If x14/13 be the solution to the inequality bx-(a+b)

Question

If x>\frac{14}{13} be the solution to the inequality bx-(a+b)<0 , where a and b are constants, then what is the solution to the inequality ax+2a+b<0 ? A. x < -15 B. x > -15 C. x < 15 D. x > 15 E. None of the above

TestPrepUnlimited · Accepted Answer

We can solve the inequality in terms of a and b, we have bx < (a + b) . Knowing the solution has a change in sign , we can confirm b is negative. Then x > \frac{a + b}{b} is the solution, and by matching the numbers we have \frac{a + b}{b} = \frac{14}{13} , which is \frac{a}{b} = \frac{1}{13} . Although there are many combinations of a and b that work, we may simply pick a pair since this is a PS question. b = -13 and a = -1 for example (again b must be negative, which means a is negative as well!). Then plug in the question: -x + -15 < 0 and x > -15 . Ans: B

Bunuel · Answer

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.

pratiksha1998 · Answer

My Solution:- bx-(a+b)<0 bx14/13, which is +ve) Dividing both sides by -b gives, x>(a+b)/b Therefore, a+b = 14 and b=13 => a = 1 Substituting these values in the second inequality, we get: x+2+13<0 x+15<0 x<-15 => Ans. Note:- This is my approach, not sure if this is the correct way or not. Kindly upvote if you think the approach is correct

ValariKapital · Answer

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?

pratiksha1998 · Answer

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

ROY13 · Answer

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

Kinshook · Answer

If x>\frac{14}{13} be the solution to the inequality bx-(a+b)<0 , where a and b are constants, then what is the solution to the inequality ax+2a+b<0 ? bx < a+b x < (a+b)/b; but x > 14/13; therefore b < 0 \frac{a+b}{b}= \frac{14}{13} \frac{a}{b} = \frac{1}{13} a/b = 1/13; a<0 ax < -2a-b x > \frac{-2a-b}{a} = -2 - b/a = -2 -13 = -15 x > -15 IMO B

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

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