= 10f - 9\\20f - 15 &= 10f - 9\,\\\qquad\qquad\qquad\text{(Opening }& \text{the brackets)}\\\\20f - 10f &= - 9 + 15\\\quad \,\,\,\,\,\,\,10f &= 6\\\quad \,\,\,\,\,\,\,\,\,\,\,\,f &= \frac{3}{5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.6\end{align}

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