Question

Решить систему уравнений: x^2-y^2=21 x^2+2xy+y^2=49 Заранее спасибо!

Answer 1

X²-y²=21                (x-y)(x+y)=21     (x-y)(x+y)=21    (x-y)*7=21      x-y=3
x²+2xy+y²=49       (x+y)²=7²            x+y=7                 x+y=7            x+y=7

Складываем первое уравнение со вторым
x+x-y+y=3+7
2x=10
x=5                y=7-5=2

Answer 2

Solving the System of Equations

To solve the system of equations \(x^2 - y^2 = 21\) and \(x^2 + 2xy + y^2 = 49\), we can use the following steps:

1. We can start by adding the two equations together to simplify the process.

2. After adding the equations, we can use the resulting equation to solve for one variable in terms of the other.

3. Once we have the value of one variable, we can substitute it back into one of the original equations to solve for the other variable.

Let's proceed with these steps to solve the given system of equations.

Solution Steps

1. Adding the Equations: - Adding the equations \(x^2 - y^2 = 21\) and \(x^2 + 2xy + y^2 = 49\), we get: \[2x^2 + 2xy = 70\]

2. Solving for \(y\) in terms of \(x\): - We can solve for \(y\) in terms of \(x\) using the equation \(2x^2 + 2xy = 70\). - Factoring out \(2x\) from the left-hand side, we get: \[2x(x + y) = 70\] - Solving for \(y\), we have: \[y = \frac{70}{2x} - x\]

3. Substituting Back to Solve for \(x\) or \(y\): - We can substitute the expression for \(y\) back into one of the original equations to solve for \(x\) or \(y.

Conclusion

By following these steps, we can solve the given system of equations and find the values of \(x\) and \(y\).