How Do You Transpose the Data Shown in These Pictures?

[vb14]

Hi Everyone,
need some help to transpose (attached picts)Thanks very much in advance.

 

[Sudharaka]

Hi Everyone,
need some help to transpose (attached picts)Thanks very much in advance.

Hi vb14, :)

Welcome to MHB! The first equation can be simplified to,

$$q=\sqrt{6p^2-4}$$

Now square both sides to get,

$$q^2=6p^2-4$$

Add $4$ to both sides,

$$q^2+4=6p^2$$

Divide both sides by $6$,

$$\frac{q^2+4}{6}=p^2$$

Take the square root of both sides,

$$p=\pm\sqrt{\frac{q^2+4}{6}}$$

Can you give it a try for the second one?

 

[vb14]

can you give a hand with second one as well?
please:)

 

[MarkFL]

Start by cubing both sides to get rid of the rational exponent on the right side...what do you have now?

 

[vb14]

2x-5x²=(6x²+5)/t³

 

[MarkFL]

Where did the $x$ on the left side come from?

I would first just write:

\(\displaystyle t^3=\frac{6x^2+5}{2-5x^2}\)

Next, multiply through by $2-5x^2$ to get:

\(\displaystyle 2t^3-5t^3x^2=6x^2+5\)

Now, what you want to do is arrange the equation with all the terms involving $x$ on one side and everything else on the other, the factor out $x^2$ and divide through by the other factor to solve for $x^2$ and then take the square root of both sides.

edit: I see you have also posted this at MMF and have been given the solution.

 

[vb14]

got this, is it correct?

 

[MarkFL]

Yes, that is correct. Do you understand how that was obtained?