# Ex.4.3 Q3 Simple-Equations Solutions-Ncert Maths Class 7

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## Question

Solve the following equations.

(a) $$4 = 5(p \,– 2)$$

(b) $$– 4 = 5(p\, – 2)$$

(c) $$16 = 4 + 3(t + 2)$$

(d) $$4 + 5(p - 1) =34$$

(e) $$0 = 16 + 4(m \,– 6)$$

Video Solution
Simple Equations
Ex 4.3 | Question 3

## Text Solution

What is Known?

Equations.

What is unknown?

The value of the variable.

Reasoning:

Transpose the variables on the one side and constants on the other side, then simplify them and get the value of variable.

Steps:

(a) $$4 = 5(p\, – 2)$$

\begin{align}4 &= 5(p\, – 2)\\4& = 5p \,– 10\\5p &= 4 + 10\end{align}

Therefore, \begin{align} p = \frac{{14}}{5}\end{align}

(b) $$– 4 = 5(p \,– 2)$$

\begin{align}– 4&= 5(p \,– 2)\\– 4 &= 5p \,– 10\\– 4 + 10 &= 5p\\6&= 5p\end{align}

Therefore, \begin{align} p = \frac{6}{5}\end{align}

(c) $$16 = 4 + 3(t + 2)$$

\begin{align}16 &= 4 + 3(t + 2)\\16 &= 4 + 3t + 6\\16 \,– 10 &= 3t\\6 &=3t\\ t &= \frac{6}{3} = 2\end{align}

(d) $$4 + 5(p - 1) =34$$

\begin{align} 4 + 5(p - 1) &=34\\4 + 5p \,– 5 &= 34\\5p \,– 1 &= 34\\5p &= 35\\p &= \frac{{35}}{5}\\p &= 7\end{align}

(e) $$0 = 16 + 4(m \,– 6)$$

\begin{align}0 &= 16 + 4(m \,– 6)\\0 &= 16 + 4m \,– 24\\8 &= 4m\\ m &= \frac{8}{4}\\m &= 2\end{align}

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