An algebra problem by Djordje Veljkovic

1) Adding the two equations gives us:

$x^2 + 4xy + 4y^2 = 169$

2) This can be written as:

$(x+2y)^2 = 169$ , meaning that

$x+2y = +-13$

3) From here onward, there are two cases:

• $x+2y = 13$

This means that $x = 13 - 2y$

Replacing $x$ in the equation we got in the first step, we get the following quadratic equation:

$2y^2 + 13y - 155 = 0$ ,

whose solutions are $y = 5$ , and $y = -\frac{23}{2}$ .

So, in this case we get the ordered pairs $(x_1, y_1)$ , and $(x_2, y_2)$ , which are (3, 5) and (36, $\frac{23}{2}$ ).

• $x+2y = -13$

This means that $x = -13 - 2y$

Again, we replace $x$ in the equation from the first step, and get the following quadratic equation:

$2y^2 - 13y - 155 = 0$ ,

whose solutions are $y = -5$ , and $y = (\frac{23}{2}$ ).

In this case, we get the ordered pairs $(x_3, y_3)$ , and $(x_4, y_4)$ , which are (-3, -5) and (-36, $\frac{23}{2}$ ).

The task states that the inputed answer should be:

$(x_1 * y_1) + (x_2 * y_2) + (x_3 * y_3) + (x_4 * y_4)$ ,

so, we get,

$15 - 414 + 15 - 414 = -798$

We eliminate xy from the equations, then we have 12 y^2 – x^2 = 291. We divide two equations above by y^2, we have (x/y)^2 + 3 (x/y) = 54/y^2 , and 4 + (x/y) = 115/y^2 , then we eliminate RHS, we got, 115(x/y)^2 +291 (x/y) – 261 = 0, we obtain (x/y) = 69/115, or -72/23 , or y = 115/69 x, or y = (-23)/72 x. Substitute each to 12 y^2 – x^2 = 291.Then we have the solutions: (3,5), (-3,-5), (36,-23/2), (-36,23/2). Then (x1y1)+ (x2y2)+(x3 y3)+(x4y4) = -798