Suppose the operations “#” and “*t*” are defined so that:

6 # = 20; 2 # = 4; 10 # = 36; and 5 *t* = 4.5; 10 *t* = 7; 8 *t* = 6.

Then what is *n* if 3 # # *t* = *n*?

We see that:

6 # = 20

2 # = 4

10 # = 36

2 # = 4

10 # = 36

*x*# = 4*x*– 4 = 4(*x*– 1)Also:

5 *t* = 4.5

10 *t* = 7

8 *t* = 6

*x* *t* = *x*/2 + 2

So if 3 # #

8 #

28

16 =

*t*=*n*, then8 #

*t*=*n*28

*t*=*n*16 =

*n*