# Graphing inequalities

• Aug 16th 2006, 10:32 PM
foofergutierrez
Graphing inequalities
Can somebody please show me how to graph inequalities? I don't get it and my homework is way late so please HELP! :eek:

x>4 - graph the solution for each set of inequalities

3x[U]>2x-4 - solve and graph the solution set of each inequalities

I just need for someone to show and explain how to do these and I can try to do the rest on my own. Any help will be greatly appreciated :o THANKS
• Aug 17th 2006, 04:36 AM
Quick
Do you mean on a graph or number line?
• Aug 17th 2006, 04:54 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
Do you mean on a graph or number line?

Yes
---
In the first problem just, draw a circle on top of 4 (to show you exclude 4) and draw a lined arrow to the right.

In the second one, solve for all x,
$3x>2x-4$
$3x-2x>2x-2x-4$
$x>-4$
That means a circle above -4 and draw an arrowed line to the right.
• Aug 17th 2006, 06:50 AM
Soroban
Hello, foofergutierrez!

You should be familiar with Solving Inequalities.

Solve for the variable as you would with an equation except:
. . when multiplying or dividing by a negative, reverse the inequality.

Quote:

Graph the solution for: . $x > 4$

This is the set of all numbers greater than 4.

. . . $- - - - \circ\!\!=\!=\!=\!=\!=\!=$
. . . . . . . . . $4$

The circle $(\circ)$ indicates that the $4$ is not included.

Quote:

Graph the solution for: . $3x \,\geq \,2x - 4$

Solve for $x:\;\;3x\:\geq \:2x - 4\quad\Rightarrow\quad x \:\geq \:-4$

This is the set of all numbers greater than or equal to $-4.$

. . . $- - \bullet\!\!=\!=\!=\!=\!=\!=$
. . . . . $-4$

The solid dot $(\bullet)$ indicates that the $-4$ is included.

• Aug 17th 2006, 12:04 PM
foofergutierrez
THANKS but I need more help
Ok I kind of understand what to do but here are some other practice equations I would like to see done before my attempt to work my problems so that I know that I am doing them right. :confused:

7(x-3>5x-14

2(x+28)<6(x)

I also have this weird word problem that I cannot make any sense of.

The cost for a long distance call is $.36 for the first minute and$.21 for each additional minute. Write an equality representing the number of minutes a person could talk without exceeding $3.00. Again, I thank all who can help me and if I could I would pay ou back somehow! You guys are awesome!!! :cool: • Aug 17th 2006, 12:10 PM ThePerfectHacker Quote: Originally Posted by foofergutierrez 7(x-3)>5x-14 Open parantheses, $7x-21>5x-14$ Remove the x's $2x-21>-14$ Add 21 to both sides, $2x>7$ Divide by 2, $x>3.5$ --- Quote: 2(x+28)<6(x)] Open parantheses, $2x+56<6x$ Remove the x's (subtract 2x), $56<4x$ Divide by 4, $14 • Aug 17th 2006, 12:36 PM foofergutierrez Thank you Perfect Hacker Thanks a million! It makes perfect sense to me now, you are a lifesaver :rolleyes:. • Aug 17th 2006, 12:40 PM ThePerfectHacker Quote: Originally Posted by foofergutierrez you are a lifesaver I know. From where I come from everyone call me "God". • Aug 17th 2006, 02:32 PM Soroban Hello again, foofergutierrez! Quote: The cost for a long distance call is$0.36 for the first minute
and $0.21 for each additional minute. Write an equality representing the number of minutes a person could talk without exceeding$3.00.

Let $x$ be the number of minutes we can talk for $3.00 or less. We are charged $36\!\!\not\!c$ for the first minute. We are charged $21\!\!\not\!c$ each for the other $x - 1$ minutes. . . This costs us: . $21(x - 1)$ cents. Hence, the total charge is: . $36 + 21(x - 1)$ cents Since we do not want to exceed$3 $(300\!\!\not\!c)$
. . we have: . $36 + 21(x - 1)\:\leq \:300$

Solve for $x:\;\;36 + 21x - 21 \:\leq\:300$

. . . . . . . . . . . . . $21x + 15\:\leq\:300$

. . . . . . . . . . . . . . . . $21x \:\leq \:285$

. . . . . . . . . . . . . . . . . . $x\:\leq \:13.5714...$

Therefore, we can talk up to 13 minutes without going over the \$3 limit.