# Thread: Inequality Help.

1. ## Inequality Help.

[These are FRACTIONS. The Slash signs are for FRACTIONS]

$\displaystyle 1/6x + 1/2 < 1/12x + 2/3$

There are four choices to what this could be.

a: $\displaystyle x < 14/3$
b: $\displaystyle x > 14/3$
c: $\displaystyle x < 2$
d. $\displaystyle x > 2$

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$\displaystyle -4 (3x - 6) < 3 (x + 3)$

There are four choices to this also.

a. x < -1
b. x > 1
c. x < 1
d. x > -1

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$\displaystyle 3 |2x - 4| - 5 < 7$

There are four choices to this.

a. 0 < x < 4
b. x < - 4 or x > 0
c. - 4 < x < 0
d. x < 0 or x > 4

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$\displaystyle |x + 1| - 2 > 5$

There are four choices.

a. 6 < x < -8
b. x > 8 or x < -6
c. 8 < x < -6
d. x > 6 or x < -6

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Dave's grades on four tests were 95, 76, 82, 65. What must his lowest grade on the next test be to have an average greater than 80.

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Six less than a number is between (-3) and 11. What real numbers are possible soulutions ?

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Thank you for your help !!!

2. Originally Posted by omgitzbella
[These are FRACTIONS. The Slash signs are for FRACTIONS]

$\displaystyle 1/6x + 1/2 < 1/12x + 2/3$

There are four choices to what this could be.

a: $\displaystyle x < 14/3$
b: $\displaystyle x > 14/3$
c: $\displaystyle x < 2$
d. $\displaystyle x > 2$
Do you mean:

$\displaystyle \frac{1}{6x} + \frac{1}{2} < \frac{1}{12x} + \frac{2}{3}$

or

$\displaystyle \frac{1}{6}~x + \frac{1}{2} < \frac{1}{12}~x + \frac{2}{3}$ ?

RonL

3. Originally Posted by omgitzbella;83389--------------------------------------------------------------------

[tex
-4 (3x - 6) < 3 (x + 3)[/tex]

There are four choices to this also.

a. x < -1
b. x > 1
c. x < 1
d. x > -1
$\displaystyle -4 (3x - 6) < 3 (x + 3)$

expand brackets:

$\displaystyle -12x + 24 < 3 x + 9$

add $\displaystyle 12x$ to both sides:

$\displaystyle 24 < 15 x + 9$

Now subtract $\displaystyle 9$ from both sides:

$\displaystyle 15 < 15 x$

and divide by $\displaystyle 15$, to get?

RonL

4. i meant the second suggestion that you posted. Where x isn't included in the fraction.

5. Originally Posted by omgitzbella

$\displaystyle 3 |2x - 4| - 5 < 7$

There are four choices to this.

a. 0 < x < 4
b. x < - 4 or x > 0
c. - 4 < x < 0
d. x < 0 or x > 4
$\displaystyle 3\cdot |2x-4|-5<7\Leftrightarrow 3\cdot |2x-4|<12\Leftrightarrow$
$\displaystyle \Leftrightarrow|2x-4|<4\Leftrightarrow -4<2x-4<4\Leftrightarrow 0<2x<8\Leftrightarrow 0<x<4$

6. Originally Posted by omgitzbella
$\displaystyle |x + 1| - 2 > 5$

There are four choices.

a. 6 < x < -8
b. x > 8 or x < -6
c. 8 < x < -6
d. x > 6 or x < -6
$\displaystyle |x+1|-2>5\Leftrightarrow |x+1|>7\Leftrightarrow x+1<-7 \ \text{or} \ x+1>7\Leftrightarrow x<-8 \ \text{or} \ x>6$

7. Originally Posted by omgitzbella
[These are FRACTIONS. The Slash signs are for FRACTIONS]
I believe we get the general idea
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Dave's grades on four tests were 95, 76, 82, 65. What must his lowest grade on the next test be to have an average greater than 80.
$\displaystyle \frac{95 + 76 + 82 + 65 + x}{5} > 80$ (Remember there will be 5 tests...)

$\displaystyle = 95 + 76 + 82 + 65 + x > 400$

$\displaystyle = 318 + x > 400$

$\displaystyle = x > 82$
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Six less than a number is between (-3) and 11. What real numbers are possible soulutions ?
$\displaystyle -3 < x - 6 < 11$ (Get rid of the -6 by adding six to the left and right side)

$\displaystyle = 3 < x < 17$

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Thank you for your help !!!

8. Originally Posted by omgitzbella
Dave's grades on four tests were 95, 76, 82, 65. What must his lowest grade on the next test be to have an average greater than 80.

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Six less than a number is between (-3) and 11. What real numbers are possible soulutions ?

-------------------------------------------------------------------

Thank you for your help !!!
$\displaystyle \displaystyle\frac{95+76+82+65+x}{5}>80\Leftrighta rrow 318+x>400\Leftrightarrow x>82$

$\displaystyle -3<x-6<11\Rightarrow 3<x<17$