2007-05-17 16:16:16 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Answer: Y = - 2 or (+ 5/3)
STEPS:
Given equation is
3 y^2 + y - 10 = 0
Factorizing the equation, we get
3 y^2 +6y-5y - 10 = 0
=>. 3y ( y + 2) - 5 ( y + 2) = 0
=> (3y - 5) * ( y + 2) = 0
Either 3y - 5 should be = 0; in that case y = 5/3...Answer (1)
Or y+2 must be = 0 In that case y = - 2 .... Answer (2)
Answer: Y = - 2 or (+ 5/3)
2007-05-17 16:21:28 · answer #1 · answered by Anonymous · 1⤊ 0⤋
3y^2 + y -10 = 0 (factorization)
(3y - 5) (y + 2) = 0
3y - 5 = 0 => y = 5/3
y + 2 = 0 => y = -2
2007-05-17 16:24:13 · answer #2 · answered by frank 7 · 0⤊ 0⤋
3y^2 +y - 10 = 0
factor
(3y-5)( y +2) = 0
3y -5 = 0
3y = 5
y = 5/3
y + 2 = 0
y = -2
2007-05-17 16:24:09 · answer #3 · answered by lizzie 3 · 0⤊ 0⤋
3y^2+y-10=0
3y^2+6y-5y-10=0
3y(y+2)-5(y+2)=0
(y+2)(3y-5)=0
therefore, y+2=0 or 3y-5=0
ans:y= -2 or 5/3
2007-05-17 16:38:49 · answer #4 · answered by priya 2 · 0⤊ 0⤋
3y² + y - 10 = 0
(3y - 5).(y + 2) = 0
y = 5 / 3, y = - 2
Can be checked using quadratic formula:-
y = [- 1 ± â(1 + 120)] / 6
y = [- 1 ± â(121)] / 6
y = [- 1 ± 11)] / 6
y = 10/6 , y = - 12 / 6
y = 5 / 3 , y = - 2 (as above)
2007-05-18 02:36:53 · answer #5 · answered by Como 7 · 0⤊ 0⤋
3y^2+y-10=0
3y^2+6y-5y-10=0
3y(y+2)-5(y+2)=0
(y+2)(3y-5)=0
Either y+2=0 or 3y-5=0
y= -2 or 5/3 ans
2007-05-17 16:25:50 · answer #6 · answered by alpha 7 · 0⤊ 0⤋
(3y-5)(y+2)
y=-2, 5/3
2007-05-17 16:26:13 · answer #7 · answered by Rob 2 · 0⤊ 0⤋
first subract 10 from both sides and then factor.
3y^2 +y -10 = 0
(3y -5)(y+2) =0 So
3y - 5 = 0 ....................... y = 5/3
y+2 =0 .......................... y = -2
2007-05-17 16:24:04 · answer #8 · answered by Trini 3 · 0⤊ 0⤋
3y^2+y=10
3y^2+y-10=0
(3y-5)(y+2)=0
y=5/3
y=-2
2007-05-17 16:23:04 · answer #9 · answered by misshahila 2 · 0⤊ 0⤋
3y^2+y-10=0
a=3
b=1
c=-10
y= - b+ or - sqrt ( b^2 - 4 a c )/ 2a.
just replace
2007-05-17 16:46:38 · answer #10 · answered by Anonymous · 0⤊ 0⤋