# Excercise 2.1 Linear Equations in One Variable- NCERT Solutions Class 8

Rational Numbers

## Chapter 2 Ex.2.1 Question 1

Solve the equation:

\[x- 2= 7\]

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

In an equation values of left-hand side (LHS) and right-hand side (RHS) are equal. The two sides of the equation are balanced. We perform mathematical operations so that the balance is not disturbed.

**Steps:**

\[x- 2= 7\]

Transposing (\(-2\)) to RHS we get:

\begin{align}x &= {\rm{7}} + {\rm{ 2 }}\\\text{ }\!\!~\!\!\text{ }x&=9\end{align}

## Chapter 2 Ex.2.1 Question 2

Solve the equation:

\[y{\rm{ }} + {\rm{ 3 }} = {\rm{ 1}}0\]

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

In an equation values of left-hand side (LHS) and right-hand side (RHS) are equal. The two sides of the equation are balanced. We perform mathematical operations so that the balance is not disturbed.

**Steps:**

\[y{\rm{ }} + {\rm{ 3 }} = {\rm{ 1}}0\]

Transposing \(3\) to RHS we get,

\[\begin{align}&y{\rm{ }}= 10-{\rm{3}}\\&y = 7\end{align}\]

## Chapter 2 Ex.2.1 Question 3

Solve the equation:

\({\rm{6}} = {\rm{ }}z + {\rm{2}}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

In an equation values of left-hand side (LHS) and right-hand side (RHS) are equal. The two sides of the equation are balanced. We perform mathematical operations so that the balance is not disturbed.

**Steps:**

\[{\rm{6}} = {\rm{ }}z + {\rm{2}}\]

Transposing \(2\) to LHS we get,

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

## Chapter 2 Ex.2.1 Question 4

Solve the equation:

\(\begin{align}\frac{3}{7}\,\, + x\,\, = \,\,\frac{{17}}{7}\end{align}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

\[\frac{3}{7}\,\, + \,\,x\,\, = \,\,\frac{{17}}{7}\]

Transposing \(\begin{align}\frac{3}{7}\end{align}\) to RHS,

we get

\[\begin{align}x &=\frac{17}{7} - \frac{3}{7}\\x&= \,\,\frac{14}{7}\\x &= 2\end{align}\]

## Chapter 2 Ex.2.1 Question 5

Solve the equation:

\(\,\left( {{\rm{6}}x = {\rm{12}}} \right)\)

**Solution**

**Video Solution**

**What is known?**

Equation

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

We have \(6x = 12,\) dividing expressions with \(6\) both sides

\[\begin{align}\frac{6x}{6}&=\frac{12}{6}\\x &= 2\end{align}\]

## Chapter 2 Ex.2.1 Question 6

Solve the equation:

\(\begin{align}\frac{t}{5}\, = \,10\end{align}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

We have \(\begin{align}\frac{t}{5}\,\, = \,\,10, \end{align}\) multiplying expressions with \(5\) both sides

\[~\frac{t}{5}\times 5\,\,=\,\,10\times 5\]

We get \(t=50\)

## Chapter 2 Ex.2.1 Question 7

Solve the equation:

\(\begin{align}\frac{{2x}}{3}\,\, = 18\end{align}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

By multiplying with \(\begin{align}\frac{3}{2}\end{align}\) in both sides,

\[\frac{3}{2} \times \frac{{2x}}{3}\,\, = \,\,18 \times \frac{3}{2}\]

We get \(x = 27\)

## Chapter 2 Ex.2.1 Question 8

Solve the equation:

\(\begin{align}1.6\,\, = \,\,\frac{y}{{1.5}}\end{align}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

Multiplying with \(1.5\) both sides

\[\,\,\frac{y}{{1.5}} \times 1.5 = 1.6 \times 1.5\,\,\]

We get \(y = 2.4\)

## Chapter 2 Ex.2.1 Question 9

Solve the equation:

\({\rm{ 7}}x-{\rm{9}} = {\rm{16}}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

Transposing \((- 9)\) to RHS, we get

\[\begin{align}{\rm{7}}x &= {\rm{16 }} + {\rm{ 9 }}\\{\rm{7}}x &= {\rm{25}}\end{align}\]

Now dividing both sides by \(7\):

\[\begin{align}\frac{{7x}}{7}\,\, &= \,\,\frac{{25}}{7}\end{align}\]

We get \(\begin{align}x = 3.57\end{align}\)

## Chapter 2 Ex.2.1 Question 10

Solve the equation:

\({\rm{14}}y-{\rm{8}} = {\rm{13}}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

Transposing \((- 8)\) to RHS, we get

\[\begin{align}14y &= {\rm{13 }} + {\rm{ 8 }}\\14y &= 21\end{align}\]

Now dividing both sides by \(14\):

\[\frac{{14y}}{{14}}\,\, = \,\,\frac{{21}}{{14}}\]

We get, \(y\) \(= 1.5\)

## Chapter 2 Ex.2.1 Question 11

Solve the equation:

\({\rm{17}} + {\rm{6}}p{\rm{ }} = {\rm{ 9}}\)

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

Transposing \(17\) to RHS, we get:

\[\begin{align}{\rm{6}}p &= {\rm{9}}-{\rm{17}}\\{\rm{6}}p &= - {\rm{8}}\end{align}\]

Now dividing both sides by \(6\).

\[\begin{align}\frac{{6p}}{6}\,\, = \,\, - \frac{8}{6}\\p\,\, = \,\, - \frac{4}{3}\end{align}\]

## Chapter 2 Ex.2.1 Question 12

Solve the equation:

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

**Solution**

**Video Solution**

**What is known?**

Equations

**What is unknown?**

Value of the variable

**Reasoning: **

**Steps:**

Transposing \(1\) to RHS, we get_{ ,}

\[\begin{align}\frac{x}{3}\,\,\, &= \,\,\frac{7}{{15}}\,\, - \,\,1\\\frac{x}{3}\,\, &= \,\,\frac{{7 - 15}}{{15}}\\\frac{x}{3}\,\, &= \,\,\frac{{ - 8}}{{15}}\end{align}\]

Now multiplying both sides by \(3\), we get

\[\begin{align}\frac{x}{3} \times 3\,\, &= \,\, - \frac{8}{{15}} \times 3\\x\,\, &= \,\,\frac{{ - 8}}{5}\end{align}\]