Question

Как решить систему уравнений ху=8 х+у+3=0

Answer 1

Ну океееей:) 
Давайте выразим из второго уравнения x. Мне лично так больше нравится. 
xy=8
x=-y-3 

y(-y-3)=8
-y^2-3y-8=0
Теперь давайте вспомним такую вещь как дискриминант
D=9-32=-23
К сожалению наш дискриминант меньше нуля. И я могу с уверенность сказать что ДЕЙСТВИТЕЛЬНЫХ корней нет у этого уравнения. А значит у системы уравнений тоже нет корней. Возможно в условии ошибка. Или поправьте мою ошибку, если я не прав.
P.S. Не сдавайтесь и учите математику:) 

Answer 2

Solving the System of Equations

To solve the system of equations \(xu = 8\) and \(x + uy = 3\), we can use the substitution method or the elimination method. Let's use the substitution method to solve this system.

1. Solving by Substitution: We can solve the first equation for \(x\) and then substitute that value into the second equation to solve for \(y\).

From the first equation, we have \(x = \frac{8}{u}\).

Substituting \(x\) in the second equation, we get \(\frac{8}{u} + uy = 3\).

Multiplying through by \(u\) to clear the fraction, we get \(8 + u^2y = 3u\).

Rearranging the terms, we have the quadratic equation \(u^2y - 3u + 8 = 0\).

We can solve this quadratic equation for \(u\) and \(y\).

2. Quadratic Equation Solution: The quadratic equation \(u^2y - 3u + 8 = 0\) can be solved using the quadratic formula: \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Here, \(a = y\), \(b = -3\), and \(c = 8\).

Plugging these values into the quadratic formula, we can find the solutions for \(u\) and \(y\).

3. Finding \(x\) and \(y\): Once we have the values of \(u\) and \(y\), we can use the equation \(x = \frac{8}{u}\) to find the corresponding values of \(x\).

This will give us the solutions for the system of equations \(xu = 8\) and \(x + uy = 3\).

Would you like to proceed with solving the quadratic equation to find the values of \(u\) and \(y\)?