Q5 of 107 Page 104

The equation which cannot be solved in integers is

(a) Given equation is 5y – 3 = –18


5y = – 18 + 3 [transposing 3 to RHS]


5y = –15


[dividing both sides by 5]


y = –5 (integer)


(b) Given equation is 3x – 9 = 0


3x = 9 [Transposing 9 to RHS]


[Dividing both sides by 3]


x = 3 (integer)


(c) Given equation is 3z + 8 = 3 + z


On transposing z and 8 to LHS and RHS respectively, we get


3z – z = 3 – 8


2z = –5


[dividing both sides by 2]


(neither a positive fraction nor an integer)


(d) Given equation is 9y + 8 = 4y – 7


On transposing 4y and 8 to LHS and RHS respectively, we get


9y – 4y = –7 – 8


5y = –15


[Dividing both sides by 5]


y = – 3 (integer)

More from this chapter

All 107 →