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!
+128406 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)
+128406 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)
