153. $\left\{\begin{array}{c}x^{2}+y^{2}=a^{2} \\ x y=b\end{array}\right.$ and $\left\{\begin{array}{c}x^{2}+y^{2}=13 \\ x y=6 .\end{array}\right.$
153. Doubling the second equation term by term, we have:

$$
(x+y)^{2}=a^{2}+2 b,(x-y)^{2}=a^{2}-2 b
$$

therefore,

$$
x+y=\sqrt{a^{2}+2 b} \quad \text { and } \quad x-y=\sqrt{a^{2}-2 b}
$$

from here we find $x$ and $y$. For the numerical equation $x=3 ; y=2$.