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Ex.4.2 Q4 Simple-Equations Solution - NCERT Maths Class 7

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Question

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

 Video Solution
Simple Equations
Ex 4.2 | Question 4

Text 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,

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

  
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