Solve {l}{y=frac{k(x_{1}-1)}{x_{1}+2}(x+2)}{y=frac{k(x_{2}-1)}{x_{2}-2}(x-2)}

$$\left\{ \begin{array} { l } { y = \frac { k ( x _ { 1 } - 1 ) } { x _ { 1 } + 2 } ( x + 2 ) } \\ { y = \frac { k ( x _ { 2 } - 1 ) } { x _ { 2 } - 2 } ( x - 2 ) } \end{array} \right.$$

Solve for x, y

$$\left\{\begin{matrix}x=-\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{4-3x_{2}-x_{1}}\text{, }y=\frac{4k\left(1-x_{2}\right)\left(x_{1}-1\right)}{4-3x_{2}-x_{1}}\text{, }&x_{1}\neq 4-3x_{2}\text{ and }k\neq 0\text{ and }x_{1}\neq -2\text{ and }x_{2}\neq 2\\x\in \mathrm{R}\text{, }y=\frac{k\left(x_{1}-1\right)\left(x+2\right)}{x_{1}+2}\text{, }&\left(k=0\text{ and }x_{1}\neq -2\text{ and }x_{2}\neq 2\right)\text{ or }\left(x_{2}=1\text{ and }x_{1}=1\right)\end{matrix}\right.$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by x_{1}+2.

$$y\left(x_{1}+2\right)=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply y by x_{1}+2.

$$yx_{1}+2y=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply k by x_{1}-1.

$$yx_{1}+2y=\left(kx_{1}-k\right)\left(x+2\right)$$

Use the distributive property to multiply kx_{1}-k by x+2.

$$yx_{1}+2y=kx_{1}x+2kx_{1}-kx-2k$$

Subtract kx_{1}x from both sides.

$$yx_{1}+2y-kx_{1}x=2kx_{1}-kx-2k$$

Add kx to both sides.

$$yx_{1}+2y-kx_{1}x+kx=2kx_{1}-2k$$

Reorder the terms.

$$-kxx_{1}+kx+x_{1}y+2y=2kx_{1}-2k$$

Combine all terms containing x,y.

$$\left(-kx_{1}+k\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k$$

Consider the second equation. Multiply both sides of the equation by x_{2}-2.

$$y\left(x_{2}-2\right)=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply y by x_{2}-2.

$$yx_{2}-2y=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply k by x_{2}-1.

$$yx_{2}-2y=\left(kx_{2}-k\right)\left(x-2\right)$$

Use the distributive property to multiply kx_{2}-k by x-2.

$$yx_{2}-2y=kx_{2}x-2kx_{2}-kx+2k$$

Subtract kx_{2}x from both sides.

$$yx_{2}-2y-kx_{2}x=-2kx_{2}-kx+2k$$

Add kx to both sides.

$$yx_{2}-2y-kx_{2}x+kx=-2kx_{2}+2k$$

Reorder the terms.

$$-kxx_{2}+kx+x_{2}y-2y=-2kx_{2}+2k$$

Combine all terms containing x,y.

$$\left(-kx_{2}+k\right)x+\left(x_{2}-2\right)y=-2kx_{2}+2k$$

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.

$$\left(k-kx_{1}\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k,\left(k-kx_{2}\right)x+\left(x_{2}-2\right)y=2k-2kx_{2}$$

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

$$\left(k-kx_{1}\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k$$

Subtract \left(x_{1}+2\right)y from both sides of the equation.

$$\left(k-kx_{1}\right)x=\left(-\left(x_{1}+2\right)\right)y+2k\left(x_{1}-1\right)$$

Divide both sides by -kx_{1}+k.

$$x=\frac{1}{k\left(1-x_{1}\right)}\left(\left(-\left(x_{1}+2\right)\right)y+2k\left(x_{1}-1\right)\right)$$

Multiply \frac{1}{k\left(-x_{1}+1\right)} times -\left(x_{1}+2\right)y+2k\left(-1+x_{1}\right).

$$x=\left(-\frac{x_{1}+2}{k\left(1-x_{1}\right)}\right)y-2$$

Substitute \frac{-yx_{1}-2y+2kx_{1}-2k}{\left(-x_{1}+1\right)k} for x in the other equation, \left(k-kx_{2}\right)x+\left(x_{2}-2\right)y=2k-2kx_{2}.

$$\left(k-kx_{2}\right)\left(\left(-\frac{x_{1}+2}{k\left(1-x_{1}\right)}\right)y-2\right)+\left(x_{2}-2\right)y=2k-2kx_{2}$$

