Question 726814: Stuck and need help on solving this
7x-8y=24
xy^2=1
Found 2 solutions by Alan3354, DrBeeee:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 7x-8y=24
xy^2=1
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xy^2=1
x = 1/y^2
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7x-8y=24
7/y^2 - 8y = 24
Multiply by y^2
7 - 8y^3 = 24y^2
8y^3 + 24y - 7 = 0
The zeroes can be found by graphical or numerical methods.
y =~ 0.284 (the only real # solution, by graphical methods)
The other 2 zeroes are complex numbers.
Are you in a class that is familiar with those?
Answer by DrBeeee(684)  (Show Source):
You can put this solution on YOUR website! Given;
(1) x*y^2 = 1 and
(2) 7x - 8y = 24
Solve by substitution. Solve (1) for x and get
(3) x = 1/(y^2); y not equal to zero.
Place (3) into (2) and get
(4) 7/(y^2) - 8y - 24 = 0
Multiply (4) by y^2 and get
(5) 7 - 8y^3 - 24y^2 =0 or
(6) 8y^3 + 24y^2 - 7 = 0
We know that (6) has one real root because it is an odd power equation in y. To get the real root I set the last two terms equal to zero and solve for y. This gives us
(7) 24y^2 - 7 = 0 or
(8) y^2 = 7/24 which is approximately
(9) y = 1/2
To check this trial value of y, place it into (6) and get
(10) 8*(1/2)^2 + 24*(1/2)^2 - 7 = 0 or
(11) 1 + 6 - 7 = 0
Therefore
(12) (y-1/2)
is a factor of (6). Now use long division of (6) divided by (12) to get
(13) 8y^2 + 28y + 14 = 0 or
(14) 4y^2 + 14y + 7 = 0
The roots of (14) are
(15) y = {(-7+sqrt(21))/4,(-7-sqrt(21))/4}
To get the corresponding values of x, substitute each of the three values of y given by (9), and (14) into (1) and get
(16) x = 1/(1/2)^2 or
(17) x = 4 and
(18) x = {8/(35-7*sqrt(21)), 8/(35 + 7*sqrt(21))}
I'll check (4,1/2) in (2).
Is (7*4 - 8*(1/2) = 24)?
Is (28 - 4 = 24)?
Is (24 = 24)? Yes
Answer: The three solution pairs are
1) (4,1/2),
2) (8/(35-7*sqrt(21)),((-7+sqrt(21))/4), and
3) (8/(35+7*sqrt(21)),((-7-sqrt(21))/4})
P.S. I checked 2) and 3) off line. They're OK.
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