{x кв+y кв=25 {x+y=7

$\left \{ {{x^{2}+y^{2}=25} \atop {x+y=7}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {(x+y)^{2}=7^{2}}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {x^{2}+y^{2}+2yx=49}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {25+2yx=49}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {2yx=24}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {yx=12}} \right.$

$\left \{ {{x^{2}+y^{2}=25} \atop {x= \frac{12}{y}}} \right.$

$\left \{ {{(\frac{12}{y})^{2}+y^{2}=25} \atop {x= \frac{12}{y}}} \right.$

x>0, y>0
$\frac{144}{y^{2}}+y^{2}=25$
$\frac{144+y^{4}}{y^{2}}=25$
$144+y^{4}=25y^{2}$
$y^{4}-25y^{2}+144=0$

Замена: $y^{2}=t\ \textgreater \ 0$

$t^{2}-25t+144=0, D=25^{2}-4*144=49$
$t_{1}= \frac{25-7}{2}=9\ \textgreater \ 0$
$t_{2}= \frac{25+7}{2}=16\ \textgreater \ 0$

Вернемся к замене:
1) $y^{2}=9$
$y_{1}=3$
$x_{1}= \frac{12}{y_{1}}=4$

$y_{2}=-3$ - посторонний корень

2) $y^{2}=16$
$y_{3}=4$
$x_{3}= \frac{12}{y_{1}}=3$

$y_{4}=-4$ - посторонний корень

Ответ: (3;4), (4;3)