# Excercise 4.2 Simple-Equations - NCERT Solutions Class 7

Go back to  'Simple Equations'

## Chapter 4 Ex.4.2 Question 1

Give the steps you will use to separate the variable and then solve the equation:

(a) $$x \,– 1 = 0$$

(b) $$x + 1 = 0$$

(c) $$x\, – 1 = 5$$

(d) $$x + 6 = 2$$

(e) $$y \,– 4 = \,– 7$$

(f)  $$y\, – 4 = 4$$

(g) $$y + 4 = 4$$

(h) $$y + 4 = \,– 4$$

### Solution

What is Known?

Equations.

What is unknown?

The first step which should be used to separate the variable in order to solve the equation.

Reasoning:

First try to reduce the equation by adding or subtracting by the same number and get the value of variable.

Steps:

(a) $$x \,– 1 = 0$$

Adding one to both sides of the equation we get,

\begin{align}x-1 + 1 &= 0 + 1\\x &= 1\end{align}

(b) $$x + 1 = 0$$

Subtracting one from both sides of the equation we get,

\begin{align}x + 1 -1 &= 0 -1\\x &= - 1\end{align}

(c) $$x \,– 1 = 5$$

Adding one to both sides of the equation we get,

\begin{align}x-1 + 1&= 5 + 1\\x &= 6\end{align}

(d) $$x + 6 = 2$$

Subtracting $$6$$ from both sides of the equation we get,

\begin{align}x + 6 -6 &= 2 -6\\x &= -4\end{align}

(e) $$y \,– 4 = – 7$$

Adding $$4$$ to both sides of the equation we get,

\begin{align}y-4 + 4 &= -7 + 4\\y &= -3\end{align}

(f) $$y \,– 4 = 4$$

Adding $$4$$ to both sides of the equation we get,

\begin{align}y-4 + 4 &= 4 + 4\\y &= 8\end{align}

(g) $$y + 4 = 4$$

Subtracting $$4$$ from both sides of the equation we get,

\begin{align}y + 4 -4 &= 4 -4\\y &= 0\end{align}

(h) $$y + 4 = – 4$$

Subtracting $$4$$ from both sides of the equation we get,

\begin{align}y + 4 -4&= -4 -4\\y &=-8\end{align}

## Chapter 4 Ex.4.2 Question 2

Give the steps you will use to separate the variable and then solve the equation:

(a) $$3l = 42$$

(b)\begin{align}\frac{b}{2} = 6\end{align}

(c)\begin{align}\frac{p}{7}=4\end{align}

(d) $$4x = 25$$

(e) $$8y = 36$$

(f)\begin{align}\frac{z}{3} = \frac{5}{4}\end{align}

(g)\begin{align}\frac{a}{5} = \frac{7}{{15}}\end{align}

(h) $$20t = - 10$$

### Solution

What is known?

Equation.

What is the unknown?

The first step we use to separate the variable in order to solve the equations.

Reasoning:

First try to reduce the equation by multiplying or dividing both sides of the equation by the same number to obtain the value of variable.

Steps:

(a) $$3l = 42$$

Divide both the sides by $$3$$ we get,

\begin{align}\frac{{3l}}{3} &= \frac{{42}}{3}\\l &= 14\end{align}

(b)\begin{align}\frac{b}{2} = 6\end{align}

Multiplying both sides by $$2,$$

\begin{align}\frac{b}{2}\times 2&=~6\times 2\\b{\rm{ }}&= {\rm{ }}12\end{align}

(c)\begin{align}~\frac{p}{7}=4\end{align}

Multiplying both sides by $$7,$$

\begin{align}\frac{p}{7} \times 7 &= 4 \times 7\\p &= 28\end{align}

(d) $$4x = 25$$

Dividing both the sides by $$4$$ we get,

\begin{align}\frac{4}{4}x &= \frac{{25}}{4}\\x &= \frac{{25}}{4}\end{align}

(e) $$8y{\rm{ }} = {\rm{ }}36$$

Dividing both the sides by $$8$$ we get,

\begin{align}\frac{8}{8}y &= \frac{{36}}{8}\\x &= \frac{9}{2}\end{align}

(f)\begin{align}\frac{z}{3} = \frac{5}{4}\end{align}

Multiplying both sides by $$3$$ we get,

\begin{align}\frac{z}{3} \times 3 &= \frac{5}{4} \times 3\\z &= \frac{{15}}{4}\end{align}

(g)\begin{align}\frac{a}{5} = \frac{7}{{15}}\end{align}

Multiplying both sides by $$5$$ we get,

\begin{align}\frac{a}{5} \times 5 &= \frac{7}{{15}} \times 5\\z &= \frac{7}{3}\end{align}

(h) $$20t = -10$$

Multiplying both sides by $$20$$ we get,

\begin{align}\frac{{20}}{{20}} \times t &= \frac{{ - 10}}{{20}}\\t &= \frac{{ - 1}}{2}\end{align}

## Chapter 4 Ex.4.2 Question 3

Give the steps you will use to separate the variable and then solve the equation:

(a) \begin{align}3n-2 = 46\end{align}

(b) \begin{align}5m + 7 = 17\end{align}

(c) \begin{align}\frac{{20p}}{3} = 40\end{align}

(d) \begin{align}\frac{{3p}}{{10}} = 6\end{align}

### Solution

What is Known?

Equations

What is unknown?

The first step we use to separate the variable in order to solve the equations.

Reasoning:

First try to reduce the equation by adding, subtracting, multiplying or dividing both sides of the equation by the same number to get the value of variable.

