# How to solve this in a mathematical way

Help

#### Attachments

• 47.3 KB Views: 26

#### Dr.Peterson

##### Elite Member
What "non-mathematical" way have you tried? You may be expecting something unrealistic.

It's perfectly "mathematical" to just think logically, without following a routine "method". Show us your thoughts, and we may tell you you're doing it in the best possible way. (I can think of several approaches that are equally good, so I don't want to just give you a method that may not be natural for you.)

#### MethMath11

##### Junior Member
I
What "non-mathematical" way have you tried? You may be expecting something unrealistic.

It's perfectly "mathematical" to just think logically, without following a routine "method". Show us your thoughts, and we may tell you you're doing it in the best possible way. (I can think of several approaches that are equally good, so I don't want to just give you a method that may not be natural for you.)
Well, I tried every option and see if it fits the equation, it took me approximately 5 min to get the answer, so I was hoping to get a mathematical way to solve it so if there's something like this in my test soon, I could use a mathematical way rather than a guessing game

#### pka

##### Elite Member
Well, I tried every option and see if it fits the equation, it took me approximately 5 min to get the answer, so I was hoping to get a mathematical way to solve it so if there's something like this in my test soon, I could use a mathematical way rather than a guessing game .
I would almost always advise students to look to comparisons to zero.
Do you see that the question is equivalent to $$\displaystyle (a-3)x<b-2~?$$
Because $$\displaystyle x<0$$ we see at once that $$\displaystyle a>3$$ must be the case. Now the inequality is strict so $$\displaystyle a\ne 3$$, thus the cases are fewer.

#### ksdhart2

##### Senior Member
Here's the way I solved it, although I discovered in doing so that two of the answer choices are correct, so that's annoying. Begin with the given inequality and isolate the terms a bit:

$$ax + 2 < 3x + b$$
$$ax - 3x + 2 - 2 < 3x - 3x + b - 2$$
$$(a - 3)x < b - 2$$

Next we'll consider two cases. If $$a > 3$$ then $$a - 3 > 0$$ and dividing by that quantity will not change the direction of the inequality:

$$\displaystyle x < \frac{b - 2}{a - 3}$$

Now let's make a sign argument. We're given that $$x < 0$$ so we really have:

$$\displaystyle \text{(negative)} < \frac{b - 2}{\text{(positive)}}$$

The only way for this to be true for every negative number is if $$b - 2 \ge 0$$ (why?), so that gives us one possible solution $$a > 3, \: b \ge 2$$. This doesn't exactly match one of the given answer choices, but it does imply that answer choice (a) is valid (why?).

Going back a step, we can consider another case. If $$a < 3$$ then $$a - 3 < 0$$ and dividing by that quantity will change the direction of the inequality:

$$\displaystyle x > \frac{b - 2}{a - 3}$$

Making our sign argument again:

$$\displaystyle \text{(negative)} > \frac{b - 2}{\text{(negative)}}$$

From this we can see that $$b > 2$$ is a requirement, but we can also see that the right-hand side of the inequality will always be some fixed number, and there's no possible way we can guarantee that every negative number will be greater than some fixed negative number (why?). Therefore, we can conclude that if $$a < 3$$ there is no solution.

Finally, we can consider the last case. If $$a = 3$$ then $$a - 3 = 0$$ and we cannot divide by that quantity. However, we can see that the inequality then becomes:

$$0 < b - 2$$

And this is true so long as $$b - 2 > 0 \implies b > 2$$. This gives us another possible solution $$a = 3, \: b > 2$$ which is answer choice (e).

Edit: Whoops! I made a sign error of my own. I've fixed this now, and also inserted a caveat to make my meaning more explicit. Sorry about any confusion this may have caused.

Last edited:

#### Dr.Peterson

##### Elite Member
My thinking closely parallels ksdhart2, but focused on the graph.

Consider the line y = (a-3)x - (b-2). We want this to be below the x-axis for all x<0. So either the slope has to be positive and the y-intercept non-positive, or the slope could be zero and the y-intercept negative. These lead to answers A (and more) and E respectively. Less formal, but quicker to work through, I think.

@MethMath11, what solution did you get? And what topics have you been learning that might suggest what method is expected?

EDIT: I made some small corrections in the conditions.

Last edited: