Question

(x + y = 6,xy = -7;​

Answer 1

$\left \{ {{x+y=6} \atop {xy=-7}} \right. =>x=6-y\\\\(6-y)y=-7\\6y-y^2=-7\\y^2-6y-7=0\\D=36+28=64=8^2\\y_1=(6-8)/2=-1\\y_2=(6+8)/2=7\\\\\left \{ {{x_1=6-(-1)} \atop {x_2=6-7}} \right. =>\left \{ {{x_1=7} \atop {x_2=-1}} \right.$

ОТВЕТ: $x_1=7\\x_2=-1\\y_1=-1\\y_2=7$

Answer 2

To solve the system of equations $x + y = 6$ and $xy = -7$, you can use the method of substitution or elimination. Let's use substitution in this case:

First, solve one of the equations for one variable in terms of the other. We can solve the first equation for $x$:

$x = 6 - y$

Now, substitute this expression for $x$ into the second equation:

$(6 - y)y = -7$

Multiply both sides by $y$ to simplify:

$6y - y^2 = -7$

Rearrange the equation:

$y^2 - 6y + 7 = 0$

Now, you have a quadratic equation in terms of $y$. You can factor it:

$(y - 7)(y - 1) = 0$

Now, set each factor equal to zero and solve for $y$:

1. $y - 7 = 0$ $y = 7$

2. $y - 1 = 0$ $y = 1$

So, you have two possible values for $y$: $y = 7$ and $y = 1$.

Now that you have the values of $y$, you can find the corresponding values of $x$ using the equation $x = 6 - y$:

For $y = 7$: $x = 6 - 7$ $x = -1$

For $y = 1$: $x = 6 - 1$ $x = 5$

Therefore, the solution to the system of equations is:

1. $x = -1$ and $y = 7$
2. $x = 5$ and $y = 1$