question_answer 27)

Complete the entries in the third column of the table.

S. No.	Equation	Value of variable	Equation satisfied Yes/No
(a)	\[10y=80\]	\[y=10\]
(b)	\[10y=80\]	\[y=8\]
(c)	\[10y=80\]	\[y=5\]
(d)	\[4l=20\]	\[l=20\]
(e)	\[4l=20\]	\[l=80\]
(f)	\[4l=20\]	\[l=5\]
(g)	\[b+5=9\]	\[b=5\]
(h)	\[b+5=9\]	\[b=9\]
(i)	\[b+5=9\]	\[b=4\]
(j)	\[h-8=5\]	\[h=13\]
(k)	\[h-8=5\]	\[h=8\]
(l)	\[h-8=5\]	\[h=0\]
(m)	\[p+3=1\]	\[p=3\]
(n)	\[p+3=1\]	\[p=1\]
(o)	\[p+3=1\]	\[p=0\]
(p)	\[p+3=1\]	\[p=-1\]
(q)	\[p+3=1\]	\[p=-2\]

TIPS Firstly, put the given value of variable in LHS of given equation and simplify. If value of LHS is equal to RHS, then that value of variable will satisfy the equation otherwise not.                

Answer:

               

S. No.	Equation	Value of variable	Equation satisfied Yes/No
(a)	\[10y=80\]	\[y=10\]	No, on putting \[y=10\] in LHS equation \[10y=80,\]we get  \[LHS=10\times 10=100\ne 80\]i.e. \[LHS\ne RHS\]
(b)	\[10y=80\]	\[y=8\]	Yes, on putting \[y=8\] in LHS of equation \[10y=80,\]we get\[LHS=10\times 8=80=80\]i.e. LHS = RHS
(c)	\[10y=80\]	\[y=5\]	No, on putting \[y=5\] in LHS of equation \[10y=80,\] we get \[LHS=10\times 5=50\ne 80\] i.e. \[LHS\ne RHS\]
(d)	\[4l=20\]	\[l=20\]	No, on putting \[l=20\] in LHS of equation \[4l=20,\]we get  \[LHS=4\times 20=80\ne 20\]i.e. \[LHS\ne RHS\]
(e)	\[4l=20\]	\[l=80\]	No, on putting \[l=80\] in LHS of equation \[4l=20,\]we get \[LHS=4\times 80=320\ne 20\]i.e.  LHS= RHS
(f)	\[4l=20\]	\[l=5\]	Yes, on putting \[l=5\] in LHS of equation \[4l=20,\]we get \[LHS=4\times 5=20=20\] i.e. LHS = RHS
(g)	\[b+5=9\]	\[b=5\]	No, on putting \[b=5\] in LHS of equation \[b+5=9,\] we get \[LHS=5+5=10\ne 9\] i.e. LHS = RHS
(h)	\[b+5=9\]	\[b=9\]	No, on putting \[b=9\] in LHS of equation \[b+5=9,\]we get \[LHS=9+5=14\ne 9\]i.e. \[LHS\ne RHS\]
(i)	\[b+5=9\]	\[b=4\]	Yes, on putting \[b=4\] in LHS of equation \[b+5=9,\] we get \[LHS=4+5=9=9\]i.e. LHS = RHS
(j)	\[h-8=5\]	\[h=13\]	Yes, on putting \[h=13\] in LHS of equation \[h-8=5,\] we get \[LHS=138=5=5\] i.e. LHS = RHS
(k)	\[h-8=5\]	\[h=8\]	No, on putting \[h=8\] in LHS of equation \[h-8=5,\]we get \[LHS=8-8=0\ne 5\]i.e. \[LHS\ne RHS\]
(l)	\[h-8=5\]	\[h=0\]	No, on putting \[h=0\] in LHS of equation \[h-8=5,\]we get \[LHS=0-8=-8-\ne 5\]i.e. \[LHS\ne RHS\]
(m)	\[p+3=1\]	\[p=3\]	No, on putting \[p=3\] in LHS of equation \[p+3=1,\]we get \[LHS=3+3=6\ne 1\]i.e. \[LHS\ne RHS\]
(n)	\[p+3=1\]	\[p=1\]	No, on putting \[p=1\] in LHS of equation \[p+3=1,\]we get \[LHS=1+3=4\times 1\] 1 i.e. \[LHS\ne RHS\]
(o)	\[p+3=1\]	\[p=0\]	No, on putting \[p=0\] in LHS of equation \[p+3=1,\]we get \[LHS=0+3=3\ne 1\] i.e. \[LHS\ne RHS\]
(p)	\[p+3=1\]	\[p=-1\]	No, on putting \[p=1\] in LHS of equation \[p+3=1,\]we get \[LHS=-1+3=2\ne 1\] i.e. \[LHS\ne RHS\]
(q)	\[p+3=1\]	\[p=-2\]	Yes, on putting \[p=-2\] in LHS of equation \[p+3=1,\] we get \[LHS=-2+~3=1\]i.e. \[LHS\ne RHS\]