Question

Решите систему уравнений методом алгебраического сложения { x^2-3y^2=22 x^2+3y^2=28очень нужно, заранее спасибо​

Answer 1

Объяснение:

складываеим и получаем 2х²=50

х²=25

х=+-5

подставим в первое уравнение системы

25-3у²=22

-3у²=-3

у²=1

у=+-1

Answer 2

Solving the System of Equations using Algebraic Addition Method

To solve the system of equations using the algebraic addition method, we can add the two equations together to eliminate one of the variables. Let's start by adding the two equations:

Step 1: Add the Equations

The given equations are: - \(x^2 - 3y^2 = 22\) - \(x^2 + 3y^2 = 28\)

Adding these two equations together, we get: \[2x^2 = 50\]

Step 2: Solve for x

Now, we can solve for \(x\) by dividing both sides by 2: \[x^2 = 25\]

Taking the square root of both sides, we get two possible values for \(x\): \[x = \pm 5\]

Step 3: Substitute x into one of the original equations

Let's substitute the values of \(x\) back into one of the original equations to solve for \(y\). We'll use the equation \(x^2 - 3y^2 = 22\).

For \(x = 5\): \[5^2 - 3y^2 = 22\] \[25 - 3y^2 = 22\] \[3y^2 = 25 - 22\] \[3y^2 = 3\] \[y^2 = 1\] \[y = \pm 1\]

For \(x = -5\): \[-5^2 - 3y^2 = 22\] \[25 - 3y^2 = 22\] \[3y^2 = 25 - 22\] \[3y^2 = 3\] \[y^2 = 1\] \[y = \pm 1\]

Conclusion

The solutions to the system of equations are: - When \(x = 5\), \(y = \pm 1\) - When \(x = -5\), \(y = \pm 1\)

Therefore, the system of equations has four solutions: \((5, 1)\), \((5, -1)\), \((-5, 1)\), and \((-5, -1)\).