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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 $$\begin{cases} 2(a^3-b^3)-7ab(a-b)=10,\\ 3a^2 -2b^2=-5. \end{cases}$$

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  • $\begingroup$ Interesting idea is to solve the equation using a system of equations but it does greatly simplify calculations. $\endgroup$
    – medicu
    Apr 28, 2014 at 18:44

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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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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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