<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] A "classical" solution:
1) Conditions for the existence of radicals give us $x\geq \frac{3}{2}.$
2) The condition for the existence of equality $(5X-4)\sqrt{2x-3} >(4X-5)\sqrt{3x-2}$ give us: $x>2.$
3)We write the equation in the form $(5X-4)\sqrt{2x-3}-2=(4X-5)\sqrt{3x-2}$and raise a squared:
$$2x^3-3x^2-3x+6=4(5x-4)\sqrt{2x-3}$$
4) Note $\sqrt{2x-3}=t>1 $ and consequently $x=\frac{t^2+3}{2}.$
5) We obtain the equation in $t$:$$t^6+6t^4-40t^3+3t^2-56t+6=0<=>(t-3)(t^5+3t^4+15t^3+5t^2+3t-2)=0$$
with the solution $t=3$ and then $x=6.$
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