Solve {l}{y-y1=m(x-x1)}{P-2=-1(1000y-2000)}

$$\left. \begin{array} { l } { y - y 1 = m ( x - x 1 ) } \\ { P - 2 = - 1 ( 1000 y - 2000 ) } \end{array} \right.$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}\text{, }y=\frac{2002-P}{1000}\text{, }&m\neq 0\\x\in \mathrm{C}\text{, }y=\frac{2002-P}{1000}\text{, }&y_{1}=-\frac{P}{1000}+\frac{1001}{500}\text{ and }m=0\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. To find the opposite of 1000y-2000, find the opposite of each term.

$$P-2=-1000y+2000$$

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

$$-1000y+2000=P-2$$

Subtract 2000 from both sides.

$$-1000y=P-2-2000$$

Subtract 2000 from -2 to get -2002.

$$-1000y=P-2002$$

Consider the first equation. Use the distributive property to multiply m by x-x_{1}.

$$y-y_{1}=mx-mx_{1}$$

Subtract mx from both sides.

$$y-y_{1}-mx=-mx_{1}$$

Add y_{1} to both sides.

$$y-mx=-mx_{1}+y_{1}$$

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.

$$-1000y=P-2002,y+\left(-m\right)x=y_{1}-mx_{1}$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$-1000y=P-2002$$

Divide both sides by -1000.

$$y=-\frac{P}{1000}+\frac{1001}{500}$$

Substitute -\frac{P}{1000}+\frac{1001}{500} for y in the other equation, y+\left(-m\right)x=y_{1}-mx_{1}.

$$-\frac{P}{1000}+\frac{1001}{500}+\left(-m\right)x=y_{1}-mx_{1}$$

Subtract -\frac{P}{1000}+\frac{1001}{500} from both sides of the equation.

$$\left(-m\right)x=-mx_{1}+\frac{P}{1000}+y_{1}-\frac{1001}{500}$$

Divide both sides by -m.

$$x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}$$

The system is now solved.

$$y=-\frac{P}{1000}+\frac{1001}{500},x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}$$

Solve for x, y

$$\left\{\begin{matrix}x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}\text{, }y=\frac{2002-P}{1000}\text{, }&m\neq 0\\x\in \mathrm{R}\text{, }y=\frac{2002-P}{1000}\text{, }&y_{1}=-\frac{P}{1000}+\frac{1001}{500}\text{ and }m=0\end{matrix}\right.$$

Steps Using Substitution

Consider the second equation. To find the opposite of 1000y-2000, find the opposite of each term.

$$P-2=-1000y+2000$$

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

$$-1000y+2000=P-2$$

Subtract 2000 from both sides.

$$-1000y=P-2-2000$$

Subtract 2000 from -2 to get -2002.

$$-1000y=P-2002$$

Consider the first equation. Use the distributive property to multiply m by x-x_{1}.

$$y-y_{1}=mx-mx_{1}$$

Subtract mx from both sides.

$$y-y_{1}-mx=-mx_{1}$$

Add y_{1} to both sides.

$$y-mx=-mx_{1}+y_{1}$$

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.

$$-1000y=P-2002,y+\left(-m\right)x=y_{1}-mx_{1}$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$-1000y=P-2002$$

Divide both sides by -1000.

$$y=-\frac{P}{1000}+\frac{1001}{500}$$

Substitute -\frac{P}{1000}+\frac{1001}{500} for y in the other equation, y+\left(-m\right)x=y_{1}-mx_{1}.

$$-\frac{P}{1000}+\frac{1001}{500}+\left(-m\right)x=y_{1}-mx_{1}$$

Subtract -\frac{P}{1000}+\frac{1001}{500} from both sides of the equation.

$$\left(-m\right)x=-mx_{1}+\frac{P}{1000}+y_{1}-\frac{1001}{500}$$

Divide both sides by -m.

$$x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}$$

The system is now solved.

$$y=-\frac{P}{1000}+\frac{1001}{500},x=-\frac{-1000mx_{1}+1000y_{1}+P-2002}{1000m}$$