a. y-5 / y-2
b. y+5 / y+2
c. 5/2
d. y-10 / y-4
2007-03-21 03:07:34 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y^2+y-20 / y^2-2y-8
= (y+5)(y-4) / (y-4)(y+2)
= (y+5) / (y+2) where y <> 4
(answer b)
2007-03-21 03:14:04 · answer #1 · answered by seah 7 · 1⤊ 0⤋
Factorise first...
y^2 + y - 20 = (y-4) (y+5)
Y^2 - 2y - 8 = (y+2) (y-4)
(y-4)'s will cancel out, leaving....
(y+5) / (y+2) - so answer B
2007-03-21 10:15:17 · answer #2 · answered by Doctor Q 6 · 0⤊ 0⤋
= (y+5)(y-4)/(y-4)(y+2)
the y+4 in the numerator and denominator cancel out leaving (y+5)/(y+2) as the answer, which is b
2007-03-21 10:14:48 · answer #3 · answered by sunflower 2 · 0⤊ 0⤋
(y+5)(y-4)/ (y-4)(y+2)
= (y+5)/(y-4)
2007-03-21 10:15:26 · answer #4 · answered by bignose68 4 · 0⤊ 0⤋
[y^2 + y - 20] / [y^2 - 2y - 8]
(y + 5)(y - 4) / (y - 4)(y + 2)
(y + 5) / (y + 2)
Answer: b
2007-03-21 10:13:03 · answer #5 · answered by Bhajun Singh 4 · 0⤊ 0⤋
y^2+y-20 / y^2-2y-8
Break numerator and denominator into component factors:
(y-4)(y+5) / (y-4)(y+2)
Cancel (y-4) term in numerator and denominator:
(y+5)/(y+2)
B is correct.
2007-03-21 10:13:48 · answer #6 · answered by MamaMia © 7 · 0⤊ 0⤋
y^2+y-20
=y^2+5y-4y-20
=y(y+5)-4(y+5)
=(y-4)(y+5)
y^2-2y-8
=y^2-4y+2y-8
=y(y-4)+2(y-4)
=(y+2)(y-4)
therefore,
y^2+y-20/y^2-2y-8
={(y-4)(y+5)}/{(y+2)(y-4)}
=(y+5)/(y+2)
So the answer is B
2007-03-21 10:25:20 · answer #7 · answered by Bubblez 3 · 0⤊ 0⤋