[tex] {(x + y - 1)}^{2} + {(4x - 6y + 1)}^{2} = 0[/tex]

Предыдущий
вопрос Следующий
вопрос

Since the sum of the squares of two real numbers is always non-negative, the only way for the sum of the squares in this equation to be equal to zero is if each term in the sum is equal to zero.

So we get
$$ (x+y-1)^2 = 0 $
$$ (4x - 6y + 1)^2 = 0 $$

Solving the first equation
$$ x + y - 1 = 0 $
$$ y = 1 - x $$

Substitute for y in the second equation
$$ (4x - 6(1-x) + 1)^2 = 0 $$

Simplify
$$ (4x - 6 + 6x + 1)^2 = 0 $
$$ (10x - 5)^2 = 0 $$

Taking the square root of both sides, we get
$$ 10x - 5 = 0 $
$$ 10x = 5 $
$$ x = \frac{1}{2} $$

Now substitute x back in y = 1 - x
$$ y = 1 - \frac{1}{2} $
$$ y = \frac{1}{2} $$

Therefore, the solution to the equation is
$$ x = \frac{1}{2} $
$$ y = \frac{1}{2} $$