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) | \[p+3=1\] | \[p=-2\] |
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. \[LHS\ne RHS\]
You need to login to perform this action.
You will be redirected in
3 sec