# Ex.4.1 Q2 Simple-Equations Solution - NCERT Maths Class 7

## Question

Check whether the value given in the brackets is a solution to the given equation or not:

(a) \(n + 5 = 19\left( {n = 1} \right)\)

(b) \(7n + 5 = 19\left( {n = -2} \right)\)

(c) \(7n + 5 = 19\left( {n = 2} \right)\)

(d) \(4p-3 = 13\left( {p = 1} \right)\)

(e) \(4p-3 = 13\left( {p = -4} \right)\)

(f) \(4p-3 = 13\left( {p = 0} \right)\)

## Text Solution

**What is unknown?**

Whether the given value is a solution of the equation or not.

**What is Known?**

Equation and the value of the variable.

**Reasoning:**

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

**Steps:**

a) Here, \(n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= 1\) (given)

L.H.S = \(n+ 5,\)

By putting, \(n= 1,\)

L.H.S \(= 1 + 5 = 6 ≠\) R.H.S

L.H.S \(≠\) R.H.S, so \(n= 1\) is not a solution of the equation.

b) Here, \(7n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= – 2\) (given)

L.H.S = \(7n+ 5\),

By putting, \(n= \,– 2\),

L.H.S\( =7 \times (−2) + 5= −9 ≠\) R.H.S

As, L.H.S \(≠\) R.H.S, so \(n= \,– 2\) is not a solution of the equation.

c) Here, \(7n+ 5\) is L.H.S, \(19\) is R.H.S and \(n= 2\) (given)

L.H.S = \(7n+ 5\),

By putting, \(n= 2\),

L.H.S \(= 7 \times (2) + 5 = 19 = \)R.H.S

As, L.H.S = R.H.S, so \(n= 2\) is a solution of the equation.

d) Here, \(4p\,– 3\) is L.H.S, \(13\) is R.H.S and \(p= 1\) (given)

L.H.S = \(4p\,– 3,\)

By putting, \(p= 1\),

L.H.S \(=4 \times (1)\, – 3 = 1 ≠ \) R.H.S

As, L.H.S \(≠\) R.H.S, so \(p= 1\) is not a solution of the equation.

e) Here, \(4p– 3\) is L. H.S, \(13\) is R.H.S and \(p= -4\) (given)

L.H.S \(=4p– 3\),

By putting, \(p= 1\),

L.H.S \(=4 \times (-4) – 3 = -19 ≠\) R.H.S

As, L.H.S \(≠\) R.H.S, so \(p= 1\) is not a solution.

f) Here, \(4p– 3\) is L. H.S, \(13\) is R.H.S and \(p= 0\) (given data)

L.H.S \(=\) \(4p\,– 3\),

By putting, \(p= 0\),

L.H.S \(=4 \times(0)–3=−3≠ \)R.H.S

As, L.H.S\(≠\)R.H.S, so \(p= 0\) is not a solution of the equation.