Interesting inequality question

[Qwertyuiop[]]

Hi, I have this question concerning an inequality bx + 3< 0 and 5 options. They are asking for the set of solution of this inequality. Initially i just solved for x which gives x < -3 /b and that is the option a) but it's not correct : P. The correct answer if b). Usually what i do with inequalities is i just replace the inequality sign with = sign and solve for x. Then draw a number line and substitute suitable values in the original equation to find what values make the inequality true. But in this case we have a variable b so don't know what values i need to substitute in LOL .



EDIT: While writing this question I just realized that b could be negative, they didn't say b has to be greater than 0 . So if b is -ve, dividing by a negative values flips the sign of the inequality so we get x < -3/b OR x > -3/b. Is that correct ?

 

[Zermelo]

Hi, I have this question concerning an inequality bx + 3< 0 and 5 options. They are asking for the set of solution of this inequality. Initially i just solved for x which gives x < -3 /b and that is the option a) but it's not correct : P. The correct answer if b). Usually what i do with inequalities is i just replace the inequality sign with = sign and solve for x. Then draw a number line and substitute suitable values in the original equation to find what values make the inequality true. But in this case we have a variable b so don't know what values i need to substitute in LOL .



EDIT: While writing this question I just realized that b could be negative, they didn't say b has to be greater than 0 . So if b is -ve, dividing by a negative values flips the sign of the inequality so we get x < -3/b OR x > -3/b. Is that correct ?

Yup, you figured it out ? (the edit part)
I would not reccomend solving inequalities by just solving an equation and looking at the number line, because of this example! Multypling by negative values changes the inequality

 

[Steven G]

I am sorry but I disagree with your answer.
The correct answer is:

x < -3/b if b>0 OR x > -3/b when b<0 and no solution if b=0 (since 3<0 is false--what if the problem had been bx + 3>0??)

 

[Qwertyuiop[]]

I am sorry but I disagree with your answer.
The correct answer is:

x < -3/b if b>0 OR x > -3/b when b<0 and no solution if b=0 (since 3<0 is false--what if the problem had been bx + 3>0??)

Well the question says b cannot equal 0. But I was thinking, if b < 0 doesn't the - sign cancel out ? It should become x > 3/b ? Or am i missing something : D

 

[topsquark]

Well the question says b cannot equal 0. But I was thinking, if b < 0 doesn't the - sign cancel out ? It should become x > 3/b ? Or am i missing something : D

Yes, the negative sign would cancel out... once you made the substitution for the numerical value of b. If you leave it as b then you can't cancel anything. For example, if b < 0 then is -3b = 3b? But if we know that b = -2 then -3b = -3 (-2) = 6.

-Dan

 

[Qwertyuiop[]]

Yes, the negative sign would cancel out... once you made the substitution for the numerical value of b. If you leave it as b then you can't cancel anything. For example, if b < 0 then is -3b = 3b? But if we know that b = -2 then -3b = -3 (-2) = 6.

-Dan

ok got it thanks