System of equations

$$\\ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } =\lambda p_1\\\\ \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\lambda p_2$$

$$\dfrac{ \dfrac{1}{2}* x_1^{-\dfrac{1}{2}} *x_2^{\dfrac{1}{3}} } { \dfrac{1}{3}* x_1^{\dfrac{1}{2}}*x_2^{-\dfrac{2}{3}} } =\dfrac{\lambda p_1}{\lambda p_2}$$

$$\Rightarrow (1) \quad \dfrac {p_1}{p_2}=\dfrac{3}{2}\dfrac{x_2}{x_1}$$

$$\begin{array}{rcccc} &I.&&II.&\\\\ (2)\qquad p_1x_1+p_2x_2=m \quad | \quad &\times&\dfrac{1}{p_2} \quad | \quad &\times&\dfrac{1}{p_1} \end{array}$$

$$I.\quad \dfrac{p_1}{p_2}x_1+x_2 = \dfrac{m}{p_2} \quad \Rightarrow \quad \dfrac{3}{2}\dfrac{x_2}{x_1}x_1+x_2=\dfrac{m}{p_2}\quad \Rightarrow \quad \dfrac{5}{2}x_2=\dfrac{m}{p_2}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_2=\dfrac{2}{5}\dfrac{m}{p_2}}}$$

$$II.\quad x_1+\dfrac{p_2}{p_1}x_2 = \dfrac{m}{p_1}\quad\Rightarrow \quad x_1+ \dfrac{2}{3}\dfrac{x_1}{x_2}x_2=\dfrac{m}{p_1} \quad \Rightarrow \quad \dfrac{5}{3}x_1=\dfrac{m}{p_1}$$

$$\textcolor[rgb]{1,0,0}{\boxed{x_1=\dfrac{3}{5}\dfrac{m}{p_1}} }$$

Multiply the first equation by x₁^(1/2), and the second by x₂^(2/3).

Move the negative terms to the other side in each equation, and then align them so that both x₁ terms are on the same side, (the lhs say, the x₂ terms on the rhs).

The product of the two lhs's will equal the product of the two rhs's, do this and simplify.

That should get you 3x₂p₂ = 2x₁p₁

Now solve in conjunction with the third equation.

Here's my take on this:

Thank you guys,

Since I'm doing a resit it took me a while to get back to this question, but your answers are really helpful

I hope I  will pass this time