$\frac{1}{5}{x}_1^{-\frac{4}{5}}{x}_2^{\frac{4}{5}}-p_1{x}_3=0$

$\frac{4}{5}{x}_1^{\frac{1}{5}}{x}_2^{-\frac{1}{5}}-p_2{x}_3=0$

$I-p_1{x}_1-p_2{x}_2=0$

Hint $1$: Use the eLecture alternative approach

Hint $2$: While I included the values for $x_3,$ you don't need to explicitly check those. In other words, if your $x_1$ and $x_2$ is correct, the $x_3$ option will also be correct. In a few weeks, we'll see why we don't care about $x_3$ as much as $x_1$ and $x_2.$

a$.x_1=\frac{\text{I}}{5p_1}$;$x_2=\frac{4\text{I}}{5p_2}$;$x_3=\frac{\frac{1}{5}(\frac{\text{I}}{5p_1}{)}^{-\frac{4}{5}}(\frac{4\text{I}}{5p_2}{)}^{\frac{4}{5}}}{p_1}=\frac{\frac{4}{5}(\frac{\text{I}}{5p_1}{)}^{\frac{1}{5}}(\frac{4\text{I}}{5p_2}{)}^{-\frac{1}{5}}}{p_2}$

b$.x_1=\frac{\text{I}}{5p_2}$;$x_2=\frac{4\text{I}}{5p_1}$;$x_3=\frac{\frac{1}{5}(\frac{\text{I}}{5p_2}{)}^{-\frac{4}{5}}(\frac{4\text{I}}{5p_1}{)}^{\frac{4}{5}}}{p_2}=\frac{\frac{4}{5}(\frac{\text{I}}{5p_2}{)}^{\frac{1}{5}}(\frac{4\text{I}}{5p_1}{)}^{-\frac{1}{5}}}{p_1}$

c$.x_1=\frac{4\text{I}}{5p_1}$;$x_2=\frac{\text{I}}{5p_2}$;$x_3=\frac{\frac{1}{5}(\frac{4\text{I}}{5p_1}{)}^{-\frac{4}{5}}(\frac{\text{I}}{5p_2}{)}^{\frac{4}{5}}}{p_1}=\frac{\frac{4}{5}(\frac{4\text{I}}{5p_1}{)}^{\frac{1}{5}}(\frac{\text{I}}{5p_2}{)}^{-\frac{1}{5}}}{p_2}$

d$.$ No solution exists