Question

Решите систему уравнений x^2+xy=12 y-x=2 (под одной скобкой)

Answer 1

$\left \{ {{x^{2}+xy=12} \atop {y-x=2}} \right. \\ \left \{ {{x^{2}+x(2+x)=12} \atop {y=2+x}} \right.$
Решим первое уравнение:
$x^{2}+x(2+x)=12 \\ x^{2} +2x+ x^{2} =12 \\ 2 x^{2} +2x-12=0 |:2 \\ x^{2} +x-6=0$
Найдём дискриминант.
$D=1^{2}-4*1*(-6)=1+24=25 \\ \sqrt{D} = \sqrt{25} =5 \\ x_{1,2} = \frac{-1+/-5}{2} \\ x= \frac{-1-5}{2} = \frac{-6}{2} =-3 \\ x= \frac{-1+5}{2} = \frac{4}{2} =2$
При x=-3
y=2+(-3)=-1
При x=2
y=2+2=4
Ответ: (-3;-1); (2;4)

Answer 2

Solving the System of Equations

To solve the system of equations x^2 + xy = 12 and y - x = 2, we can use the substitution method. First, let's solve the second equation for y in terms of x:

y = x + 2

Now, we can substitute this expression for y into the first equation:

x^2 + x(x + 2) = 12

Simplifying this equation gives us a quadratic equation in terms of x. Let's solve for x.

x^2 + x^2 + 2x - 12 = 0

This simplifies to:

2x^2 + 2x - 12 = 0

We can solve for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Where a = 2, b = 2, and c = -12.

Let's calculate the values of x using the quadratic formula.