Помогите пожалуйста решить.

$1)(\frac{8}{7})^{x}*(\frac{7}{8})^{-y}=\frac{64}{49}\\\\(\frac{8}{7})^{x}*(\frac{8}{7})^{y}=\frac{64}{49}\\\\(\frac{8}{7})^{x+y}=(\frac{8}{7})^{2}\\\\x+y=2$

5^{y}\\\\x>y" alt="21^{-8}=(21^{x})^{y}\\\\21^{xy}=21^{-8} \\\\xy=-8\\\\3)5^{x}>5^{y}\\\\x>y" align="absmiddle" class="latex-formula">

$4)\left \{ {{x+y=2} \atop {xy=-8}} \right.\\\\\left \{ {{x=2-y} \atop {(2-y)y=-8}} \right.\\\\\left \{ {{x=2-y} \atop {y^{2}-2y-8=0 }} \right.\\\\\left \{ {{y_{1}=4 } \atop {x_{1}=2-4=-2 }} \right.\\\\\left \{ {{y_{2}=-2 } \atop {x_{2}=2-(-2)=4 }} \right.$

Так как x > y , то ответ : (4 ; - 2)