We have two cases to consider:
Case 1: $x \ge 0$ and $y \ge 0$ In this case, the equations become: $2x + y = 15$ $x = 20$
Solving these equations, we get $x = 20$ and $y = -35$. This case doesn't satisfy the condition $y \ge 0$.
Case 2: $x < 0$ and $y < 0$ In this case, the equations become: $y = 15$ $x = 20$
This case also doesn't satisfy the conditions $x < 0$ and $y < 0$.
Case 3: $x \ge 0$ and $y < 0$ In this case, the equations become: $2x + y = 15$ $x - 2y = 20$
Solving these equations, we get $x = 10$ and $y = -5$.
Case 4: $x < 0$ and $y \ge 0$ In this case, the equations become: $y = 15$ $x + 2y = 20$
Solving these equations, we get $x = -10$ and $y = 15$.
From the above cases, only the third case satisfies both equations. Therefore, $x - y = 10 - (-5) = 15$.
So, the value of $(x - y)$ is 15.
If the set of all values of \( a \), for which the equation \( 5x^3 - 15x - a = 0 \) has three distinct real roots, is the interval \( (\alpha, \beta) \), then \( \beta - 2\alpha \) is equal to
If the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) has equal roots, where \( a + c = 15 \) and \( b = \frac{36}{5} \), then \( a^2 + c^2 \) is equal to .