Multiply -kx_{2}+k times \frac{-yx_{1}-2y+2kx_{1}-2k}{\left(-x_{1}+1\right)k}.

$$\left(-\frac{\left(1-x_{2}\right)\left(x_{1}+2\right)}{1-x_{1}}\right)y+2kx_{2}-2k+\left(x_{2}-2\right)y=2k-2kx_{2}$$

Add -\frac{\left(1-x_{2}\right)\left(x_{1}+2\right)y}{-x_{1}+1} to \left(x_{2}-2\right)y.

$$\frac{x_{1}+3x_{2}-4}{1-x_{1}}y+2kx_{2}-2k=2k-2kx_{2}$$

Subtract 2k\left(-1+x_{2}\right) from both sides of the equation.

$$\frac{x_{1}+3x_{2}-4}{1-x_{1}}y=4k-4kx_{2}$$

Divide both sides by \frac{x_{1}+3x_{2}-4}{-x_{1}+1}.

$$y=\frac{4k\left(1-x_{1}\right)\left(1-x_{2}\right)}{x_{1}+3x_{2}-4}$$

Substitute \frac{4k\left(1-x_{2}\right)\left(-x_{1}+1\right)}{x_{1}+3x_{2}-4} for y in x=\left(-\frac{x_{1}+2}{k\left(1-x_{1}\right)}\right)y-2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\left(-\frac{x_{1}+2}{k\left(1-x_{1}\right)}\right)\times \frac{4k\left(1-x_{1}\right)\left(1-x_{2}\right)}{x_{1}+3x_{2}-4}-2$$

Multiply -\frac{x_{1}+2}{k\left(-x_{1}+1\right)} times \frac{4k\left(1-x_{2}\right)\left(-x_{1}+1\right)}{x_{1}+3x_{2}-4}.

$$x=-\frac{4\left(1-x_{2}\right)\left(x_{1}+2\right)}{x_{1}+3x_{2}-4}-2$$

Add -2 to -\frac{4\left(x_{1}+2\right)\left(1-x_{2}\right)}{x_{1}+3x_{2}-4}.

$$x=\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{x_{1}+3x_{2}-4}$$

The system is now solved.

$$x=\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{x_{1}+3x_{2}-4},y=\frac{4k\left(1-x_{1}\right)\left(1-x_{2}\right)}{x_{1}+3x_{2}-4}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by x_{1}+2.

$$y\left(x_{1}+2\right)=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply y by x_{1}+2.

$$yx_{1}+2y=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply k by x_{1}-1.

$$yx_{1}+2y=\left(kx_{1}-k\right)\left(x+2\right)$$

Use the distributive property to multiply kx_{1}-k by x+2.

$$yx_{1}+2y=kx_{1}x+2kx_{1}-kx-2k$$

Subtract kx_{1}x from both sides.

$$yx_{1}+2y-kx_{1}x=2kx_{1}-kx-2k$$

Add kx to both sides.

$$yx_{1}+2y-kx_{1}x+kx=2kx_{1}-2k$$

Reorder the terms.

$$-kxx_{1}+kx+x_{1}y+2y=2kx_{1}-2k$$

Combine all terms containing x,y.

$$\left(-kx_{1}+k\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k$$

Consider the second equation. Multiply both sides of the equation by x_{2}-2.

$$y\left(x_{2}-2\right)=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply y by x_{2}-2.

$$yx_{2}-2y=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply k by x_{2}-1.

$$yx_{2}-2y=\left(kx_{2}-k\right)\left(x-2\right)$$

Use the distributive property to multiply kx_{2}-k by x-2.

$$yx_{2}-2y=kx_{2}x-2kx_{2}-kx+2k$$

Subtract kx_{2}x from both sides.

$$yx_{2}-2y-kx_{2}x=-2kx_{2}-kx+2k$$

Add kx to both sides.

$$yx_{2}-2y-kx_{2}x+kx=-2kx_{2}+2k$$

Reorder the terms.

$$-kxx_{2}+kx+x_{2}y-2y=-2kx_{2}+2k$$

Combine all terms containing x,y.

