Ex.4.2 Q2 Simple-Equations Solution - NCERT Maths Class 7

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Question

Give first the step 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) \(20t = - 10\)

Text 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,

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

  
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