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1.One ordered pair $$(a,b)$$satisfies the two equations $$ab^4=384$$and $$a^2b^5=4608$$ . What is the value of in this ordered pair?

2.Real numbers x and y satisfy \begin{align*} x + xy^2 &= 250y, \\ x - xy^2 &= -240y. \end{align*}
Enter all possible values of separated by commas.

Thank You!

Mar 18, 2020
edited by Guest  Mar 18, 2020

#1
+111360
+1

1.

ab^4  = 384     ⇒   a =  384/b^4 ⇒  a^2  = 384^2  / b^8     (1)

a^2b^5  = 4608     (2)

Sub  (1)  into (2)  for a^2   and we have  that

[(384^2) / b^8]  b^5 =  4608    simplify

384^2 / b^3   = 4608    rearrange  as

b^3   = 384^2 / 4608

b^3  = 32

b = ∛32   = ∛[8*4]  = 2∛4

And

a = 384/ [ 2∛4]^4  =  384 / [16 * 4^(4/3) =   24 /  [ 4 * 4^(1/3) =    6 / ∛4  =  6∛4^2 / 4  = (3/2)∛16  =

(3/2)*2 * ∛2  = 3∛2

So

(a,b)   =   ( 3∛2 , 2∛4)

Mar 19, 2020
#2
+111360
+1

2.

x + xy^2  =  250y

x - xy^2   = -240y       add these  and we get that

2x =  10y

x = 5y

So we have  that

5y  + (5y)y^2 = 250y

5y^3  = 245y

5y^3  - 245y  =  0

5y  (y^2  - 49)   = 0

5y ( y - 7) ( y + 7)   = 0

Setting each factor to  0 and solving for y produces

y =0   y =  7   and  y   =-7

So   using  x = 5y....the solutions are

(x,y )  =   (0, 0)    ( 35, 7)   and (-35, - 7)

Mar 19, 2020