question_answer 1)

Fill in the blanks with >, < or = sign. (a) \[(3)+(6)\_\_\_(3)(6)\] (b)\[(21)(10)\_\_\_(31)+(11)\] (c) \[45(11)\_\_\_57+(4)\] (d) \[(25)(42)\_\_(42)(25)\] TIPS Firstly, find the value of LHS (left hand side) and RHS (right hand side). Then, put the sign > (greater than), < (less than) or = (equal to) by using these rules (i) Every positive integer > 0 then negative integer. (ii) Every negative integer < 0 then positive integer. (iii) If LHS and RHS give same value, then use equal to (=)

Answer:

(a) We have, \[(3)+(-6)\_\_(3)(6)\] LHS \[=(3)+(-6)=-9\] RHS \[=(3)(6)=(3)+\](Additive inverse of ?6) \[=3+6=3\] Since, ?9 is a negative integer and 3 is a positive integer. \[\therefore \] \[-93\] (b) We have, \[(21)(10)\_\_(31)+(11)\] LHS \[=(21)(10)=21+\](Additive inverse of ?10) \[=21+10=11\] RHS \[=(31)+(11)=42\] Here, both are negative integers but ?42 is to the left of ?11. \[\therefore \] \[-11-42\] (c) We have, \[45(11)\_\_57+(4)\] LHS \[=45(11)=45+11\] [\[\because \]Additive inverse of ?11 is 11] = 56 \[\text{RHS}=57+(-4)=53+4+(-4)=53+0=53\] Here, both are positive integers but 56 is to the right of 53. \[\therefore \] \[5653\] (d) We have, \[(-25)-(-42)\_\_\_\_-42-<-25)\] \[LHS=(-25)-(-42)=-25+\](Additive inverse of ?42) \[=-25+42=17\] \[RHS=-42-(-25)=-42+\](Additive inverse of ?25) \[=-42+25=-17\] Here, 17 is a positive integer and ? 7 is a negative integer. \[\therefore \] \[17-17\]