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x^2 + y^2 &= 10 \\xy &= 6\end{cases} \]

Question

x^2 + y^2 &= 10 \xy &= 6\end{cases} ]

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Solution

This is a system of two equations. Let's solve it step by step.

  1. We have two equations here:

    • x^2 + y^2 = 10
    • xy = 6
  2. From the second equation, we can express y in terms of x:

    • y = 6/x
  3. Now, we can substitute y in the first equation with 6/x:

    • x^2 + (6/x)^2 = 10
    • x^2 + 36/x^2 = 10
    • Multiply through by x^2 to clear the denominator:
    • x^4 + 36 = 10x^2
    • Rearrange to form a quadratic equation:
    • x^4 - 10x^2 + 36 = 0
  4. Let's make a substitution to solve this quadratic equation. Let u = x^2. Then the equation becomes:

    • u^2 - 10u + 36 = 0
  5. This can be factored into:

    • (u - 6)(u - 4) = 0
  6. Setting each factor equal to zero gives the solutions u = 6 and u = 4.

  7. But remember, u = x^2. So x^2 = 6 and x^2 = 4. Therefore, the solutions for x are x = sqrt(6), -sqrt(6), 2, -2.

  8. Substitute x back into the equation y = 6/x to find the corresponding y values. For x = sqrt(6), y = 6/sqrt(6) = sqrt(6). For x = -sqrt(6), y = -sqrt(6). For x = 2, y = 3. For x = -2, y = -3.

So the solutions to the system of equations are (sqrt(6), sqrt(6)), (-sqrt(6), -sqrt(6)), (2, 3), and (-2, -3).

This problem has been solved

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