# Ex.4.1 Q1 Simple-Equations Solution - NCERT Maths Class 7

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## Question

Complete the last column of the table.

 S. No Equation Value Say, whether the equation is satisfied. (Yes/No) (i) $$x + 3 = 0$$ $$x= 3$$ (ii) $$x + 3 = 0$$ $$x= 0$$ (iii) $$x + 3 = 0$$ $$x= -3$$ (iv) $$x - 7 = 1$$ $$x= 7$$ (v) $$x - 7 = 1$$ $$x= 8$$ (vi) $$5x= 25$$ $$x= 0$$ (vii) $$5x= 25$$ $$x= 5$$ (viii) $$5x= 25$$ $$x= -5$$ (ix) \begin{align}\frac{m}{3} = 2\end{align} $$m= -6$$ (x) \begin{align}\frac{m}{3} = 2\end{align} $$m= 0$$ (xi) \begin{align}\frac{m}{3} = 2\end{align} $$m= 6$$

## Text Solution

What is Known?

Equation and the value of the variable.

What is unknown?

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

Reasoning:

Put the value 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 given value is not a solution of the equation.

Steps:

(i) \begin{align}x + 3 = 0,x = 3\end{align}

L.H.S $$=$$ $$x+ 3$$, R.H.S $$= 0$$

By putting, $$x= 3$$

L.H.S $$=3 + 3= 6 ≠$$R.H.S

Therefore, “No”, the equation is not satisfied.

(ii) \begin{align}x+ 3 = 0,x= 0\end{align}

L.H.S $$=$$ $$x+ 3$$, R.H.S$$= 0$$

By putting, $$x= 0;$$

L.H.S $$=0+3= 3 ≠$$R.H.S

Therefore, “No”, the equation is not satisfied.

(iii) \begin{align}x+ 3 = 0, x= -3\end{align}

L.H.S $$=$$ $$x+ 3$$, R.H.S $$= 0$$

By putting, $$x= -3$$

L.H.S $$= -3 + 3 = 0 =$$ R.H.S

Therefore, “Yes”, the equation is satisfied.

(iv) \begin{align}x– 7 = 1, x= 7\end{align}

L.H.S $$=$$ $$x\,– 7$$, R.H.S $$= 1$$

By putting, $$x= 7$$;

L.H.S $$= 7 \,– 7 = 0 ≠$$ R.H.S

Therefore, “No”, the equation is not satisfied.

(v) \begin{align}x– 7 = 1, x= 8\end{align}

L.H.S $$=$$ $$x– 7$$, R.H.S $$= 1$$

By putting, $$x= 8;$$

L.H.S $$= 8 \,– 7 = 1 =$$ R.H.S

Therefore, “Yes”, the equation is satisfied.

(vi) \begin{align}5x= 25, x= 0,\end{align}

L.H.S $$=$$ $$5x$$, R.H.S$$= 25$$

By putting, $$x= 0;$$

L.H.S $$=$$ $$5 \times 0=0 ≠$$R.H.S

Therefore, “No”, the equation is not satisfied.

(vii) \begin{align}5x= 25, x= 5,\end{align}

L.H.S $$= 5x$$, R.H.S $$= 25$$

By putting, $$x= 5;$$

L.H.S $$=$$ $$5 \times 5 = 25$$= R.H.S

Therefore, “Yes”, the equation is satisfied.

(viii) \begin{align}5x = 25, x= – 5,\end{align}

L.H.S $$=$$ $$5x$$, R.H.S $$= 25$$

By putting, $$x= – 5;$$

L.H.S $$=$$ $$5 \times \left( { - 5} \right) = - 25≠$$R.H.S

Therefore, “No”, the equation is not satisfied.

(ix) \begin{align}\frac{m}{3} = 2, m= – 6,\end{align}

L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$

By putting, $$m$$$$= – 6;$$

L.H.S $$=$$ \begin{align}\frac{{ - 6}}{3}=−2≠\end{align} R.H.S

Therefore, “No”, the equation is not satisfied.

(x) \begin{align}\frac{m}{3} = 2, m=0\end{align}

L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$

By putting, $$m= 0;$$

L.H.S $$=$$ \begin{align}\frac{0}{3}=0≠\end{align}R.H.S

Therefore, “No”, the equation is not satisfied.

(xi) \begin{align}\frac{m}{3} = 2 m= 6,\end{align}

L.H.S $$=$$ \begin{align}\frac{m}{3}\end{align}, R.H.S $$= 2$$

By putting, $$m$$ $$= 6;$$

L.H.S $$=$$ \begin{align}\frac{6}{3}= 2 =\end{align} R.H.S

Therefore, “Yes”, the equation is satisfied

 S. No Equation Value Say, weather the equation is satisfied. (Yes/No) (i) $$x + 3 = 0$$ $$x= 3$$ No (ii) $$x + 3 = 0$$ $$x= 0$$ No (iii) $$x + 3 = 0$$ $$x= -3$$ Yes (iv) $$x - 7 = 1$$ $$x= 7$$ No (v) $$x - 7 = 1$$ $$x= 8$$ Yes (vi) $$5x= 25$$ $$x= 0$$ No (vii) $$5x= 25$$ $$x= 5$$ Yes (viii) $$5x= 25$$ $$x= -5$$ No (ix) \begin{align}\frac{m}{3} = 2\end{align} $$m= -6$$ No (x) \begin{align}\frac{m}{3} = 2\end{align} $$m= 0$$ No (xi) \begin{align}\frac{m}{3} = 2\end{align} $$m= 6$$ Yes

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