$$\left(-kx_{2}+k\right)x+\left(x_{2}-2\right)y=-2kx_{2}+2k$$

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

$$\left(k-kx_{1}\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k,\left(k-kx_{2}\right)x+\left(x_{2}-2\right)y=2k-2kx_{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-kx_{1}+k&x_{1}+2\\-kx_{2}+k&x_{2}-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right))\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right))\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right))\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}k-kx_{1}&x_{1}+2\\k-kx_{2}&x_{2}-2\end{matrix}\right))\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{x_{2}-2}{\left(k-kx_{1}\right)\left(x_{2}-2\right)-\left(x_{1}+2\right)\left(k-kx_{2}\right)}&-\frac{x_{1}+2}{\left(k-kx_{1}\right)\left(x_{2}-2\right)-\left(x_{1}+2\right)\left(k-kx_{2}\right)}\\-\frac{k-kx_{2}}{\left(k-kx_{1}\right)\left(x_{2}-2\right)-\left(x_{1}+2\right)\left(k-kx_{2}\right)}&\frac{k-kx_{1}}{\left(k-kx_{1}\right)\left(x_{2}-2\right)-\left(x_{1}+2\right)\left(k-kx_{2}\right)}\end{matrix}\right)\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{x_{2}-2}{k\left(x_{1}+3x_{2}-4\right)}&-\frac{x_{1}+2}{k\left(x_{1}+3x_{2}-4\right)}\\-\frac{1-x_{2}}{x_{1}+3x_{2}-4}&\frac{1-x_{1}}{x_{1}+3x_{2}-4}\end{matrix}\right)\left(\begin{matrix}2k\left(x_{1}-1\right)\\2k-2kx_{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{x_{2}-2}{k\left(x_{1}+3x_{2}-4\right)}\times 2k\left(x_{1}-1\right)+\left(-\frac{x_{1}+2}{k\left(x_{1}+3x_{2}-4\right)}\right)\left(2k-2kx_{2}\right)\\\left(-\frac{1-x_{2}}{x_{1}+3x_{2}-4}\right)\times 2k\left(x_{1}-1\right)+\frac{1-x_{1}}{x_{1}+3x_{2}-4}\left(2k-2kx_{2}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{x_{1}+3x_{2}-4}\\\frac{4k\left(x_{1}-1\right)\left(x_{2}-1\right)}{x_{1}+3x_{2}-4}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{x_{1}+3x_{2}-4},y=\frac{4k\left(x_{1}-1\right)\left(x_{2}-1\right)}{x_{1}+3x_{2}-4}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by x_{1}+2.

$$y\left(x_{1}+2\right)=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply y by x_{1}+2.

$$yx_{1}+2y=k\left(x_{1}-1\right)\left(x+2\right)$$

Use the distributive property to multiply k by x_{1}-1.

$$yx_{1}+2y=\left(kx_{1}-k\right)\left(x+2\right)$$

Use the distributive property to multiply kx_{1}-k by x+2.

$$yx_{1}+2y=kx_{1}x+2kx_{1}-kx-2k$$

Subtract kx_{1}x from both sides.

$$yx_{1}+2y-kx_{1}x=2kx_{1}-kx-2k$$

Add kx to both sides.

$$yx_{1}+2y-kx_{1}x+kx=2kx_{1}-2k$$

Reorder the terms.

$$-kxx_{1}+kx+x_{1}y+2y=2kx_{1}-2k$$

Combine all terms containing x,y.

$$\left(-kx_{1}+k\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k$$

Consider the second equation. Multiply both sides of the equation by x_{2}-2.

$$y\left(x_{2}-2\right)=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply y by x_{2}-2.

$$yx_{2}-2y=k\left(x_{2}-1\right)\left(x-2\right)$$

Use the distributive property to multiply k by x_{2}-1.

$$yx_{2}-2y=\left(kx_{2}-k\right)\left(x-2\right)$$

Use the distributive property to multiply kx_{2}-k by x-2.

$$yx_{2}-2y=kx_{2}x-2kx_{2}-kx+2k$$

Subtract kx_{2}x from both sides.

$$yx_{2}-2y-kx_{2}x=-2kx_{2}-kx+2k$$

Add kx to both sides.

$$yx_{2}-2y-kx_{2}x+kx=-2kx_{2}+2k$$

Reorder the terms.

$$-kxx_{2}+kx+x_{2}y-2y=-2kx_{2}+2k$$

Combine all terms containing x,y.

$$\left(-kx_{2}+k\right)x+\left(x_{2}-2\right)y=-2kx_{2}+2k$$

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.

$$\left(k-kx_{1}\right)x+\left(x_{1}+2\right)y=2kx_{1}-2k,\left(k-kx_{2}\right)x+\left(x_{2}-2\right)y=2k-2kx_{2}$$

