Question

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Answer 1

is it

y + 2 1
----- = ------
y y - 5 ???

Answer 2

yes thats how it is

Answer 3

y + 2 1
----- = ------
y y - 5

cross multiply to get

(y - 5)(y + 2) = 1*y
y² - 5y + 2y - 10 = y
y² - 3y - y - 10 = 0
y² - 4y - 10 = 0

so u get a quadratic equation...

do u know how to solve this n get value of x ?????

Answer 4

no i dont

Answer 5

ok, i'll show u...

Answer 6

thanks

Answer 7

- b ± sqroot(b² - 4ac)
y = ----------------------
2a
where a = coefficient of y² = 1
b = coefficient of y = -4
c = constant terms = - 10

Answer 8

soo the answer is?

Answer 9

so
- (-4) ± sqroot[(-4)² - 4(1)(-10) 4 ± sqroot[16 + 40] 4 ± sqroot[56]
y = ----------------------------- = -------------------- = -----------------
2(1) 2 2

4 ± 2sqroot[14] 2 { 2 ± sqroot[14]}
y = --------------- = ------------------
2 2

y = 2 + sqroot[14] or 2 - sqroot[14]

Answer 10

o.o

Answer 11

if it is clear, pls click GOOD ANSWER button for me ☺☻☺

Answer 12

is the answer 14?
lol

Answer 13

i gave you a medal
but i dont have my answer -.-

Answer 14

sorry I had to go to sleep and logged out before your post reached me....

Since it is a quadratic equation, you get two values of y
y = 2 + sqroot[14]
or
y = 2 - sqroot[14]