Solve {l}{x-2/x+2=y-7/y-5}{x+1/x-1=y+3/y-5}

$$\left. \begin{array} { l } { \frac { x - 2 } { x + 2 } = \frac { y - 7 } { y - 5 } } \\ { \frac { x + 1 } { x - 1 } = \frac { y + 3 } { y - 5 } } \end{array} \right.$$

Solve for x, y

$$x = \frac{10}{7} = 1\frac{3}{7} \approx 1.428571429$$

$$y = \frac{47}{7} = 6\frac{5}{7} \approx 6.714285714$$

Steps Using Substitution

Consider the first equation. Variable x cannot be equal to -2 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(y-5\right)\left(x+2\right), the least common multiple of x+2,y-5.

$$\left(y-5\right)\left(x-2\right)=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply y-5 by x-2.

$$yx-2y-5x+10=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply x+2 by y-7.

$$yx-2y-5x+10=xy-7x+2y-14$$

Subtract xy from both sides.

$$yx-2y-5x+10-xy=-7x+2y-14$$

Combine yx and -xy to get 0.

$$-2y-5x+10=-7x+2y-14$$

Add 7x to both sides.

$$-2y-5x+10+7x=2y-14$$

Combine -5x and 7x to get 2x.

$$-2y+2x+10=2y-14$$

Subtract 2y from both sides.

$$-2y+2x+10-2y=-14$$

Combine -2y and -2y to get -4y.

$$-4y+2x+10=-14$$

Subtract 10 from both sides.

$$-4y+2x=-14-10$$

Subtract 10 from -14 to get -24.

$$-4y+2x=-24$$

Consider the second equation. Variable x cannot be equal to 1 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(y-5\right), the least common multiple of x-1,y-5.

$$\left(y-5\right)\left(x+1\right)=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply y-5 by x+1.

$$yx+y-5x-5=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply x-1 by y+3.

$$yx+y-5x-5=xy+3x-y-3$$

Subtract xy from both sides.

$$yx+y-5x-5-xy=3x-y-3$$

Combine yx and -xy to get 0.

$$y-5x-5=3x-y-3$$

Subtract 3x from both sides.

$$y-5x-5-3x=-y-3$$

Combine -5x and -3x to get -8x.

$$y-8x-5=-y-3$$

Add y to both sides.

$$y-8x-5+y=-3$$

Combine y and y to get 2y.

$$2y-8x-5=-3$$

Add 5 to both sides.

$$2y-8x=-3+5$$

Add -3 and 5 to get 2.

$$2y-8x=2$$

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.

$$-4y+2x=-24,2y-8x=2$$

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

$$-4y+2x=-24$$

Subtract 2x from both sides of the equation.

$$-4y=-2x-24$$

Divide both sides by -4.

$$y=-\frac{1}{4}\left(-2x-24\right)$$

Multiply -\frac{1}{4} times -2x-24.

$$y=\frac{1}{2}x+6$$

Substitute \frac{x}{2}+6 for y in the other equation, 2y-8x=2.

$$2\left(\frac{1}{2}x+6\right)-8x=2$$

Multiply 2 times \frac{x}{2}+6.

$$x+12-8x=2$$

Add x to -8x.

$$-7x+12=2$$

Subtract 12 from both sides of the equation.

$$-7x=-10$$

Divide both sides by -7.

$$x=\frac{10}{7}$$

Substitute \frac{10}{7} for x in y=\frac{1}{2}x+6. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{1}{2}\times \frac{10}{7}+6$$

Multiply \frac{1}{2} times \frac{10}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{5}{7}+6$$

Add 6 to \frac{5}{7}.

$$y=\frac{47}{7}$$

The system is now solved.

$$y=\frac{47}{7},x=\frac{10}{7}$$

Steps Using Matrices

Consider the first equation. Variable x cannot be equal to -2 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(y-5\right)\left(x+2\right), the least common multiple of x+2,y-5.

$$\left(y-5\right)\left(x-2\right)=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply y-5 by x-2.

$$yx-2y-5x+10=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply x+2 by y-7.

$$yx-2y-5x+10=xy-7x+2y-14$$

Subtract xy from both sides.

$$yx-2y-5x+10-xy=-7x+2y-14$$

Combine yx and -xy to get 0.

$$-2y-5x+10=-7x+2y-14$$

Add 7x to both sides.

$$-2y-5x+10+7x=2y-14$$

Combine -5x and 7x to get 2x.

$$-2y+2x+10=2y-14$$

Subtract 2y from both sides.

$$-2y+2x+10-2y=-14$$

Combine -2y and -2y to get -4y.

$$-4y+2x+10=-14$$

Subtract 10 from both sides.

$$-4y+2x=-14-10$$

Subtract 10 from -14 to get -24.

$$-4y+2x=-24$$

Consider the second equation. Variable x cannot be equal to 1 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(y-5\right), the least common multiple of x-1,y-5.

$$\left(y-5\right)\left(x+1\right)=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply y-5 by x+1.

$$yx+y-5x-5=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply x-1 by y+3.

