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

Go back to  'Linear Equations in One Variable'

Chapter 2 Ex.2.1 Question 1

Solve the equation:

$x- 2= 7$

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

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

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

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:

$\frac{3}{7}\,\, + \,\,x\,\, = \,\,\frac{{17}}{7}$

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

What is known?

Equation

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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

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:

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}

Related Sections
Related Sections

math teachers and top

Personalized Curriculum
Instant Doubts clarification
Cover latest CBSE Syllabus
Unlimited Mock & Practice tests
Covers CBSE, ICSE, IB curriculum

Instant doubt clearing with Cuemath Advanced Math Program
0