Steps:

(a) $$3n-2 = 46$$

Adding $$2$$ to both sides of the equation, we get

\begin{align}3n-2 + 2 &= 46 + 2\\3n &= 48\end{align}

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3n}}{3} = \frac{{48}}{3}\\n = 16\end{align}

(b) $$5m + 7 = 17$$

Subtracting $$7$$ from both sides of the equation, we get

\begin{align}5m + 7 - 7 &= 17 - 7\\5m& = 10\end{align}

Dividing both the sides by $$5$$ we get,

\begin{align}\frac{{5m}}{5} &= \frac{{10}}{5}\\m &= 2\end{align}

(c) \begin{align}\frac{{20p}}{3} = 40\end{align}

Multiplying both the sides by $$3$$ we get,

\begin{align}\frac{{20p}}{3} \times 3 &= 40 \times 3\\20p &= 120\end{align}

Dividing both the sides by $$20$$ we get,

\begin{align}\frac{{20p}}{{20}} &= \frac{{120}}{{20}}\\p &= 6\end{align}

(d) \begin{align}\frac{{3p}}{{10}} = 6\end{align}

Multiplying both the sides by $$10$$ we get,

\begin{align}\frac{{3p}}{{10}} \times 10 &= 6 \times 10\\3p &= 60\end{align}

Dividing both the sides by $$20$$ we get,

\begin{align}\frac{{3p}}{3} &= \frac{{60}}{3}\\p &= 20\end{align}

## Chapter 4 Ex.4.2 Question 4

Solve the following equations:

(a)  \begin{align}10p = 100\end{align}

(b)  \begin{align}10p + 10 = 100\end{align}

(c)   \begin{align}\frac{p}{4} = 5\end{align}

(d)  \begin{align}\frac{{ - p}}{3} = 5\end{align}

(e)  \begin{align}\frac{{3p}}{4} = 6\end{align}

(f)   \begin{align}3s = -9\end{align}

(g)   \begin{align}3s + 12 = 0\end{align}

(h)   \begin{align}3s = 0\end{align}

(i)   \begin{align}2q = 6\end{align}

(j)   \begin{align}2q-6 = 0\end{align}

(k)   \begin{align}2q + 6 = 0\end{align}

(l)   \begin{align}2q + 6 = 12\end{align}

### Solution

What is Known?

Equations

What is unknown?

Value of the variable.

Reasoning:

First try to reduce the equation by adding, subtracting, multiplying or dividing by the same number and get the value of variable.

Steps:

(a) $$10p = 100$$

Dividing both the sides by $$10$$ we get,

\begin{align}\frac{{10p}}{{10}}&= \frac{{100}}{{10}}\\p&= 10\end{align}

(b) $$10p + 10 = 100$$

Subtracting $$10$$ from both sides we get,

\begin{align}10p + 10 - 10&= 100 - 10\\10p&= 90\end{align}

Dividing both the sides by $$10$$ we get,

\begin{align}\frac{{10p}}{{10}} &= \frac{{90}}{{10}}\\p &= 9\end{align}

(c) \begin{align}\frac{p}{4} = 5\end{align}

Multiplying both the sides by $$4$$ we get,

\begin{align}\frac{p}{4} \times 4 &= 5 \times 4\\p& = 20\end{align}

(d) \begin{align}\frac{{ - p}}{3} = 5\end{align}

Multiplying both the sides by $$3$$,

\begin{align}\frac{{ - p}}{3} \times 3 &= 5 \times 3\\ - p &= 15\\p &= - 15\end{align}

(e) \begin{align}\frac{{3p}}{4} = 6\end{align}

Multiplying both the sides by $$4,$$

\begin{align}\frac{{3p}}{4} \times 4& = 6 \times 4\\3p&= 24\end{align}

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3p}}{3}&= \frac{{24}}{3}\\p&= 8\end{align}

(f) $$3s = -9$$

Dividing both the sides by $$3$$,

\begin{align}\frac{{3s}}{3}&= \frac{{ - 9}}{3}\\p&= - 3\end{align}

(g) $$3s + 12 = 0$$

Subtracting $$12$$ from both the sides of the equation we get,

\begin{align}3s + 12 - 12 &= 0 - 12\\3s &= - 12\end{align}

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3s}}{3}&= \frac{{ - 12}}{3}\\p&= - 4\end{align}

(h) $$3s = 0$$

Dividing both the sides by $$3$$ we get,

\begin{align}\frac{{3s}}{3} &= \frac{0}{3}\\p& = 0\end{align}

(i) $$2q = 6$$

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2} &= \frac{6}{2}\\q &= 3\end{align}

(j) $$2q-6 = 0$$

Adding $$6$$ to both sides of the equation we get,

\begin{align}2q - 6 + 6 &= 0 + 6\\2q &= 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2} = \frac{6}{2}\\\end{align}

(k) $$2q + 6 = 0$$

Subtracting $$6$$ from both the sides of the equation we get,

\begin{align}2q + 6 - 6&= 0 - 6\\2q&= - 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2}&= \frac{{ - 6}}{2}\\p&= - 3\end{align}

(l) $$2q + 6 = 12$$

Subtracting $$6$$ from both the sides of the equation we get,

\begin{align}2q + 6 - 6&= 12 - 6\\2q&= 6\end{align}

Dividing both the sides by $$2$$ we get,

\begin{align}\frac{{2q}}{2}&= \frac{6}{2}\\p&= 3\end{align}

Related Sections
Related Sections