Question

{х-у=2{ху=15это одна система решите умоляю)))

Answer 1

Замена из 1 уравнения y = x - 2
x(x - 2) = 15
x^2 - 2x - 15 = 0
(x - 5)(x + 3) = 0
x1 = -3; y1 = -3 - 2 = -5
x2 = 5; y2 = 5 - 2 = 3

Answer 2

$\left \{ {{x-y=2} \atop {xy=15}} \right.$
Из первого уравнения выразим х.
х-у = 2  ⇒  х = у+2
Подставим полученное значение х во второе уравнение.

(у+2)у=15
у²+2у=15
у²+2у-15=0
D=2²-4(-15)=4+60=64
√D=±8
у₁=(-2+8)/2=3
у₂=(-2-8)/2=-5

у₁=3
х₁=у₁+2=3+2=5

у₂=-5
х₂=у₂+2=-5+2=-3

Ответ: (5;3), (-3;-5)

Answer 3

Solving the System of Equations

To solve the system of equations x - y = 2 and xy = 15, we can use various methods such as substitution, elimination, or graphing. Let's solve it using the substitution method.

1. Start with the equation x - y = 2. 2. Solve this equation for x in terms of y: - Add y to both sides: x = y + 2.

Now, substitute this value of x into the second equation xy = 15:

y(y + 2) = 15

Simplify the equation:

y^2 + 2y = 15

Rearrange the equation to bring all terms to one side:

y^2 + 2y - 15 = 0

Now, we have a quadratic equation in terms of y. We can solve this equation by factoring, completing the square, or using the quadratic formula. Let's factor the equation:

y^2 + 5y - 3y - 15 = 0

y(y + 5) - 3(y + 5) = 0

(y - 3)(y + 5) = 0

From this equation, we can find two possible values for y:

1. y - 3 = 0 → y = 3 2. y + 5 = 0 → y = -5

Now, substitute these values of y back into the equation x = y + 2 to find the corresponding values of x:

1. For y = 3, x = 3 + 2 = 5. 2. For y = -5, x = -5 + 2 = -3.

Therefore, the solutions to the system of equations x - y = 2 and xy = 15 are x = 5, y = 3 and x = -3, y = -5.

Please let me know if you need any further assistance!