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Choose the solutions of the following system of equations: x² + y² = 6 and x² + y = 6

To solve the system of equations:

  • x² + y² = 6
  • x² + y = 6

We can start by isolating one variable in one of the equations. Let’s take the second equation:

y = 6 - x²

Now we can substitute this expression for y into the first equation:

x² + (6 - x²)² = 6

Next, expand the square:

x² + (36 - 12x² + x⁴) = 6

This simplifies to:

x⁴ - 11x² + 30 = 0

Letting u = x², we can rewrite the equation as:

u² - 11u + 30 = 0

Next, we can factor this quadratic equation:

(u - 6)(u - 5) = 0

This gives us two solutions for u

  • u = 6
  • u = 5

Substituting back for x², we have:

  • x² = 6 → x = ±√6
  • x² = 5 → x = ±√5

Now we will use these values of x to find corresponding y values using y = 6 - x²:

  • If x² = 6, then y = 6 - 6 = 0
  • If x² = 5, then y = 6 - 5 = 1

Therefore, the solutions to the system of equations are:

  • (√6, 0)
  • (-√6, 0)
  • (√5, 1)
  • (-√5, 1)

In conclusion, the solutions to the given system of equations are:

(√6, 0), (-√6, 0), (√5, 1), (-√5, 1)

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