question_answer 28)

Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation. (a) \[5m=60\]                    (10, 5, 12, 15) (b) \[n+12=20\]                (12, 8, 20, 0) (c) \[p-5=5\]                       (0, 10, 5, ?5) (d) \[\frac{q}{2}=7\]                                       (7, 2, 10, 14) (e) r ? 4 = 0                         (4, ?4, 8, 0) (f) \[x+4=2\]                      (?2, 0, 2, 4) TIPS Firstly, write the LHS and RHS of given equation. Then, put given values one by one in LHS. If value of LHS is equal to RHS, then that value of variable is the solution of given equation otherwise not.                   

Answer:

                (a) Given, equation is \[5m=60\] Here,     LHS = 5 m and RHS = 60 Now, for \[m=10,\] \[LHS=5\times 10=50\ne HS\] So, \[m=10\] does not satisfy the equation.                              For         \[m=5,\]                 \[LHS=5\times 5=25\ne HS\] So, \[m=5\] does not satisfy the equation. For \[m=12\]      \[LHS=5\times 12=60=RHS\] So, \[m=12\] is the solution of given equation. Form = 15,           \[LHS=5\times 15=75\ne RHS\] So, \[m=15\] does not satisfy the equation. Hence, 12 is a solution of equation \[5m=60\]. (b) Given, equation is \[n+12=20\] Here,     \[LHS=n+12\] and \[RHS=20\] Now, for \[n=12\]            \[LHS=12+12=24\ne RHS\] So, \[n=12\] does not satisfy the equation. For \[n=8,\]        \[LHS=8+12=20=RHS\] So, \[n=8\] is the solution of given equation. Now, for \[n=20,\] \[LHS=20+12=32\ne RHS\] So, \[n=20\] does not satisfy the equation. For \[n=0,\]        \[LHS=0+12=12\ne RHS\] So, \[n=0\] does not satisfy the equation. Hence, \[n=8\] is a solution of equation \[n+12=20\]. (c) Given, equation is \[p-5=5\] Here, \[LHS=p-5\] and \[RHS=5\] Now, for \[p=0\],             \[LHS=0-5=-5\ne RHS\] So, \[p=0\] does not satisfy the equation. For \[p=10,\]      \[LHS=10-5=5=RHS\] So, \[p=10\] is the solution of given equation. Now, for \[p=5,\]             \[LHS=5-5=0\ne RHS\] So, \[p=5\] does not satisfy the equation. Now, for \[p=-5,\] \[LHS=-5-5=-10,\ne RHS\] So, \[p=-5\] does not satisfy the equation. Hence, \[p=10\] is a solution of equation \[p-5=5\]. (d) Given equation is \[\frac{q}{2}=7.\] Here,     \[LHS=\frac{q}{2}\] and \[RHS=7\] Now, for \[q=7,\]             \[LHS=\frac{7}{2}=3\frac{1}{2}\ne RHS\] So, \[q=7\] does not satisfy the equation. Now, for \[q=2,\] \[LHS=\frac{2}{2}=1\ne RHS\] So, \[q=2\] does not satisfy the equation. Now, for\[q=10,\] \[LHS=\frac{10}{2}=5\ne RHS\] So, \[q=10\] does not satisfy the equation. For \[q=14,\]      \[LHS=\frac{14}{2}=7=RHS\] So, \[q=14\]is the solution of given equation. Hence, \[q=14\] is a solution of equation \[\frac{q}{2}=7.\] (e) Given equation is \[r-4=0\] Here, \[LHS=r4\] and \[RHS=0\] Now, for \[r=-4,\] \[LHS=-4-4=-8\ne RHS\] So, \[r=-4\] is the solution of given equation. Now, for \[r=8,\]  \[LHS=8-4=4\ne RHS\] So, \[r=8\] does not satisfy the equation. Now, for \[r=0,\]  \[LHS=0-4=-4\ne RHS\] So, \[r=0\] does not satisfy the equation. Hence, \[r=4\] is a solution of equation \[r-4=0\]. (f) Given equation is \[x+4=2\] Here, \[LHS=x+4\] and \[RHS=2\] Now, for \[x=-2,\] \[LHS=-2+4=2=RHS\] So, \[x=-2\] is the solution of given equation. Now, for \[x=0,\]  \[LHS=0+4=4\ne RHS\] So, does not satisfy the equation. Now, for \[x=2,\] \[LHS=2+4=6\ne RHS\] So, does not satisfy the equation. Now, for \[x=4,\]             \[LHS=4+4=8\ne RHS\] So, \[x=4\] does not satisfy the equation. Hence, \[x=-2\]is a solution of equation \[x+4=2\].