algebra: solve x^(-3) = 1/27; 2/(x-1) > 3/(x+2)

[dericteong]

I'm not sure this is Beginning algebra or Advanced algebra.
Pls help for these equation.. Thanks.

 

[dericteong]

I tried this.. for Q1.
Pls help to check..

 

[dericteong]

Stuck at 2nd question. Can anyone help? I noticed that we need to find the value for numerator is zero and denominator is zero, then to construct a table.

 

[stapel]

You ended up with the following inequality:

. . . . .\(\displaystyle \L \frac{x\, -\, 10}{(x\, -\, 2)(x\, +\, 2)}\, <\, 0\)

(By the way, thank you for showing your work so nicely!!)

It is, of course, simpler to find where the rational expression is equal to zero (namely, where the numerator is zero) and where the rational expression is undefined (namely, where the denominator is zero). These values are x = -2, x = 2, and x = 10. These values split the number line into intervals:

. . . . .\(\displaystyle \L (-\infty,\, -2),\, (-2, 2),\, \mbox{ and }\, \(2, +\infty)\)

To find the sign of the rational expression on each interval, pick some x-value inside each interval, and evaluate. The particular value isn't important; it only matters whether the value is positive or negative.

. . . . .\(\displaystyle \L x\,=\,-3:\, \frac{x\, -\, 10}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{-3\, -\, 10}{(-3\, -\, 2)(-3\, +\, 2)}\, =\, \frac{-13}{(-5)(-1)}\, =\, \frac{-13}{5}\, <\, 0\)

. . . . .\(\displaystyle \L x\, =\,\: 0:\,\: \frac{x\, -\, 10}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{0\, -\, 10}{(0\, -\, 2)(0\, +\, 2)}\, =\, \frac{-10}{(-2)(2)}\, =\, \frac{5}{2}\, >\, 0\)

. . . . .\(\displaystyle \L x\, =\,\: 3:\,\: \frac{x\, -\, 10}{(x\, -\, 2)(x\, +\, 2)}\, =\, \frac{3\, -\, 10}{(3\, -\, 2)(3\, +\, 2)}\, =\, \frac{-7}{(1)(5)}\, =\, -\frac{7}{5}\, <\, 0\)

The solution is any interval(s) on which the rational expression is, as you observed, less than zero.

Eliz.

 

[wjm11]

Find the solution set of

2/(x-1) > 3/(x+2)

(I think a denominator inadvertently got changed from "x-1" to "x-2" in your solution.)

First of all, x cannot equal 1 or –2, as the expressions are undefined for these values of x.
Also note that the left and right sides of the expression are equal at x = 7, as follows:
Cross multiply, then rearrange to solve:

2(x+2) = 3(x-1)
2x + 4 = 3x – 3
x = 7

Examine the various intervals:
(-infinity, -2); statement is TRUE
(-2, 1); statement is FALSE
(1, 7); statement is TRUE
(7, infinity): statement is FALSE

 

[stapel]

(I think a denominator inadvertently got changed from "x-1" to "x-2" in your solution.)

I didn't even notice that. Thank you for catching the error!

Note to dericteong: The method outlined will work for any exercise of this sort. Just use the corrected denominator to get the corrected intervals.

Eliz.