Solve {l}{x-8=y-6}{xy=(x-8)(y-6)+104}

$$\left\{ \begin{array} { l } { x - 8 = y - 6 } \\ { x y = ( x - 8 ) ( y - 6 ) + 104 } \end{array} \right.$$

Solve for x, y

$$x=12$$

$$y=10$$

Steps Using Substitution

Consider the first equation. Subtract y from both sides.

$$x-8-y=-6$$

Add 8 to both sides.

$$x-y=-6+8$$

Add -6 and 8 to get 2.

$$x-y=2$$

Consider the second equation. Use the distributive property to multiply x-8 by y-6.

$$xy=xy-6x-8y+48+104$$

Add 48 and 104 to get 152.

$$xy=xy-6x-8y+152$$

Subtract xy from both sides.

$$xy-xy=-6x-8y+152$$

Combine xy and -xy to get 0.

$$0=-6x-8y+152$$

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

$$-6x-8y+152=0$$

Subtract 152 from both sides. Anything subtracted from zero gives its negation.

$$-6x-8y=-152$$

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.

$$x-y=2,-6x-8y=-152$$

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

$$x-y=2$$

Add y to both sides of the equation.

$$x=y+2$$

Substitute y+2 for x in the other equation, -6x-8y=-152.

$$-6\left(y+2\right)-8y=-152$$

Multiply -6 times y+2.

$$-6y-12-8y=-152$$

Add -6y to -8y.

$$-14y-12=-152$$

Add 12 to both sides of the equation.

$$-14y=-140$$

Divide both sides by -14.

$$y=10$$

Substitute 10 for y in x=y+2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=10+2$$

Add 2 to 10.

$$x=12$$

The system is now solved.

$$x=12,y=10$$

Steps Using Matrices

Consider the first equation. Subtract y from both sides.

$$x-8-y=-6$$

Add 8 to both sides.

$$x-y=-6+8$$

Add -6 and 8 to get 2.

$$x-y=2$$

Consider the second equation. Use the distributive property to multiply x-8 by y-6.

$$xy=xy-6x-8y+48+104$$

Add 48 and 104 to get 152.

$$xy=xy-6x-8y+152$$

Subtract xy from both sides.

$$xy-xy=-6x-8y+152$$

Combine xy and -xy to get 0.

$$0=-6x-8y+152$$

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

$$-6x-8y+152=0$$

Subtract 152 from both sides. Anything subtracted from zero gives its negation.

$$-6x-8y=-152$$

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

$$x-y=2,-6x-8y=-152$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\-152\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right))\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right))\left(\begin{matrix}2\\-152\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right))\left(\begin{matrix}2\\-152\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-6&-8\end{matrix}\right))\left(\begin{matrix}2\\-152\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{-8-\left(-\left(-6\right)\right)}&-\frac{-1}{-8-\left(-\left(-6\right)\right)}\\-\frac{-6}{-8-\left(-\left(-6\right)\right)}&\frac{1}{-8-\left(-\left(-6\right)\right)}\end{matrix}\right)\left(\begin{matrix}2\\-152\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}&-\frac{1}{14}\\-\frac{3}{7}&-\frac{1}{14}\end{matrix}\right)\left(\begin{matrix}2\\-152\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{7}\times 2-\frac{1}{14}\left(-152\right)\\-\frac{3}{7}\times 2-\frac{1}{14}\left(-152\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12,y=10$$

Steps Using Elimination

Consider the first equation. Subtract y from both sides.

$$x-8-y=-6$$

Add 8 to both sides.

$$x-y=-6+8$$

Add -6 and 8 to get 2.

$$x-y=2$$

Consider the second equation. Use the distributive property to multiply x-8 by y-6.

$$xy=xy-6x-8y+48+104$$

Add 48 and 104 to get 152.

$$xy=xy-6x-8y+152$$

Subtract xy from both sides.

$$xy-xy=-6x-8y+152$$

Combine xy and -xy to get 0.

$$0=-6x-8y+152$$

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

$$-6x-8y+152=0$$

Subtract 152 from both sides. Anything subtracted from zero gives its negation.

$$-6x-8y=-152$$

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.

$$x-y=2,-6x-8y=-152$$

To make x and -6x equal, multiply all terms on each side of the first equation by -6 and all terms on each side of the second by 1.

$$-6x-6\left(-1\right)y=-6\times 2,-6x-8y=-152$$

Simplify.

$$-6x+6y=-12,-6x-8y=-152$$

Subtract -6x-8y=-152 from -6x+6y=-12 by subtracting like terms on each side of the equal sign.

$$-6x+6x+6y+8y=-12+152$$

Add -6x to 6x. Terms -6x and 6x cancel out, leaving an equation with only one variable that can be solved.

$$6y+8y=-12+152$$

Add 6y to 8y.

$$14y=-12+152$$

Add -12 to 152.

$$14y=140$$

Divide both sides by 14.

$$y=10$$

Substitute 10 for y in -6x-8y=-152. Because the resulting equation contains only one variable, you can solve for x directly.

$$-6x-8\times 10=-152$$

Multiply -8 times 10.

$$-6x-80=-152$$

Add 80 to both sides of the equation.

$$-6x=-72$$

Divide both sides by -6.

$$x=12$$

The system is now solved.

$$x=12,y=10$$