Complete the last column of the table.
S.No.EquationValueSay
AcademicMathematicsNCERTClass 7

To do:

We have to complete the given table.

Solution:

(i) $x+3=0$ , $x=3$

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) $x+3=0$, $x=0$

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) $x+3=0$, $x=-3$

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) $x-7=1$, $x=7$

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) $x-7=1$, $x=8$

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) $5x=25,\ x=0$,

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

By putting, $x=0$;

L.H.S$=5(0)=0$

$≠$R.H.S

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

(vii) $5x=25,\ x=5$,

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

By putting, $x=5$;

L.H.S$=5(5)=25$

$=$R.H.S

Therefore, “Yes”, the equation is satisfied.

(viii) $5x=25,\ x=-5$,

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

By putting, $x=-5$;

L.H.S$=5(-5)=-25$

$≠$R.H.S

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

(ix) $\frac{m}{3}=2,\ m=-6$,

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

By putting, $m=-6$;

L.H.S$=-\frac{6}{3}=-2$

$≠$R.H.S

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

(x) $\frac{m}{3}=2,\ m=0$,

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

By putting, $m=0$;

L.H.S$=\frac{0}{3}=0$

$≠$R.H.S

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

(xi) $\frac{m}{3}=2,\ m=6$,

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

By putting, $m=6$;

L.H.S$=\frac{6}{3}=2$

$=$R.H.S

Therefore, “Yes”, the equation is satisfied.

Therefore, the complete table is as follows-


raja
Updated on 10-Oct-2022 13:33:38

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