CHALLENGE PROBLEMS!!!

[Guest]

Let r be the number that results when both the base and the exponent of a^b are tripled, where a and b are positive. If r=a^bX^b, where x is positive, then x equals

a) 3 b) 3a^2 c) 27a^s d) 2s^3b e) 3a^2b



Solve for x is the recipical of (1/x-1) is -2


Find all positive interger values of x and y that satisfy the equation 1/x+x/y+1/xy=1




that's it for now...if you get any of them tell me how plz.
Thanks a bunch
-Anna

 

[soroban]

Hello, anna!

Find all positive interger values of x and y that satisfy the equation: 1/x + x/y + 1/xy = 1

Multiply through by xy: . y + x² + 1 .= .xy

. . . . . . . . . . . . . . . . x² + 1 . . . . . . . . . . . .2
Solve for y: . y . = . --------- . = . x + 1 + ------
. . . . . . . . . . . . . . . . .x - 1 . . . . . . . . . . . x - 1

Since y is a positive integer, then (x-1) must divide into 2.
. . . There are only two possible values: .x = 2, 3

Answers: . (x,y) .= .(2,5), (3,5)

 

[Denis]

"Solve for x is the recipical of (1/x-1) is -2"

That should be RECIPROCAL; and your "is" should be "if";
would you like as careless an answer as your question is?

Rule: b/a is the reciprocal of a/b

Now TRY!

 

[soroban]

Hello, anna!

There are some typos in #1 . . .

1) Let r be the number that results when both the base and the exponent of a^b are tripled,

where a and b are positive.

If r = a^bX^b, where x is positive, then x equals

a) 3 . . b) 3a² . . c) 27a^s . . d) 2s³b . . e) 3a²b
.

We have: . (3a)^3b . = . a^bX^b

. . . . . . (3^3b)(a^3b) . = . a^bX^b

. . . . . . (3³)^b(a^b)³ . = . a^bX^b

. . . . . . (27^b)(a^b)³ . = . a^bX^b

Divide both sides by a^b: . 27^b(a^b)² . = . X^b

We have: . X^b . = . (27^b)(a²)^b . = . (27a²)^b

Therefore: . X . = . 27a²