To make k\left(-x_{1}+1\right)x and k\left(-x_{2}+1\right)x equal, multiply all terms on each side of the first equation by -kx_{2}+k and all terms on each side of the second by -x_{1}k+k.

$$\left(k-kx_{2}\right)\left(k-kx_{1}\right)x+\left(k-kx_{2}\right)\left(x_{1}+2\right)y=\left(k-kx_{2}\right)\left(2kx_{1}-2k\right),\left(k-kx_{1}\right)\left(k-kx_{2}\right)x+\left(k-kx_{1}\right)\left(x_{2}-2\right)y=\left(k-kx_{1}\right)\left(2k-2kx_{2}\right)$$

Simplify.

$$\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}x+k\left(1-x_{2}\right)\left(x_{1}+2\right)y=2\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2},\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}x+k\left(1-x_{1}\right)\left(x_{2}-2\right)y=2\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}$$

Subtract \left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}x+k\left(1-x_{1}\right)\left(x_{2}-2\right)y=2\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2} from \left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}x+k\left(1-x_{2}\right)\left(x_{1}+2\right)y=2\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2} by subtracting like terms on each side of the equal sign.

$$\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}x+\left(-\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}\right)x+k\left(1-x_{2}\right)\left(x_{1}+2\right)y+\left(-k\left(1-x_{1}\right)\left(x_{2}-2\right)\right)y=2\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2}-2\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}$$

Add \left(-x_{2}+1\right)\left(-x_{1}+1\right)k^{2}x to -\left(-x_{2}+1\right)\left(-x_{1}+1\right)k^{2}x. Terms \left(-x_{2}+1\right)\left(-x_{1}+1\right)k^{2}x and -\left(-x_{2}+1\right)\left(-x_{1}+1\right)k^{2}x cancel out, leaving an equation with only one variable that can be solved.

$$k\left(1-x_{2}\right)\left(x_{1}+2\right)y+\left(-k\left(1-x_{1}\right)\left(x_{2}-2\right)\right)y=2\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2}-2\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}$$

Add k\left(-x_{2}+1\right)\left(x_{1}+2\right)y to -k\left(-x_{1}+1\right)\left(x_{2}-2\right)y.

$$k\left(4-3x_{2}-x_{1}\right)y=2\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2}-2\left(1-x_{1}\right)\left(1-x_{2}\right)k^{2}$$

Add 2\left(1-x_{2}\right)\left(-1+x_{1}\right)k^{2} to -2\left(-x_{1}+1\right)\left(-x_{2}+1\right)k^{2}.

$$k\left(4-3x_{2}-x_{1}\right)y=4\left(1-x_{2}\right)\left(x_{1}-1\right)k^{2}$$

Divide both sides by k\left(4-3x_{2}-x_{1}\right).

$$y=\frac{4k\left(1-x_{2}\right)\left(x_{1}-1\right)}{4-3x_{2}-x_{1}}$$

Substitute \frac{4\left(-1+x_{1}\right)\left(1-x_{2}\right)k}{4-x_{1}-3x_{2}} for y in \left(k-kx_{2}\right)x+\left(x_{2}-2\right)y=2k-2kx_{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$\left(k-kx_{2}\right)x+\left(x_{2}-2\right)\times \frac{4k\left(1-x_{2}\right)\left(x_{1}-1\right)}{4-3x_{2}-x_{1}}=2k-2kx_{2}$$

Multiply x_{2}-2 times \frac{4\left(-1+x_{1}\right)\left(1-x_{2}\right)k}{4-x_{1}-3x_{2}}.

$$\left(k-kx_{2}\right)x+\frac{4k\left(1-x_{2}\right)\left(x_{1}-1\right)\left(x_{2}-2\right)}{4-3x_{2}-x_{1}}=2k-2kx_{2}$$

Subtract \frac{4\left(x_{2}-2\right)\left(-1+x_{1}\right)\left(1-x_{2}\right)k}{4-x_{1}-3x_{2}} from both sides of the equation.

$$\left(k-kx_{2}\right)x=\frac{2k\left(x_{2}-1\right)\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{4-3x_{2}-x_{1}}$$

Divide both sides by -kx_{2}+k.

$$x=-\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{4-3x_{2}-x_{1}}$$

The system is now solved.

$$x=-\frac{2\left(2x_{1}x_{2}-3x_{1}+x_{2}\right)}{4-3x_{2}-x_{1}},y=\frac{4k\left(1-x_{2}\right)\left(x_{1}-1\right)}{4-3x_{2}-x_{1}}$$