If x>14/13 be the solution to the inequality bx-(a+b)<0, where a and b

Hi pratiksha1998,

The only problem with this solution is that you assumed b=13. But, we know that b is negative. Hence b=-13, which gives a=-1.
Using these values of a and b will give:
-x-2-13<0
-x-15<0
x+15>0
x>-15 (B)

What do you think?

Yes I agree with you, as I am dividing both sides by -b, so b should be taken as -13
The answer choice should be B x>-15 IMO as well

Can somebody please explain why is b negative? How are you assuming that b is negative?

 
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\(bx-(a+b)<0\);

\(bx < a+b\).

If b were positive, we'd divide the above by positive b, keep the sign and would get \(x < \frac{a+b}{b}\). However, since we are given that the solution is \(x>\frac{14}{13}\), with > sign instead of the < sign, then b must be negative for the sign of \(bx < a+b\) to flip when we divide by b. Hence, we get \(x > \frac{a+b}{b}\) instead and thus \(\frac{a+b}{b} = \frac{14}{13}\).

Next, \(\frac{a+b}{b} = \frac{14}{13}\) simplifies to \(\frac{a}{b} = \frac{1}{13}\). Since we already established that b is negative, then a must also be negative for their ratio to equal a positive number.

Now, let's work with \(ax+2a+b<0\):

\(ax+2a+b<0\)

\(ax < -2a-b\)

Divide by a, which is negative and flip the sing:

\(x > -2-\frac{b}{a}\)

Since \(\frac{a}{b} = \frac{1}{13}\), then \(\frac{b}{a}=13\). Thus, we get:

\(x > -2-13\)

\(x > -15\)

Answer: B.
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