MHB How Do You Transpose the Data Shown in These Pictures?

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The discussion focuses on transposing mathematical equations shared in attached pictures. One participant simplifies the first equation to derive a formula for p in terms of q. They also guide another user on how to manipulate a second equation by cubing both sides and rearranging terms to isolate x. The conversation confirms the correctness of the provided solutions and checks for understanding of the methods used. Overall, the thread emphasizes collaborative problem-solving in algebraic transposition.
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Hi Everyone,
need some help to transpose (attached picts)Thanks very much in advance.
 

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vb14 said:
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?
 
can you give a hand with second one as well?
please:)
 
Start by cubing both sides to get rid of the rational exponent on the right side...what do you have now?
 
2x-5x2=(6x2+5)/t3
 
Where did the $x$ on the left side come from?

I would first just write:

$$t^3=\frac{6x^2+5}{2-5x^2}$$

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

$$2t^3-5t^3x^2=6x^2+5$$

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.
 
got this, is it correct?
 

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

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