Нужна ваша помощь!!Решить системы уравнений: (фото внутри)

1) $\left \{ {{ x^{2}+2y=2 } \atop {x-y=3}} \right.$

$\left \{ {{ x^{2}+2(x-3)=2} \atop {y=x-3}} \right.$

$\left \{ {{ x^{2}+2x-6=2} \atop {y=x-3}} \right.$

$x^{2}$ +2x-8=0
D= $b^{2}$ -4ac=36

x1= $\frac{-b- \sqrt{D} }{2a}$ = $\frac{-2-6}{2} =-4$

x1= $\frac{-b+ \sqrt{D} }{2a}$ = $\frac{-2+6}{2} =2$

y1=3-(-4)=7
y2=3-2=1

2. $\left \{ {{ 2^{5x+y}= 64^{ \frac{1}{2} } } \atop {3x+2y=-1}} \right.$

$\left \{ {{ 2^{5x+y}=8 } \atop {y=- \frac{1+3x}{2} }} \right.$

$\left \{ {{ 2^{5x- \frac{1+3x}{2}}= 2^{3} } } \atop {y=- \frac{1+3x}{2} }} \right.$

$5x- \frac{1+3x}{2} =3$

$\frac{7x-1}{2} =3$

$7x-1=6$
$7x=7$
$x= \frac{7}{7} =1$

$y=- \frac{1+2}{2} =- \frac{3}{2}$

3. $\left \{ {{ log_{6}x+ log_{6}y =3 } \atop { 3^{x+y}= (\frac{1}{3}) ^{-13} }} \right.$

$\left \{ {{ log_{6}x+ log_{6}y=3 } \atop { 3^{x+y}= 3^{13} }} \right.$

$\left \{ {{ log_{6}(13-y)+ log_{6}y=3 } \atop {x=13-y}} \right.$

$\left \{ {{ log_{6}(13-y)y=3} \atop {x=13-y}} \right.$

$\left \{ {{ log_{6}13y- y^{2}=3} \atop {x=13-y}} \right.$

$\left \{ {{ 13y- y^{2}= 6^{3} } \atop {x=13-y}} \right.$

$y^{2} -13y+216=0$

$D= b^{2} -4ac=169-4*216\ \textless \ 0$ -нет корней