Solve {c}{P_{y}y=2xP_{x}}{R=P_{x}x+P_{y}y}

$$\left.\begin{array} { c } { P _ { y } y = 2 x P _ { x } } \\ { R = P _ { x } x + P _ { y } y } \end{array} \right\}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{R}{3P_{x}}\text{, }y=\frac{2R}{3P_{y}}\text{, }&P_{y}\neq 0\text{ and }P_{x}\neq 0\\x=0\text{, }y\in \mathrm{C}\text{, }&P_{x}\neq 0\text{ and }R=0\text{ and }P_{y}=0\\x\in \mathrm{C}\text{, }y=0\text{, }&R=0\text{ and }P_{x}=0\\x\in \mathrm{C}\text{, }y\in \mathrm{C}\text{, }&R=0\text{ and }P_{x}=0\text{ and }P_{y}=0\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

$$P_{y}y+\left(-2P_{x}\right)x=0$$

Add 2P_{x}x to both sides of the equation.

$$P_{y}y=2P_{x}x$$

Divide both sides by P_{y}.

$$y=\frac{1}{P_{y}}\times 2P_{x}x$$

Multiply \frac{1}{P_{y}} times 2P_{x}x.

$$y=\frac{2P_{x}}{P_{y}}x$$

Substitute \frac{2P_{x}x}{P_{y}} for y in the other equation, P_{y}y+P_{x}x=R.

$$P_{y}\times \frac{2P_{x}}{P_{y}}x+P_{x}x=R$$

Multiply P_{y} times \frac{2P_{x}x}{P_{y}}.

$$2P_{x}x+P_{x}x=R$$

Add 2P_{x}x to P_{x}x.

$$3P_{x}x=R$$

Divide both sides by 3P_{x}.

$$x=\frac{R}{3P_{x}}$$

Substitute \frac{R}{3P_{x}} for x in y=\frac{2P_{x}}{P_{y}}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{2P_{x}}{P_{y}}\times \frac{R}{3P_{x}}$$

Multiply \frac{2P_{x}}{P_{y}} times \frac{R}{3P_{x}}.

$$y=\frac{2R}{3P_{y}}$$

The system is now solved.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$

Steps Using Matrices

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Write the equations in matrix form.

$$\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\R\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right).

$$inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{P_{x}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}&-\frac{-2P_{x}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}\\-\frac{P_{y}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}&\frac{P_{y}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}\end{matrix}\right)\left(\begin{matrix}0\\R\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3P_{y}}&\frac{2}{3P_{y}}\\-\frac{1}{3P_{x}}&\frac{1}{3P_{x}}\end{matrix}\right)\left(\begin{matrix}0\\R\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3P_{y}}R\\\frac{1}{3P_{x}}R\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2R}{3P_{y}}\\\frac{R}{3P_{x}}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$

Steps Using Elimination

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Subtract P_{y}y+P_{x}x=R from P_{y}y+\left(-2P_{x}\right)x=0 by subtracting like terms on each side of the equal sign.

$$P_{y}y+\left(-P_{y}\right)y+\left(-2P_{x}\right)x+\left(-P_{x}\right)x=-R$$

Add P_{y}y to -P_{y}y. Terms P_{y}y and -P_{y}y cancel out, leaving an equation with only one variable that can be solved.

$$\left(-2P_{x}\right)x+\left(-P_{x}\right)x=-R$$

Add -2P_{x}x to -P_{x}x.

$$\left(-3P_{x}\right)x=-R$$

Divide both sides by -3P_{x}.

$$x=\frac{R}{3P_{x}}$$

Substitute \frac{R}{3P_{x}} for x in P_{y}y+P_{x}x=R. Because the resulting equation contains only one variable, you can solve for y directly.

$$P_{y}y+P_{x}\times \frac{R}{3P_{x}}=R$$

Multiply P_{x} times \frac{R}{3P_{x}}.

$$P_{y}y+\frac{R}{3}=R$$

Subtract \frac{R}{3} from both sides of the equation.

$$P_{y}y=\frac{2R}{3}$$

Divide both sides by P_{y}.

