Wolfram seems to confirm this.

I got something different. $(-3,4);(4,-3)$

Yeah, that's correct, DeathLlama9. Another approach would just to stupidely use the second equation as $a=\tfrac{15-3b}{3-b}$, plugging that into the first one from which you'd get a polynomial of degree $4$. As it's easy to guess two solutions, e.g. $(0,5)$ and $(5,0)$, it's just mere calculation to proceed.

Guys I got four solutions: (5,0), (0,5), (4,-3), (-3,4)

Kezer wrote: Yeah, that's correct, DeathLlama9. Another approach would just to stupidely use the second equation as $a=\tfrac{15-3b}{3-b}$, Yup!, that's what I did

From the second equation we get that \[ (a - 3)(b - 3) = -6 \implies a = \frac{3b - 15}{b - 3} \]Plugging this into our first equation we get \begin{align*} a^2 + b^2 &= \left(\frac{3b - 15}{b - 3}\right)^2 + b^2 \\ &= \frac{9b^2 - 90b + 225}{b^2 - 6b + 9} + b^2 \\ &= 25 \end{align*}Thus we get the equation \[ b^4 - 6b^3 - 7b^2 + 60b = 0 \]So we know that $b=0$ is a root. Then we have the cubic \[ b^3 - 6b^2 - 7b + 60 \]Using the Rational Root Theorem, we find that the other roots are $b = -3, 4, 5$. Now using these values we find that the pairs are $(a, b) = (5, 0), (4, -3), (-3, 4), (0, 5)$.

Uhm why do I have a longer solution...

a+b=(15+ab)/3 a²+b²= 25 (a+b)²-2ab=25 Which gives a²b²+12ab=0 ab=0, ab=-12 When ab=0. a+b=5 a-b=5 or -5 So, a=5, b=0 a=0, b=5 When ab=-12 a+b= 1 a-b=7 or -7 a= 4, b=-3 a=-3, b=4 So, we have (a,b)= (5,0);(0,5);(4,-3);(-3,4) # Krishijivi

Kezer wrote: Find all real pairs $(a,b)$ that solve the system of equation \begin{align*} a^2+b^2 &= 25, \\ 3(a+b)-ab &= 15. \end{align*}(German MO 2016 - Problem 1) $\color{blue}\boxed{\textbf{Answer: (0,5), (5,0), (4,-3), (-3,4)}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$a^2+b^2=(a+b)^2-2ab=25...(I)$$$$3(a+b)-ab=15...(II)$$$$\Rightarrow 6(a+b)-2ab=30...(III)$$$(III)-(I):$ $$\Rightarrow (a+b)(6-a-b)=5$$$a+b=s:$ $$\Rightarrow s(6-s)=5$$$$\Rightarrow s^2-6s+5=0$$$$\Rightarrow (s-5)(s-1)=0$$$$\Rightarrow s\in \{ 0,5 \}$$$\color{red}\boxed{\textbf{If s=5:}}$ $\color{red}\rule{24cm}{0.3pt}$ By $(I)$: $$\Rightarrow 2ab=0$$$$\Rightarrow a \text{ or }b =0$$$WLOG$ $a=0:$ $$\Rightarrow b=5$$$$\Rightarrow (0,5), (5,0) \text{ are solutions}$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If s=1:}}$ $\color{red}\rule{24cm}{0.3pt}$ By $(I):$ $$\Rightarrow 2ab=-24$$$$\Rightarrow ab=-12$$$$\Rightarrow b=\frac{-12}{a}$$$$\Rightarrow a+\frac{-12}{a}=1$$$$\Rightarrow \frac{a^2-12}{a}=1$$$$\Rightarrow a^2-12=a$$$$\Rightarrow a^2-a-12=0$$$$\Rightarrow (a+3)(a-4)=0$$$$\Rightarrow a=-3 \text{ or }a=4$$$$\Rightarrow (-3,4), (4,-3) \text{ are solutions }$$$\color{red}\rule{24cm}{0.3pt}$ $\Rightarrow \boxed{(0,5), (5,0), (4,-3), (-3,4) \textbf{ are the only solutions }}_\blacksquare$ $\color{blue}\rule{24cm}{0.3pt}$

$(a+b)^2=25+2ab \Longrightarrow 3\sqrt{25+2ab}-ab=15$. Let $ab=s$ so $3\sqrt{25+2s}-s=15 \Longrightarrow 9(25+2ab)=225+30s+s^2 \Longrightarrow -12s=s^2$. Then, $s=-12, s=0$. If $s=-12$: $ab=-12$ $(a+b)=25-24=1$ $a+b=\pm{1}$ Only solutions are $(a,b)=(-3,4), (-4,3)$. If $s=0$: $a$ or $b$ is 0. WLOG, let $a=0 \Longrightarrow b=5$. Only solutions are $(a,b)=(0,5), (5,0)$

(a + b)² - 2ab = 25 3(a + b) - ab = 15 Let (a + b) = x and ab = y then we have x² - 2y = 25 —————(1) 3x - y = 15 On plugging y = 3x - 15 in (1), we get x² - 2(3x - 15) = 25 x² - 6x + 5 = 0 (x - 1)(x - 5) = 0 x = 1,5 y = -12,0 (a + b) = 1,5 and ab = -12,0 (a,b) = (4,-3)(-3,4),(5,0) and (0,5)