marcodonzelli wrote:

walker wrote:

A

we can solve it by drawing of by algebra.

L: \(y=ax+b\) and a e \((-2,+\infty)\)

the x-intercept of line L is b

1. \(-7=a*3+b\) ==> \(b=-3*a-7\) b e \((-\infty,-1)\) suff.

2. \(-9=a*5+b\) ==> \(b=-5*a-9\) b e \((-\infty,1)\) insuff.

yep

I think C is correct

x-intercept of L is under control not only of b, but also a.

x= - b/a

1. x = (-3a -7)/a =-3 - 7/a, x-intercept wil (-) if a>0 and will (+) if -2<a<0 --> not sufficient

2. x = - 5 - 9/a --> not sufficient

1 and 2 combined: a = 1, b = - 13, so x - intercept negative --> suff.

C

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