Solve 2x^2-xy+y^2=16quadP(3,2)

$$2 x ^ { 2 } - x y + y ^ { 2 } = 16 \quad P ( 3,2 )$$

Solve for P

$$P=\frac{5\left(2x^{2}-xy+y^{2}\right)}{256}$$

Steps for Solving Linear Equation

Multiply 16 and 3,2 to get 51,2.

$$2x^{2}-xy+y^{2}=51,2P$$

Swap sides so that all variable terms are on the left hand side.

$$51,2P=2x^{2}-xy+y^{2}$$

Divide both sides of the equation by 51,2, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{51,2P}{51,2}=\frac{2x^{2}-xy+y^{2}}{51,2}$$

Dividing by 51,2 undoes the multiplication by 51,2.

$$P=\frac{2x^{2}-xy+y^{2}}{51,2}$$

Divide 2x^{2}-xy+y^{2} by 51,2 by multiplying 2x^{2}-xy+y^{2} by the reciprocal of 51,2.

$$P=-\frac{5xy}{256}+\frac{5x^{2}}{128}+\frac{5y^{2}}{256}$$

Solve for x

$$x=\frac{\sqrt{10240P-175y^{2}}}{20}+\frac{y}{4}$$

$$x=-\frac{\sqrt{10240P-175y^{2}}}{20}+\frac{y}{4}\text{, }P\geq \frac{35y^{2}}{2048}$$