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 - x²Now we can substitute this expression for y into the first equation:
x² + (6 - x²)² = 6Next, expand the square:
x² + (36 - 12x² + x⁴) = 6This simplifies to:
x⁴ - 11x² + 30 = 0Letting u = x², we can rewrite the equation as:
u² - 11u + 30 = 0Next, we can factor this quadratic equation:
(u - 6)(u - 5) = 0This gives us two solutions for u
u = 6u = 5
Substituting back for x², we have:
x² = 6→x = ±√6x² = 5→x = ±√5
Now we will use these values of x to find corresponding y values using y = 6 - x²:
- If
x² = 6, theny = 6 - 6 = 0 - If
x² = 5, theny = 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)