$$y=\frac{2R}{3P_{y}}$$

The system is now solved.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{R}{3P_{x}}\text{, }y=\frac{2R}{3P_{y}}\text{, }&P_{y}\neq 0\text{ and }P_{x}\neq 0\\x=0\text{, }y\in \mathrm{R}\text{, }&P_{x}\neq 0\text{ and }R=0\text{ and }P_{y}=0\\x\in \mathrm{R}\text{, }y=0\text{, }&R=0\text{ and }P_{x}=0\\x\in \mathrm{R}\text{, }y\in \mathrm{R}\text{, }&R=0\text{ and }P_{x}=0\text{ and }P_{y}=0\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

$$P_{y}y+\left(-2P_{x}\right)x=0$$

Add 2P_{x}x to both sides of the equation.

$$P_{y}y=2P_{x}x$$

Divide both sides by P_{y}.

$$y=\frac{1}{P_{y}}\times 2P_{x}x$$

Multiply \frac{1}{P_{y}} times 2P_{x}x.

$$y=\frac{2P_{x}}{P_{y}}x$$

Substitute \frac{2P_{x}x}{P_{y}} for y in the other equation, P_{y}y+P_{x}x=R.

$$P_{y}\times \frac{2P_{x}}{P_{y}}x+P_{x}x=R$$

Multiply P_{y} times \frac{2P_{x}x}{P_{y}}.

$$2P_{x}x+P_{x}x=R$$

Add 2P_{x}x to P_{x}x.

$$3P_{x}x=R$$

Divide both sides by 3P_{x}.

$$x=\frac{R}{3P_{x}}$$

Substitute \frac{R}{3P_{x}} for x in y=\frac{2P_{x}}{P_{y}}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{2P_{x}}{P_{y}}\times \frac{R}{3P_{x}}$$

Multiply \frac{2P_{x}}{P_{y}} times \frac{R}{3P_{x}}.

$$y=\frac{2R}{3P_{y}}$$

The system is now solved.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$

Steps Using Matrices

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Write the equations in matrix form.

$$\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\R\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right).

$$inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}P_{y}&-2P_{x}\\P_{y}&P_{x}\end{matrix}\right))\left(\begin{matrix}0\\R\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{P_{x}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}&-\frac{-2P_{x}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}\\-\frac{P_{y}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}&\frac{P_{y}}{P_{y}P_{x}-\left(-2P_{x}\right)P_{y}}\end{matrix}\right)\left(\begin{matrix}0\\R\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3P_{y}}&\frac{2}{3P_{y}}\\-\frac{1}{3P_{x}}&\frac{1}{3P_{x}}\end{matrix}\right)\left(\begin{matrix}0\\R\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3P_{y}}R\\\frac{1}{3P_{x}}R\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2R}{3P_{y}}\\\frac{R}{3P_{x}}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$

Steps Using Elimination

Consider the first equation. Subtract 2xP_{x} from both sides.

$$P_{y}y-2xP_{x}=0$$

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

$$P_{x}x+P_{y}y=R$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$P_{y}y+\left(-2P_{x}\right)x=0,P_{y}y+P_{x}x=R$$

Subtract P_{y}y+P_{x}x=R from P_{y}y+\left(-2P_{x}\right)x=0 by subtracting like terms on each side of the equal sign.

$$P_{y}y+\left(-P_{y}\right)y+\left(-2P_{x}\right)x+\left(-P_{x}\right)x=-R$$

Add P_{y}y to -P_{y}y. Terms P_{y}y and -P_{y}y cancel out, leaving an equation with only one variable that can be solved.

$$\left(-2P_{x}\right)x+\left(-P_{x}\right)x=-R$$

Add -2P_{x}x to -P_{x}x.

$$\left(-3P_{x}\right)x=-R$$

Divide both sides by -3P_{x}.

$$x=\frac{R}{3P_{x}}$$

Substitute \frac{R}{3P_{x}} for x in P_{y}y+P_{x}x=R. Because the resulting equation contains only one variable, you can solve for y directly.

$$P_{y}y+P_{x}\times \frac{R}{3P_{x}}=R$$

Multiply P_{x} times \frac{R}{3P_{x}}.

$$P_{y}y+\frac{R}{3}=R$$

Subtract \frac{R}{3} from both sides of the equation.

$$P_{y}y=\frac{2R}{3}$$

Divide both sides by P_{y}.

$$y=\frac{2R}{3P_{y}}$$

The system is now solved.

$$y=\frac{2R}{3P_{y}},x=\frac{R}{3P_{x}}$$