1. ## Please simplify for me !!!?

2 different sets of equations:

F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=1/2

4F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=2

2. $F+2T=30$
$
\frac{{(0.5F)}^ \frac{-1}{2}}{{(0.5T)}^ \frac{-1}{2}}=\frac{1}{2}
$

If the above is your question
Cancel 0.5 of numerator and denominator and your question becomes
$
\sqrt{\frac{T}{F}} =\frac{1}{2}
$

Squaring both sides
$
T=\frac{F}{4}
$

So
$
T =\frac{30-2T}{4}
$

I think you can go ahead and solve this and next question

3. ## Indices

Hello schoolhelp
Originally Posted by schoolhelp

2 different sets of equations:

F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=1/2

4F+2T=30
(0.5F^ -0.5)/(0.5T^ -0.5)=2
First of all, notice that in $\frac{0.5F^{-0.5}}{0.5T^{-0.5}}$ you can cancel the initial 0.5's to get $\frac{F^{-0.5}}{T^{-0.5}}$

Then you need to know that:

• $a^{-x} = \frac{1}{a^x}$; and

• $a^{0.5}=\sqrt{a}$

So $F^{-0.5} = \frac{1}{F^{0.5}}$

$= \frac{1}{\sqrt{F}}$

Do the same with the denominator:

$T^{-0.5}=\frac{1}{\sqrt{T}}$

and now invert and multiply, and the second equation reads

$\frac{\sqrt{T}}{\sqrt{F}}=\frac{1}{2}$

$\Rightarrow \frac{T}{F} = \frac{1}{4}$

$\Rightarrow F=4T$

Substitute into the first equation and solve for $T$, and then $F$.

Do the second problem in a similar way.

OK now?