Problems solving some inequalities questions

• March 29th 2011, 08:03 AM
Devi09
Problems solving some inequalities questions
a) $\frac{\2x+3}{3} \leq x-5$

I did this:

$\frac{\3(2x+3)}{3} \leq 3(x-5)$

$6x+9 \leq 3x-15$

$6x-3x \leq -15-9$

$3x \leq -24$

$x \leq -8$

$(inf , -8]$

however the answer should be : $[18,inf)$

b) another one

$\frac{\ x+4}{2} \geq \frac{\ x-2}{4}$

$\frac{\2(x+4)}{2} \geq \frac{\2(x-2)}{2(4)}$

$2x+8 \geq \frac{\2x-4}{8}$

$8(2x+8) \geq \frac{\8(2x-4)}{8}$

$16x+64 \geq 16x-32$

$16x-16x \geq -64 -32$

$0 \geq -96$

however the answer should be $[-10, inf)$

c) last one

$3x \leq x+1 \leq x-1$

$3x-1 \leq x \leq x-1-1$

don't know what to do after this step
• March 29th 2011, 08:12 AM
Plato
Your basic algebra skills are in need of some real work.
$\dfrac{x+3}{3} gives $x+3<3x-15$.
• March 29th 2011, 08:27 AM
Devi09
Quote:

Originally Posted by Plato
Your basic algebra skills are in need of some real work.
$\dfrac{x+3}{3} gives $x+3<3x-15$.

$x-3x \leq -3-15$

$-2x \leq -18$

$x \geq 9$

• March 29th 2011, 08:35 AM
Plato
Quote:

Originally Posted by Devi09
$x-3x \leq -3-15$
$-2x \leq -18$
$x \geq 9$

The given answer is wrong. $[9,\infty)$
• March 29th 2011, 01:03 PM
Quote:

Originally Posted by Devi09
a) $\frac{\2x+3}{3} \leq x-5$

that should be 2x+3 in the numerator which will give the answer provided

I did this:

$\frac{\3(2x+3)}{3} \leq 3(x-5)$

why didn't you multiply BOTH sides by 3 ?

$6x+9 \leq 3x-15$

Now you multiplied only the left side by 9.
If you multiply only one side by something other than 1,
you will have a different inequality.

$6x-3x \leq -15-9$

$3x \leq -24$

$x \leq -8$

$(inf , -8]$

however the answer should be : $[18,inf)$

b) another one

$\frac{\ x+4}{2} \geq \frac{\ x-2}{4}$

$\frac{\2(x+4)}{2} \geq \frac{\2(x-2)}{2(4)}$

You halved the right and did nothing to the other side,
so you now have a different inequality altogether.
You could have multiplied both sides by 4.

$2x+8 \geq \frac{\2x-4}{8}$

You multiplied the left by 4
and changed the denominator of the other side.
Any reason ?

$8(2x+8) \geq \frac{\8(2x-4)}{8}$

You multiplied the left by 8 and made a mysterious modification
to the numerator on the right again.
The rest is also haphazard.

$16x+64 \geq 16x-32$

$16x-16x \geq -64 -32$

$0 \geq -96$

however the answer should be $[-10, inf)$

c) last one

$3x \leq x+1 \leq x-1$

$3x-1 \leq x \leq x-1-1$

don't know what to do after this step

For the last one, you cannot have

$x+1\le\ x-1$ since x+1 is 2 greater than x-1
• March 29th 2011, 02:04 PM
Devi09
For b) 2nd try

$\frac{\ x+4}{2} \geq \frac{\ x-2}{4}$

$\frac{\4(x+4)}{4(2)} \geq x-2$

$\frac{\4x+16}{8} \geq x-2$

$4x + 16 \geq 8(x-2)$

$4x +16 \geq 8x-16$

$4x-8x \geq -16-16$

$-4x \geq -32$

$x \leq -32/-4$

$x \leq 8$
• March 29th 2011, 02:20 PM
Plato
Quote:

Originally Posted by Devi09
For b) 2nd try
$\dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}$
$x \leq 8$

Let's see why that solution is not correct.
According to your solution $x=-14$ works.
Plug it in an see that it does not.

According to your solution $x=10$ does not works.
Plug it in an see that it does work.

Here is simple algebra.
$\dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}$
Then we get $4x+16\ge 2x-4.$ HOW?
• March 29th 2011, 02:28 PM
Devi09
Quote:

Originally Posted by Plato
Let's see why that solution is not correct.
According to your solution $x=-14$ works.
Plug it in an see that it does not.

According to your solution $x=10$ does not works.
Plug it in an see that it does work.

Here is simple algebra.
$\dfrac{\ x+4}{2} \geq \dfrac{\ x-2}{4}$
Then we get $4x+16\ge 2x-4.$ HOW?

So when you multiply by 4 you don't multiply the denominator being 2 in this case

So its:

$4x+16 \geq 2x-4$

$4x -2x \geq -16-4$

$2x \geq -20$

$x \geq -10$
• March 29th 2011, 02:30 PM
Plato
Quote:

Originally Posted by Devi09
So when you multiply by 4 you don't multiply the denominator being 2 in this case

So its:

$4x+16 \geq 2x-4$

$4x -2x \geq -16-4$

$2x \geq -20$

$x \geq -10$

Way to go! By George you have it.
• March 29th 2011, 02:39 PM
Devi09
One more thing....

$233 + 6x \leq 8x - 75 \leq 475 + 6x$

this is what I did:

$233 + 75 + 6x \leq 8x \leq 475 + 75 + 6x$

$308 + 6x \leq 8x \leq 550 + 6x$

What do you do with the x's? move them to the middle x?