$$yx+y-5x-5=xy+3x-y-3$$

Subtract xy from both sides.

$$yx+y-5x-5-xy=3x-y-3$$

Combine yx and -xy to get 0.

$$y-5x-5=3x-y-3$$

Subtract 3x from both sides.

$$y-5x-5-3x=-y-3$$

Combine -5x and -3x to get -8x.

$$y-8x-5=-y-3$$

Add y to both sides.

$$y-8x-5+y=-3$$

Combine y and y to get 2y.

$$2y-8x-5=-3$$

Add 5 to both sides.

$$2y-8x=-3+5$$

Add -3 and 5 to get 2.

$$2y-8x=2$$

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

$$-4y+2x=-24,2y-8x=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-24\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-4&2\\2&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right))\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right))\left(\begin{matrix}-24\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right))\left(\begin{matrix}-24\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-4&2\\2&-8\end{matrix}\right))\left(\begin{matrix}-24\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{-4\left(-8\right)-2\times 2}&-\frac{2}{-4\left(-8\right)-2\times 2}\\-\frac{2}{-4\left(-8\right)-2\times 2}&-\frac{4}{-4\left(-8\right)-2\times 2}\end{matrix}\right)\left(\begin{matrix}-24\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{7}&-\frac{1}{14}\\-\frac{1}{14}&-\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}-24\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{7}\left(-24\right)-\frac{1}{14}\times 2\\-\frac{1}{14}\left(-24\right)-\frac{1}{7}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{47}{7}\\\frac{10}{7}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{47}{7},x=\frac{10}{7}$$

Steps Using Elimination

Consider the first equation. Variable x cannot be equal to -2 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(y-5\right)\left(x+2\right), the least common multiple of x+2,y-5.

$$\left(y-5\right)\left(x-2\right)=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply y-5 by x-2.

$$yx-2y-5x+10=\left(x+2\right)\left(y-7\right)$$

Use the distributive property to multiply x+2 by y-7.

$$yx-2y-5x+10=xy-7x+2y-14$$

Subtract xy from both sides.

$$yx-2y-5x+10-xy=-7x+2y-14$$

Combine yx and -xy to get 0.

$$-2y-5x+10=-7x+2y-14$$

Add 7x to both sides.

$$-2y-5x+10+7x=2y-14$$

Combine -5x and 7x to get 2x.

$$-2y+2x+10=2y-14$$

Subtract 2y from both sides.

$$-2y+2x+10-2y=-14$$

Combine -2y and -2y to get -4y.

$$-4y+2x+10=-14$$

Subtract 10 from both sides.

$$-4y+2x=-14-10$$

Subtract 10 from -14 to get -24.

$$-4y+2x=-24$$

Consider the second equation. Variable x cannot be equal to 1 since division by zero is not defined. Variable y cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(y-5\right), the least common multiple of x-1,y-5.

$$\left(y-5\right)\left(x+1\right)=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply y-5 by x+1.

$$yx+y-5x-5=\left(x-1\right)\left(y+3\right)$$

Use the distributive property to multiply x-1 by y+3.

$$yx+y-5x-5=xy+3x-y-3$$

Subtract xy from both sides.

$$yx+y-5x-5-xy=3x-y-3$$

Combine yx and -xy to get 0.

$$y-5x-5=3x-y-3$$

Subtract 3x from both sides.

$$y-5x-5-3x=-y-3$$

Combine -5x and -3x to get -8x.

$$y-8x-5=-y-3$$

Add y to both sides.

$$y-8x-5+y=-3$$

Combine y and y to get 2y.

$$2y-8x-5=-3$$

Add 5 to both sides.

$$2y-8x=-3+5$$

Add -3 and 5 to get 2.

$$2y-8x=2$$

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.

$$-4y+2x=-24,2y-8x=2$$

To make -4y and 2y equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by -4.

$$2\left(-4\right)y+2\times 2x=2\left(-24\right),-4\times 2y-4\left(-8\right)x=-4\times 2$$

Simplify.

$$-8y+4x=-48,-8y+32x=-8$$

Subtract -8y+32x=-8 from -8y+4x=-48 by subtracting like terms on each side of the equal sign.

$$-8y+8y+4x-32x=-48+8$$

Add -8y to 8y. Terms -8y and 8y cancel out, leaving an equation with only one variable that can be solved.

$$4x-32x=-48+8$$

Add 4x to -32x.

$$-28x=-48+8$$

Add -48 to 8.

$$-28x=-40$$

Divide both sides by -28.

$$x=\frac{10}{7}$$

Substitute \frac{10}{7} for x in 2y-8x=2. Because the resulting equation contains only one variable, you can solve for y directly.

$$2y-8\times \frac{10}{7}=2$$

Multiply -8 times \frac{10}{7}.

$$2y-\frac{80}{7}=2$$

Add \frac{80}{7} to both sides of the equation.

$$2y=\frac{94}{7}$$

Divide both sides by 2.

$$y=\frac{47}{7}$$

The system is now solved.

$$y=\frac{47}{7},x=\frac{10}{7}$$