# NCERT Class 8 Maths Linear Equations in One Variable

The chapter 2 begins with an introduction to Linear Equations in one variable by recalling the concepts of algebraic expressions and equations.The procedure for solving equations which have linear expressions on one side and the numbers on the other side is explained in the next section.This is followed by some applications of linear equations.After this, we have the procedure to solve equations having the variable on both sides is dealt and is followed by some more applications.The procedure to reduce an equation to simple form, equations reducible to linear form is described in detail in the later sections of the chapter.

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Linear Equations in One Variable

## 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

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