1. Graphing inequalities

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

2. Do you mean on a graph or number line?

3. Originally Posted by Quick
Do you mean on a graph or number line?
Yes
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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,
$\displaystyle 3x>2x-4$
$\displaystyle 3x-2x>2x-2x-4$
$\displaystyle x>-4$
That means a circle above -4 and draw an arrowed line to the right.

4. 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: .$\displaystyle x > 4$

This is the set of all numbers greater than 4.

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

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

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

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

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

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

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

5. 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.

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 $3.00. Again, I thank all who can help me and if I could I would pay ou back somehow! You guys are awesome!!! 6. Originally Posted by foofergutierrez 7(x-3)>5x-14 Open parantheses,$\displaystyle 7x-21>5x-14$Remove the x's$\displaystyle 2x-21>-14$Add 21 to both sides,$\displaystyle 2x>7$Divide by 2,$\displaystyle x>3.5$--- 2(x+28)<6(x)] Open parantheses,$\displaystyle 2x+56<6x$Remove the x's (subtract 2x),$\displaystyle 56<4x$Divide by 4,$\displaystyle 14<x$7. Thank you Perfect Hacker Thanks a million! It makes perfect sense to me now, you are a lifesaver . 8. Originally Posted by foofergutierrez you are a lifesaver I know. From where I come from everyone call me "God". 9. 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 $\displaystyle x$ be the number of minutes we can talk for $3.00 or less. We are charged$\displaystyle 36\!\!\not\!c$for the first minute. We are charged$\displaystyle 21\!\!\not\!c$each for the other$\displaystyle x - 1$minutes. . . This costs us: .$\displaystyle 21(x - 1)$cents. Hence, the total charge is: .$\displaystyle 36 + 21(x - 1)$cents Since we do not want to exceed$3 $\displaystyle (300\!\!\not\!c)$
. . we have: .$\displaystyle 36 + 21(x - 1)\:\leq \:300$

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

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

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

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

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