Are you kidding me?

Algebra Level 4

{ x 2 − 2 x y + 2 y 2 − 20 x + 5 y − 4 = 0 3 x + 2 y + 1 = 0 \large \begin{cases} x^{2}-2xy+2y^{2}-20x+5y-4=0 \\ 3x+2y+1=0 \end{cases}

Given that there exist two solution pairs ( x 1 , y 1 ) (x_{1},y_{1}) and ( x 2 , y 2 ) (x_{2},y_{2}) for the above system of equations, find x 1 + y 1 + x 2 + y 2 x_{1}+y_{1}+x_{2}+y_{2} correct to three decimal places.


The answer is -2.382.

Lee Dongheng by Lee Dongheng , 20, Malaysia

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1 solution

Tom Engelsman
Oct 30, 2016

Solving the second equation for y in terms x gives y = -3x/2 - 1/2, which when substituted into the first equation produces:

x^2 - 2x (-3x/2 - 1/2) + 2 (-3x/2 - 1/2)^2 - 20x + 5*(-3x/2 - 1/2) - 4 = 0;

which when multiplied out and combined yields the quadratic equation in x:

17*x^2 - 47x - 12 = (17x + 4)(x - 3) = 0, or x = 3, -4/17.

Plugging these values into y = -3x/2 - 1/2 gives y = -5, -5/34 respectively. The two intersection points are (x,y) = (3, -5) and (-4/17, -5/34) with coordinate sum equal to 3 - 5 - 4/17 - 5/34 = -81/34 = -2.382.

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