# Excercise 4.2 Simple-Equations - NCERT Solutions Class 7

## 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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,

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