Question 26

Explanation

Multiply the first equation by $x^2$ and the second equation by $y^2$

Subtract the second equation from the first equation

Factor both sides

Substitute into the equation

So there are two cases

For the first case, $x = y$. Substitute $y = x$ into the first equation

So $x = 0$ or $x^2 - x - 1 = 0$.

For $x^2 - x - 1 = 0$, the discriminant value is

The discriminant value $D = 5 > 0$, which means the quadratic equation has 2 distinct real roots.

So from the first case we obtain:

• $x = 0$, then $y = 0$, so the pair $(0, 0)$
• $x^2 - x - 1 = 0$ has 2 distinct real roots, so 2 pairs $(x, y)$ with $y = x$

For the second case $x^2 + xy + y^2 + x^2y^2 = 0$, since all terms are non-negative and $x^2y^2 \geq 0$, this equation is only satisfied if $x = 0$ and $y = 0$, which is already included in the first case.

So the system of equations has 3 distinct pairs.