Graphing inequalities

**foofergutierrez** · August 16th 2006, 10:32 PM

Can somebody please show me how to graph inequalities? I don't get it and my homework is way late so please HELP!

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

THANKS

**Quick** · August 17th 2006, 04:36 AM

Do you mean on a graph or number line?

**ThePerfectHacker** · August 17th 2006, 04:54 AM

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.

**Soroban** · August 17th 2006, 06:50 AM

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.

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.

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.

**foofergutierrez** · August 17th 2006, 12:04 PM

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.

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!!!

**ThePerfectHacker** · August 17th 2006, 12:10 PM

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,
$x>3.5$
---

2(x+28)<6(x)]

Open parantheses,
$2x+56<6x$
Remove the x's (subtract 2x),
$56<4x$
Divide by 4,
$14<x$

**foofergutierrez** · August 17th 2006, 12:36 PM

Thanks a million! It makes perfect sense to me now, you are a lifesaver

.

**ThePerfectHacker** · August 17th 2006, 12:40 PM

Originally Posted by foofergutierrez

you are a lifesaver

I know.
From where I come from everyone call me "God".

**Soroban** · August 17th 2006, 02:32 PM

Hello again, foofergutierrez!

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.

Thread: Graphing inequalities

LinkBack

Thread Tools

Display