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Complete the last column of the table.
S.No. | Equation | Value | Say
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- Advertisements
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