Question

(x-3)во 2 степени-4y =(x+2)(x+1)-6 (x-4)(y+6) = (x+3)(y-7)+3

Answer 1

Ответ:

ответ во вложении()()

Answer 2

It looks like you've provided a system of equations:

1. $(x-3)^2 - 4y = (x+2)(x+1) - 6$
2. $(x-4)(y+6) = (x+3)(y-7) + 3$

Let's work through these equations step by step:

Equation 1: Expand the left side: $(x-3)^2 - 4y = x^2 - 6x + 9 - 4y$

Expand the right side: $(x+2)(x+1) - 6 = x^2 + 3x + 2 - 6$

Now the equation becomes: $x^2 - 6x + 9 - 4y = x^2 + 3x - 4$

Simplify and solve for $y$: $-6x + 9 - 4y = 3x - 4$ $-10x + 13 = 4y + 3x$ $4y = -10x - 3x + 13$ $4y = -13x + 13$ $y = -\frac{13x}{4} + \frac{13}{4}$

Equation 2: Expand both sides: $(x-4)(y+6) = (x+3)(y-7) + 3$ $xy - 4y + 6x - 24 = xy + 3y - 7x - 21 + 3$

Simplify and solve for $y$: $-4y + 6x - 24 = 3y - 7x - 18$ $-7y + 6x = -7x + 6$ $-7y = -7x - 6x + 6$ $-7y = -13x + 6$ $y = \frac{13x}{7} - \frac{6}{7}$

Now you have two expressions for $y$ in terms of $x$:

1. $y = -\frac{13x}{4} + \frac{13}{4}$ (from Equation 1)
2. $y = \frac{13x}{7} - \frac{6}{7}$ (from Equation 2)

These equations give you a relationship between $x$ and $y$ that satisfy the given system of equations. If you need to find specific values of $x$ and $y$, you can substitute one of these equations into the other equation and solve for $x$ or $y$.