Simplfying Algebra!!

**Punch** · February 11th 2010, 09:13 PM

Simplify $(x+2)[\frac{\frac{-3}{2}}{(3x+1)^(1.5)}]+\frac{1}{\sqrt{3x+1}}$

>>>>;;;;Note that 1.5 is a power of 3x+1;;;;<<<<

**Grandad** · February 12th 2010, 02:00 AM

Hello Punch

Originally Posted by Punch

Simplify $(x+2)[\frac{\frac{-3}{2}}{(3x+1)^(1.5)}]+\frac{1}{\sqrt{3x+1}}$

>>>>;;;;Note that 1.5 is a power of 3x+1;;;;<<<<

If I re-write the question, I think it is: Simplify:

$(x+2)\left(\frac{-\frac{3}{2}}{(3x+1)^{1.5}}\right)+\frac{1}{\sqrt{3 x+1}}$

First, note that the $(x+2)$ is multiplying the first fraction, so it goes on top; the $-\tfrac32$ can be written with the $-3$ on top and the $2$ underneath. So we get:

$=\frac{-3(x+2)}{2(3x+1)^{1.5}}+\frac{1}{\sqrt{3x+1}}$

Do you see how that works?

Then $(...)^{1.5} = (...)^1\times(...)^{0.5}=(...)\sqrt{(...)}$ .

So the next stage is:

$=\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{1}{\sqrt{3x+1}}$

OK. Now find the lowest common denominator, which is $2(3x+1)\sqrt{3x+1}$ . So, writing the second fraction with this denominator, we get:

$=\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{2(3x+1)}{2(3x+1) \sqrt{3x+1}}$

Now combine the fractions over a single denominator and simplify:

$=\frac{-3(x+2)+2(3x+1)}{2(3x+1)\sqrt{3x+1}}$

$=\frac{3x-4}{2(3x+1)\sqrt{3x+1}}$

which you can write as:

$=\frac{3x-4}{2(3x+1)^{1.5}}$

if you like.

Grandad

**earboth** · February 12th 2010, 02:09 AM

Originally Posted by Punch

Simplify $(x+2)[\frac{\frac{-3}{2}}{(3x+1)^(1.5)}]+\frac{1}{\sqrt{3x+1}}$

>>>>;;;;Note that 1.5 is a power of 3x+1;;;;<<<<

$(x+2)\left[\frac{\frac{-3}{2}}{(3x+1)^{1.5}}\right]+\frac{1}{\sqrt{3x+1}}$

Split the exponent of 1.5 into 1 + 0.5:

$\left[\frac{-\frac{3}{2} \cdot (x+2)}{(3x+1) \cdot \sqrt{3x+1}}\right]+\frac{1}{\sqrt{3x+1}}$

Determine the common denominator of the fractions and combine the numerators.

**Punch** · February 12th 2010, 02:54 AM

Originally Posted by Grandad

Hello PunchIf I re-write the question, I think it is: Simplify:

$(x+2)\left(\frac{-\frac{3}{2}}{(3x+1)^{1.5}}\right)+\frac{1}{\sqrt{3 x+1}}$

First, note that the $(x+2)$ is multiplying the first fraction, so it goes on top; the $-\tfrac32$ can be written with the $-3$ on top and the $2$ underneath. So we get:

$=\frac{-3(x+2)}{2(3x+1)^{1.5}}+\frac{1}{\sqrt{3x+1}}$

Do you see how that works?

Then $(...)^{1.5} = (...)^1\times(...)^{0.5}=(...)\sqrt{(...)}$ .

So the next stage is:

$=\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{1}{\sqrt{3x+1}}$

OK. Now find the lowest common denominator, which is $2(3x+1)\sqrt{3x+1}$ . So, writing the second fraction with this denominator, we get:

$=\frac{-3(x+2)}{2(3x+1)\sqrt{3x+1}}+\frac{2(3x+1)}{2(3x+1) \sqrt{3x+1}}$

Now combine the fractions over a single denominator and simplify:

$=\frac{-3(x+2)+2(3x+1)}{2(3x+1)\sqrt{3x+1}}$

$=\frac{3x-4}{2(3x+1)\sqrt{3x+1}}$

which you can write as:

$=\frac{3x-4}{2(3x+1)^{1.5}}$

if you like.

Grandad

Hi, your workings looks alright and great!!!!!

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