Question

5)x - 1 = y2,y - x + 3 = 0;​

Answer 1

Ответ:

ВОТ ПОЖАЛУЙСТА ЕСЛИ ВАМ ПОНРАВИЛСЯ МОЙ ОТВЕТ И ОКАЗАЛСЯ ПОЛЕЗНЫЙ ТО ДЕЛАЙТЕ ЕГО ЛУЧШИМ❤❤

Answer 2

You have provided a system of two equations:

1. $x - 1 = y^2$
2. $y - x + 3 = 0$

We can solve this system of equations using various methods. Let's use the substitution method:

First, solve equation (2) for $x$:

$y - x + 3 = 0$

Add $x$ to both sides:

$y + 3 = x$

Now, substitute this expression for $x$ into equation (1):

$x - 1 = y^2$

$(y + 3) - 1 = y^2$

Now, simplify the equation:

$y + 2 = y^2$

Now, we have a quadratic equation in terms of $y$. To solve for $y$, you can either factor it or use the quadratic formula. In this case, let's use the quadratic formula:

The quadratic formula is $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

In this equation, $a = 1$, $b = -2$, and $c = -2$.

$y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$

Now, calculate the values of $y$:

$y = \frac{2 \pm \sqrt{4 + 8}}{2}$

$y = \frac{2 \pm \sqrt{12}}{2}$

$y = \frac{2 \pm 2\sqrt{3}}{2}$

Now, simplify further:

$y = 1 \pm \sqrt{3}$

So, we have two possible values for $y$:

1. $y = 1 + \sqrt{3}$
2. $y = 1 - \sqrt{3}$

Now that we have the values of $y$, we can find the corresponding values of $x$ using the equation $x = y + 3$:

For $y = 1 + \sqrt{3}$:

$x = (1 + \sqrt{3}) + 3 = 4 + \sqrt{3}$

For $y = 1 - \sqrt{3}$:

$x = (1 - \sqrt{3}) + 3 = 4 - \sqrt{3}$

So, the solutions to the system of equations are:

1. $x = 4 + \sqrt{3}$, $y = 1 + \sqrt{3}$
2. $x = 4 - \sqrt{3}$, $y = 1 - \sqrt{3}$