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Algebra Difficulty 5.1 AIME, harder Find the answer

Example 7 The system of equations {x+xy+y=1,x2+x2y2+y2=17\left\{\begin{array}{l}x+x y+y=1, \\ x^{2}+x^{2} y^{2}+y^{2}=17\end{array}\right. has the real solution (x,y)=(x, y)= \qquad
(1996, Eastern Airlines Cup Shanghai Junior High School Mathematics Competition)

Solution

Let x=a+b,y=abx=a+b, y=a-b. Then x+y=2a,xy=a2b2x+y=2a, xy=a^{2}-b^{2}.
Thus, the original system of equations becomes
{a2b2+2a=1,2a2+2b2+(a2b2)2=17{a=32,b=±172 \left\{\begin{array} { l } { a ^ { 2 } - b ^ { 2 } + 2 a = 1 , } \\ { 2 a ^ { 2 } + 2 b ^ { 2 } + ( a ^ { 2 } - b ^ { 2 } ) ^ { 2 } = 1 7 } \end{array} \Rightarrow \left\{\begin{array}{l} a=\frac{3}{2}, \\ b= \pm \frac{\sqrt{17}}{2} \end{array}\right.\right.

Therefore, the solutions to the original system of equations are
{x1=3+172,y1=3172;{x2=3172,y2=3+172. \left\{\begin{array} { l } { x _ { 1 } = \frac { 3 + \sqrt { 1 7 } } { 2 } , } \\ { y _ { 1 } = \frac { 3 - \sqrt { 1 7 } } { 2 } ; } \end{array} \quad \left\{\begin{array}{l} x_{2}=\frac{3-\sqrt{17}}{2}, \\ y_{2}=\frac{3+\sqrt{17}}{2} . \end{array}\right.\right.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.