Question

Y(12-y) +8=4|y|-y(3+(y-15))

Answer 1

Решение на фото есть
Надеюсь помогло )

Answer 2

Let's solve the given equation $Y(12-y) + 8 = 4|y| - y(3 + (y-15))$.

First, simplify both sides of the equation:

$Y(12 - y) + 8 = 4|y| - y(3 + y - 15)$

Distribute the terms:

$12Y - Yy + 8 = 4|y| - 3y - y^2 + 15y$

Combine like terms:

$-Yy - y^2 + 12Y + 8 = 4|y| + 12y$

Move all terms to one side of the equation to set it to zero:

$-y^2 - Yy + 12y - 4|y| + 12Y + 8 = 0$

Now, let's consider the absolute value. The expression $4|y|$ can be split into two cases based on the sign of $y$:

1. When $y \geq 0$: $4|y| = 4y$
2. When $y < 0$: $4|y| = -4y$

Substitute these into the equation:

$-y^2 - Yy + 12y - 4|y| + 12Y + 8 = 0$

1. For $y \geq 0$: $-y^2 - Yy + 12y - 4y + 12Y + 8 = 0$ Combine like terms: $-y^2 - (Y-8)y + 12Y + 8 = 0$

2. For $y < 0$: $-y^2 - Yy + 12y + 4y + 12Y + 8 = 0$ Combine like terms: $-y^2 + (12-Y)y + 12Y + 8 = 0$

Now, you have two quadratic equations. To solve for $y$, you can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For each of the two cases, substitute the values of $a$, $b$, and $c$ from the corresponding quadratic equation:

1. For $y \geq 0$: $y = \frac{-(Y-8) \pm \sqrt{(Y-8)^2 - 4(-1)(12Y+8)}}{2(-1)}$

2. For $y < 0$: $y = \frac{-(12-Y) \pm \sqrt{(12-Y)^2 - 4(-1)(12Y+8)}}{2(-1)}$

Solve each equation separately for $y$ to find the possible solutions. Keep in mind that the final solution may include both real and complex roots.