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Find x and y in
x^2+y^2+6xy+3x+5y+11=0
2x^2+4y^2+12xy+9x+7y+3=0


User Bombolo
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1 Answer

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Answer:

To find the values of x and y in the given equations:

Equation 1: x^2 + y^2 + 6xy + 3x + 5y + 11 = 0

Equation 2: 2x^2 + 4y^2 + 12xy + 9x + 7y + 3 = 0

We can solve this system of equations by using the method of substitution. Here are the steps:

Step 1: Rearrange the equations to isolate one variable in terms of the other.

Let's start with Equation 1. We can rewrite it as:

x^2 + (6y+3)x + (y^2 + 5y + 11) = 0

Step 2: Solve Equation 1 for x.

Using the quadratic formula, x = [-b ± √(b^2 - 4ac)] / (2a), where a, b, and c are the coefficients of the quadratic equation (ax^2 + bx + c = 0), we can substitute the values from Equation 1 into the quadratic formula:

x = [-(6y+3) ± √((6y+3)^2 - 4(1)(y^2 + 5y + 11))] / (2(1))

Simplifying the equation further will give us the value of x in terms of y.

Step 3: Substitute the value of x in Equation 2.

Replace x in Equation 2 with the expression obtained in Step 2. Then simplify the equation to solve for y.

Step 4: Substitute the value of y in Equation 1 to find the corresponding x-value.

User EpiX
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