(a) \(LHS= (-3) + (-6) = - 3 - 6 = – (3 + 6) = – 9\)
\(RHS = (-3) – (-6) = (-3) + 6 = 3\)
Here, \(– 9 < 3\)
Therefore, \((- 3) + (- 6) < (- 3) – (- 6)\)
(b) \(LHS = (-21) – (-10) = (-21) + 10 = -11\)
\(RHS = (-31) + (-11) = – (31 + 11) = – 42\)
Here,\( -11 > - 42\)
Therefore, \((-21) - (-10) > (-31) + (-11)\)
(c) \(LHS = 45 – (-11) = 45 + 11 = 56\)
\(RHS = 57 + (- 4) = 57 - 4 = 53\)
Here, \(56 > 53\)
Therefore, \(45 – (-11) > 57 + ( -4)\)
(d) \(LHS = (-25) – (-42) = - 25 + 42 = 17\)
\(RHS = (- 42) – (-25) = - 42 + 25 = -17\)
Here, \(17 > -17\)
Therefore, \((- 25) – (- 42) > (- 42) – (- 25).\)
Complete the drawing shown in Fig. 9.14 to indicate where the free ends of the two wires should be joined to make the bulb glow