x2-3xy+y2+1=0, 3x2-xy+3y2=13
I am assuming you want to solve the system of equations
x^2-3xy+y^2+1 = 0
3x^2-xy+3y^2 = 13
from the second, we have
xy = 3x^2+3y^2-13
using that in the first, we have
x^2-3(3x^2+3y^2-13)+y^2+1 = 0
x^2-9x^2-9y^2+39+y^2+1 = 0
-8x^2 - 8y^2 = -40
x^2 + y^2 = 5
Using that in either equation, we have xy = 2
So, our solutions are (±1,±2)and (±2,±1)
or, you could just solve the two equations using the quadratic formula. That would yield
x = [3y±√(5y^2-4)]/2
x = [y±√(152-35y^2)]/6
equate those two and you get y=±2
To solve the given system of equations:
Equation 1: x² - 3xy + y² + 1 = 0
Equation 2: 3x² - xy + 3y² = 13
We can apply a method called "substitution" to solve this system of equations.
Step 1: Solve Equation 1 for x in terms of y
Rearrange Equation 1 to isolate x:
x² - 3xy + y² = -1
x² - 3xy = -y² - 1
x² = 3xy - y² - 1
x = (3xy - y² - 1) / (1)
Step 2: Substitute x-value into Equation 2
Substitute the expression we found for x into Equation 2:
3[(3xy - y² - 1) / (1)]² - [(3xy - y² - 1) / (1)]y + 3y² = 13
Simplify and solve for y:
9x²y² - 6xy³ + 3y⁴ - 3xy + y³ - y + 3y² = 13
9x²y² - 6xy³ + 3y⁴ + y³ - 3xy + 3y² - y = 13 ║ Move the constant term to the right side
9x²y² - 6xy³ + 3y⁴ + y³ - 3xy + 3y² - y - 13 = 0 ║ Rearrange the equation
Now we have a polynomial equation in terms of y.
To solve this equation, you can use numerical methods (such as a numerical solver or graphing the equation) or algebraic methods (such as factoring, completing the square, or using the quadratic formula if it is a quadratic equation).