# 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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