need some help

[josh123]

Solve: (x - 2)/(x - 1) < 3

a. 2 < x < 5/2
b. 1/2 < x < 1
c. 2 < x < 7/2
d. -5/2 < x < -2
e. x < 2 or x > 5/2
f. x < 1/2 or x > 1
g. x < 2 or x > 7/2 would this be it?
h. x < - 5/2 or x > -2
i. none of these

x-2/x-1 - 3 <0
(x-2)-3 (x-1)/(x-1)

now what can i do?

 

[stapel]

Solve: (x - 2)/(x - 1) < 3
x-2/x-1 - 3 <0
(x-2)-3 (x-1)/(x-1)
now what can i do?

First, use parentheses, like the original question did, because "(x - 2)/(x - 1)" means:

. . . . .\(\displaystyle \large{\frac{x\mbox{ }-\mbox{ }2}{x\mbox{ }-\mbox{ }1}}\)

...but "x - 2/x - 1" means:

. . . . .\(\displaystyle \large{x\mbox{ }-\mbox{ }\frac{2}{x}\mbox{ }-\mbox{ }1}\)

Then convert things to a common denominator:

. . . . .[(x - 2) - 3(x - 1)] / (x - 1)

. . . . . . . = (1 - 2x) / (x - 1) < 0

Then figure out where this rational, (1 - 2x) / (x - 1), is either zero or else undefined. These zeroes and undefined points will divide the number line into intervals (three of them, for this exercise). Pick an x-value inside each interval, and plug it into the rational. The solution will be the interval(s) for which the x-value gave you a negative ("< 0") value.

Hope that helps.

Eliz.

Edit: Note correction above.

 

[josh123]

-5/2 < x < -2

 

[soroban]

Hello, josh123!

Solve: \(\displaystyle \;\frac{x\,-\,2}{x\,-\,1}\;<\;3\)

\(\displaystyle a)\; 2\,<\,x\,<\,\frac{5}{2}\;\;b)\;\frac{1}{2}\,<\,x\,<\,1\;\;c)\;2\,<\,x\,<\,\frac{7}{2}\;\;d)\;-\frac{5}{2}\,<\,x\,<\,-2\;\;e)\;x\,<\,2\) or \(\displaystyle x\,>\,\frac{5}{2}\)

\(\displaystyle f)\;x\,<\,\frac{1}{2}\) or \(\displaystyle x\,>\,1\;\;g)\;x\,<\,2\) or \(\displaystyle x\,>\,\frac{7}{2}\;\;h)\;x\,<\,-\frac{5}{2}\) or \(\displaystyle x\,>\,-2\;\;i)\;\text{none of these}\)

There is an elementary approach to these problems.
. . It may take longer, but the reasoning is simple.


First, solve the <u>equality</u>: \(\displaystyle \;\frac{x\,-\,2}{x\,-\,1}\:=\:3\)

. . We get: \(\displaystyle \;x\,-\,2\:=\:3x\,-\,3\;\;\Rightarrow\;\;x\,=\,\frac{1}{2}\)

Also, we note that "something funny" happens when \(\displaystyle x\,=\,1\).
. . We get 0 in the denominator.

We have two values of \(\displaystyle x\) to test: \(\displaystyle \;\frac{1}{2}\) and \(\displaystyle 1\).
. . The two values divide the number line into <u>three</u> intervals.
. . . - - - - + - - + - - - - - -
. . . . . . . .\(\displaystyle \frac{1}{2}\;\;\;1\)

. . Test a value from each interval in the original inequality.

On the interval \(\displaystyle \left(-\infty,\,\frac{1}{2}\right)\), pick a value, say, \(\displaystyle x = 0\)
. . We have: \(\displaystyle \;\frac{0\,-\,2}{0\,-\,1}\:=\:\frac{-2}{-1}\;=\;2\)
. . Does this satisfy the inequality? (Is it less than 3?) . . . yes!
. . Part of the answer is: \(\displaystyle \;x\,<\,\frac{1}{2}\)

On the interval \(\displaystyle \left(\frac{1}{2},\,1\right)\), pick a value, say, \(\displaystyle x = \frac{3}{4}\)
. . We have: \(\displaystyle \;\frac{\frac{3}{4}-2}{\frac{3}{4}-1}\:=\:\frac{-\frac{5}{4}}{-\frac{1}{4}}\:=\:5\)
. . Does this satisfy the inequality? (Is it less than 3?) . . . no.
. . So this interval is <u>not</u> part of the answer.

On the interval \(\displaystyle (1,\,\infty)\), pick a value, say, \(\displaystyle x=2\)
. . We have: \(\displaystyle \;\frac{2-2}{2-1}\,=\,0\)
. . Does this satisfy the inequality? (Is it less than 3?) . . . yes!
. . Part of the answer is: \(\displaystyle \;x\,>\,1\)


The answer is: \(\displaystyle \;x\,<\,\frac{1}{2}\) or \(\displaystyle x\,>\,1\) . . . answer (f)