Ex.4.3 Q2 Simple-Equations Solutions-Ncert Maths Class 7

Go back to  'Ex.4.3'

Question

Solve the following equations.

(a) \(2(x + 4) = 12\)

(b) \(3(n\, – 5) = 21\)

(c) \(\begin{align}3\left( {n-5} \right) = - 21\end{align} \)

(d) \(\begin{align}- 4\left( {2 + x} \right) = 8\end{align} \)

(e) \(4(2 \,– x) = 8\)

Text Solution

What is Known?

Equations.

What is unknown?

The value of the variable.

Reasoning:

To solve these equations, transpose the variables on the one side and constants on the other side, and simplify them and get the value of variable.

Steps:

(a) \(2(x + 4) = 12\)

\[\begin{align} 2(x + 4) &= 12\\2x + 8 &= 12\\2x &= 12 – 8\\2x &= 4\\x& = \frac{4}{2}\; \text{or} \;x = 2\end{align} \]

(b) \(3(n \,– 5) = 21\)

\[\begin{align}3(n – 5) &= 21\\3n – 15 &= 21\\3n &= 21 + 15\\3n &= 36 \\ n &= \frac{{36}}{3} \,\text{or}\, n = 12\end{align} \]

(c) \(\begin{align} 3\left( {n-{\rm{ }}5} \right) = - 21\end{align} \)

\[\begin{align} 3\left( {n-{\rm{ }}5} \right) &= - 21 \\3n \,– 15 &=\, –21\\3n &=\, –21 + 15\\3n& = \,–6\\ n &= \frac{{ - 6}}{3}\,\text{or}\, n = \,–2\end{align} \]

(d) \(\begin{align}- 4\left( {2 + x} \right) = 8\end{align} \)

\[\begin{align} - 4\left( {2 + x} \right) &= 8\\– 8 \,– 4x &= 8\\–4x &= 8 + 8\\–4x &= 16\\x = \frac{{ - 16}}{4}& = \,–\,4\end{align} \]

(e)  \(4(2 \,– x) = 8\)

\[\begin{align}4(2 – x) &= 8\\8 – 4x &= 8 \\– 4x &= 8\,– 8 = 0 \,\text{or} \,x = 0\end{align}\]

  
Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school