Need help in College Algebra

[Karter]

Hi, I need help with solving inequalities. It's been a very long time for me since I've done math, and I need help with understanding inequalities with fractions, and how to understand interval notation. I don't know how to solve inequalities that have a fraction on either side. For instance, x+3/8 - x+5/5 ≥ 3/10. I also need help understanding x+2/4 - x-1/3 = 2. Thank you.

 

[Deleted member 4993]

Hi, I need help with solving inequalities. It's been a very long time for me since I've done math, and I need help with understanding inequalities with fractions, and how to understand interval notation. I don't know how to solve inequalities that have a fraction on either side. For instance, x+3/8 - x+5/5 ≥ 3/10. I also need help understanding x+2/4 - x-1/3 = 2. Thank you.

Do you have:

$\displaystyle \frac{x+2}{4} - \frac{x-1}{3} \ = 2$ ................................................or something else?

 

[Otis]

It's been a very long time for me since I've done math … I don't know how to solve inequalities that have a fraction on either side. For instance, x+3/8 - x+5/5 ≥ 3/10.

Hi Karter. I agree with Subhotosh. I think you mean (x+3)/8 - (x+5)/5 ≥ 3/10. We use grouping symbols to show what goes in numerators and denominators when they contain more than one number.

The denominators are 5, 8 and 10. The number 80 is divisible by each of those. Multiply each side of the inequality by 80, and the denominators will cancel out. Show us what you get.



$\;$

 

[Karter]

Hi, yes that is what I meant. So sorry for the confusion. I was able to figure it out actually, just now need help with -4<3x+2≤18 and another problem. This one seems to be a word problem I can't figure out. The problem is as shown :

64. ) A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time.. The cost formula would be :

C=20+.30(x−60).C=20+.30(x−60).
If you have to keep your bill no greater than $50, what is the maximum calling minutes you can use?

 

[JeffM]

@Karter

The way to solve a constinuous inequality is to solve the related equality first.

Example: $x^2 - 4x - 5 > 0.$

That is continuous (there are no holes or jumps in its graph).

So the first thing to do is to solve the related equality, namely $x^2 - 4x - 5 = 0$.

The solutions to that equality are x = - 1 and x = 5.

Now pick arbitrary numbers to the left of the leftmost solution, to the right of the rightmost solution, and between each solution.

$$\text {If } x = - 2, \text { then } x < - 1 \text { and}\\ x^2 - 4x - 5 = (-2)^2 - 4(-2) - 5 = 4 + 8 - 5 = 7 > 0;\\ \text {If } x = 0 , \text { then } - 1 < x < 5 \text { and}\\ x^2 - 4x - 5 = (0)^2 - 4(0) - 5 = 0 + 0 - 5 = - 5 < 0.\\ \text {If } x = 6, \text { then } 5 < x \text { and}\\ x^2 - 4x - 5 = (6)^2 - 4(6) - 5 = 36 - 24 - 5 = 7 > 0.\\ \text{ Thus, } x^2 - 4x - 5 > 0 \iff x \in (- \infty, \ 1) \ \bigcup \ (5, \ \infty).$$

 

[JeffM]

Hi, yes that is what I meant. So sorry for the confusion. I was able to figure it out actually, just now need help with -4<3x+2≤18 and another problem. This one seems to be a word problem I can't figure out. The problem is as shown :

64. ) A basic cellular package costs $20/mo. for 60 min of calling, with an additional charge of $.30/min beyond that time.. The cost formula would be :

C=20+.30(x−60).C=20+.30(x−60).
If you have to keep your bill no greater than $50, what is the maximum calling minutes you can use?

You are close. The cost function is

$$x \le 60 \implies c = 20.00;\\ x > 60 \implies c = 20.00 + 0.3(x - 60) =\\ 20.00 + 0.3x - 18.00 = 0.3x + 2.$$
That function is continuous. So your inequality is $c \le 50.00$

Now what?