Ex.4.1 Q3 Simple-Equations Solution - NCERT Maths Class 7

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Question

Solve the equation given below by trial and error method:

(i) \(\begin{align}5p + 2 = 17\end{align}\)

(ii) \(\begin{align}3m-{\rm{1}}4 = 4\end{align}\)

Text Solution

What is Known?

Equations

What is unknown?

Solution of the equation or the value of the variable.

Reasoning:

Put the different values of the variable in the given equation. If LHS is equal to the RHS then the equation is satisfied and if it is not that means this variable is not a solution of the equation.

Steps:

(i) \(5p + 2 = 17\)

\(5p + 2=\) L.H.S

By putting, \(p= 0\),

\(5 \times 0+2=2≠17\)

By putting, \(p=1\),

\(5 \times (1)+2=7≠17\)

By putting, \(p= 2\),

\(5 \times (2) + 2 = 12 ≠\) R.H.S

By putting, \(p= 3\),

\( 5\times (3) + 2 = 17 =\) R.H.S

Therefore, \(p= 3\) is a solution of the equation.

(ii) \(3m-{\rm{1}}4 = 4\)

\(3m-{\rm{1}}4=\) L.H.S

By putting, \(m= 5\),

\(3 \times (5)–14=1≠6\)

By putting, \(m= 6\),

\(3 \times (6)–14 =4 =\)R.H.S.

Therefore, \(m= 6\) is a solution of the equation.

  
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