# Solve the inequality...

• February 18th 2009, 06:06 PM
rckstr4lfe
Solve the inequality...
I am working on a few corrections, if someone could help me, I would be very thankful.

The directions: Solve the inequality. Express the solution using set builder notation, interval notation, and graph the solution.

1. 10 - 2|2x - 1| < or equal to 6

Inequality Solution(s): -1/2 < or equal to X < or equal to 3/2
Set Builder Notation: {x| -1/2 < or equal to X < or equal to 3/2}
Interval Notation: [-1/2, 3/2]
and since these were wrong, so was my graph.

2. -1 < or equal to 5 - 2x < or equal to 10

Inequality Solution(s): -5/2 > or equal to x > or equal to 3
Set Builder Notation: {x| -5/2 > or equal to x > or equal to 3}
Interval Notation: [-5/3, 3]
and since these were wrong, so was my graph.

3. x (x - 1) (x - 2) > 0

Inequality Solution(s): 0 < x > 2
Set Builder Notation: {x| x<0 and x>2}
Interval Notation: (0, 2)
and since these were wrong, so was my graph.

The instructor wrote something about using a table/diagram to solve?

I am kind of lost on this concept... (Headbang)
• February 18th 2009, 06:55 PM
mollymcf2009
Quote:

Originally Posted by rckstr4lfe
I am working on a few corrections, if someone could help me, I would be very thankful.

The directions: Solve the inequality. Express the solution using set builder notation, interval notation, and graph the solution.

1. 10 - 2|2x - 1| < or equal to 6

Inequality Solution(s): -1/2 < or equal to X < or equal to 3/2
Set Builder Notation: {x| -1/2 < or equal to X < or equal to 3/2}
Interval Notation: [-1/2, 3/2]
and since these were wrong, so was my graph.
This is correct. How do you know it is wrong? Do you have the answer?

2. -1 < or equal to 5 - 2x < or equal to 10

Inequality Solution(s): -5/2 > or equal to x > or equal to 3 * You forgot to reverse your inequality signs when you divided by -2
Set Builder Notation: {x| -5/2 > or equal to x > or equal to 3}
Interval Notation: [-5/3, 3]
and since these were wrong, so was my graph.

3. x (x - 1) (x - 2) > 0

Have you learned how to use a number line and test numbers to find where these are greater than and less that zero? Look at this link if you need help with that: Solving Inequalities: An Overview

Inequality Solution(s): 0 < x > 2
Set Builder Notation: {x| x<0 and x>2}
Interval Notation: (0, 2)
and since these were wrong, so was my graph.

The instructor wrote something about using a table/diagram to solve?

I am kind of lost on this concept... (Headbang)

Come back if you still need help!!
• February 18th 2009, 07:00 PM
ixo
on the first problem you almost have it the x is greater than 3/2. Numerically your correct though.

$x<6 -10$

$x>\frac{-4}{-2}$ remember when you divide by a neg

$x>-2+1$ and $x>2+1$

$x>-\frac{1}{2}$ and $x>\frac{3}{2}$

$-\frac{1}{2}\frac{3}{2}$
• February 18th 2009, 07:18 PM
Soroban
Hello, rckstr4lfe!

With inequalities, we can do "normal" algebra ... except
. . when multiplying or dividing by a negative number, reverse the inequality.

Quote:

Solve the inequality.
Express the solution using set builder notation, interval notation, and graph the solution.

$1)\;\;10 - 2|2x - 1|\:\leq \:6$

We have: . $10 - 2|2x-1| \;\leq \;6$

Subtract 10: . $-2|2x-1| \;\leq \;-4$

Divide by -2: . $|2x-1| \;\;{\color{red}\geq} \;\;2$

We have: . $\begin{array}{ccccc}2x-1 \:\geq\:2 & \Rightarrow & 2x \:\geq\:3 & \Rightarrow & x \:\geq\:\frac{3}{2} \\ \\
2x-1 \:\leq\:\text{-}2 & \Rightarrow & 2x \:\leq \:\text{-}1 &\Rightarrow & x \:\leq\:\text{-}\frac{1}{2} \end{array}$

Set builder: . $\left\{x\,|\,x \leq \text{-}\tfrac{1}{2}\,\text{ or }\,x \geq \tfrac{3}{2}\right\}$

Interval: . $\left(-\infty,\text{-}\tfrac{1}{2}\right] \:\cup \:\left[\tfrac{3}{2},\infty\right)$

Graph: . $===\bullet - - - \bullet = = =$
. . . . . . . . . $\text{-}\tfrac{1}{2}\qquad\quad\;\tfrac{3}{2}$

Quote:

$2)\;\;-1 \:\leq \:5 - 2x \:\leq\:10$

Subtract 5: . $-6 \;\leq \;-2x\;\leq \;5$

Divide by -2: . $3 \;{\color{red}\geq}\;x\;{\color{red}\geq}\;\tfrac{ 5}{2}$

Set builder: . $\left\{x\,|\,\tfrac{5}{2} \leq x \leq 3 \right\}$

Interval: . $\left[\tfrac{5}{2},3\right]$

Graph: . $- - - \bullet === \bullet - - -$
. . . . . . . . . . $\tfrac{5}{2} \qquad\;\;\;3$

Quote:

$3)\;\;x (x - 1) (x - 2) \:>\: 0$

We see that the expression equals 0 when $x = 0, 1, 2.$

The three values divide the number line into four intervals.
Test a value of $x$ in each interval.

On $(-\infty, 0)$, test $x = \text{-}1\!:\;\;(\text{-}1)(\text{-}2)(\text{-}3) \:=\:\text{-}6$ . . . no

On $(0,1)$, test $x = \tfrac{1}{2}\!:\;\;\left(\tfrac{1}{2}\right)\left( \text{-}\tfrac{1}{2}\right)\left(\text{-}\tfrac{3}{2}\right) \:=\:+\tfrac{3}{8}$ . . . yes

On $(1,2)$, test $x = \tfrac{3}{2}\!:\;\;\left(\tfrac{3}{2}\right)\left( \tfrac{1}{2}\right)\left(\text{-}\tfrac{1}{2}\right) \:=\:\text{-}\tfrac{3}{8}$ . . . no

On $(2,\infty)$, text $x = 3\!:\;\;(3)(2)(1) \:=\:+6$ . . . yes

Set builder: . $\left\{x\,|\,(0 < x < 1)\text{ or }(x > 2)\right\}$

Interval: . $(0,1) \:\cup\: (2,\infty)$

Graph: . $- - - o = = = o - - - o = = =$
. - . - . . . . . . $0 \qquad\quad 1 \qquad\quad\; 2$