<s>[INST] How to solve this equation or system of equations? I want to solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2.$$
I tried. Put $a = \sqrt{2 x-3}\geqslant 0$ and $b =\sqrt{3 x-2}\geqslant 0 $.
Suppose
$$5x-4=m(2x-3)+n(3x-2)$$
then $m=\dfrac{2}{5}$ and $n=\dfrac{7}{5}$.
Therefore
$$5x-4=\dfrac{2}{5}(2x-3)+\dfrac{7}{5}(3x-2)=\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2.$$
Similarly, we have
$$4x-5=\dfrac{7}{5}a^2+\dfrac{2}{5}b^2.$$
The given system of equation are written
$$\begin{cases}
\left (\dfrac{2}{5}a^2 + \dfrac{7}{5}b^2\right )a -\left (\dfrac{7}{5}a^2 + \dfrac{2}{5}b^2\right )b=2,\\
3a^2 -2b^2=-5.
\end{cases}$$
Equavalent to
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
Now, I can not solve the last system of equations. How can I solve the equation
$$(5 x-4) \cdot\sqrt{2 x-3}-(4 x-5)\cdot \sqrt{3 x-2}=2$$
or solve the system of equations
$$\begin{cases}
2(a^3-b^3)-7ab(a-b)=10,\\
3a^2 -2b^2=-5.
\end{cases}$$
 [/INST] Hint: try integer values...usually in this kind of problem first try to see when you get integer result, so both $\sqrt{3x-2} \text{ and } \sqrt{2x-3}$ should be integers...solution: $x=6$
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