Question

1)y = x +6,x2 – 4y = -3;​

Answer 1

Ответ:

(-3;3) и (7;13)

Объяснение:

Подставляем значение у из 1 уравнения во второе

х2-4(х+6)=-3

х2-4х-24+3=0

х2-4х-21=0

Можно выделить скобки или через дискриминант.

Я сделаю 1 способ

х2+3х-7х-21=0

х(х+3)-7(х+3)=0

(х-3)*(х-7)=0

х= -3; х= 7

Подставляем в уравнение с у

у= -3+6 и у=7+6

у=3; у= 13

Answer 2

To solve the system of equations:

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

We can use the substitution method.

First, let's substitute the expression for $y$ from the first equation into the second equation:

$x^2 - 4(x + 6) = -3$

Now, distribute the -4 on the left side:

$x^2 - 4x - 24 = -3$

Next, add 3 to both sides of the equation to isolate the quadratic term:

$x^2 - 4x - 24 + 3 = 0$

Simplify:

$x^2 - 4x - 21 = 0$

Now, we can try to factor the quadratic equation:

$(x - 7)(x + 3) = 0$

To find the values of $x$, set each factor equal to zero and solve for $x$:

1. $x - 7 = 0$ => $x = 7$
2. $x + 3 = 0$ => $x = -3$

Now that we have found the values of $x$, we can use them to find the corresponding values of $y$ using the first equation:

For $x = 7$: $y = x + 6 = 7 + 6 = 13$

For $x = -3$: $y = x + 6 = -3 + 6 = 3$

So, the solution to the system of equations is:

$x = 7$ and $y = 13$

or

$x = -3$ and $y = 3$