Solve {r}{x+1/2+y-1/3=8}{x-1/3+y+1/2=9}

$$\left. \begin{array} { r } { \frac { x + 1 } { 2 } + \frac { y - 1 } { 3 } = 8 } \\ { \frac { x - 1 } { 3 } + \frac { y + 1 } { 2 } = 9 } \end{array} \right.$$

Solve for x, y

$$x=7$$

$$y=13$$

Steps Using Substitution

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.

$$\frac{1}{2}\left(x+1\right)+\frac{1}{3}\left(y-1\right)=8,\frac{1}{3}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=9$$

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

$$\frac{1}{2}\left(x+1\right)+\frac{1}{3}\left(y-1\right)=8$$

Multiply \frac{1}{2} times x+1.

$$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}\left(y-1\right)=8$$

Multiply \frac{1}{3} times y-1.

$$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}y-\frac{1}{3}=8$$

Add \frac{1}{2} to -\frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{2}x+\frac{1}{3}y+\frac{1}{6}=8$$

Subtract \frac{1}{6} from both sides of the equation.

$$\frac{1}{2}x+\frac{1}{3}y=\frac{47}{6}$$

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

$$\frac{1}{2}x=-\frac{1}{3}y+\frac{47}{6}$$

Multiply both sides by 2.

$$x=2\left(-\frac{1}{3}y+\frac{47}{6}\right)$$

Multiply 2 times -\frac{y}{3}+\frac{47}{6}.

$$x=-\frac{2}{3}y+\frac{47}{3}$$

Substitute \frac{-2y+47}{3} for x in the other equation, \frac{1}{3}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=9.

$$\frac{1}{3}\left(-\frac{2}{3}y+\frac{47}{3}-1\right)+\frac{1}{2}\left(y+1\right)=9$$

Add \frac{47}{3} to -1.

$$\frac{1}{3}\left(-\frac{2}{3}y+\frac{44}{3}\right)+\frac{1}{2}\left(y+1\right)=9$$

Multiply \frac{1}{3} times \frac{-2y+44}{3}.

$$-\frac{2}{9}y+\frac{44}{9}+\frac{1}{2}\left(y+1\right)=9$$

Multiply \frac{1}{2} times y+1.

$$-\frac{2}{9}y+\frac{44}{9}+\frac{1}{2}y+\frac{1}{2}=9$$

Add -\frac{2y}{9} to \frac{y}{2}.

$$\frac{5}{18}y+\frac{44}{9}+\frac{1}{2}=9$$

Add \frac{44}{9} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{5}{18}y+\frac{97}{18}=9$$

Subtract \frac{97}{18} from both sides of the equation.

$$\frac{5}{18}y=\frac{65}{18}$$

Divide both sides of the equation by \frac{5}{18}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=13$$

Substitute 13 for y in x=-\frac{2}{3}y+\frac{47}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times 13+\frac{47}{3}$$

Multiply -\frac{2}{3} times 13.

$$x=\frac{-26+47}{3}$$

Add \frac{47}{3} to -\frac{26}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=7$$

The system is now solved.

$$x=7,y=13$$

Steps Using Matrices

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

$$\frac{1}{2}\left(x+1\right)+\frac{1}{3}\left(y-1\right)=8,\frac{1}{3}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=9$$

Simplify the first equation to put it in standard form.

$$\frac{1}{2}\left(x+1\right)+\frac{1}{3}\left(y-1\right)=8$$

Multiply \frac{1}{2} times x+1.

$$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}\left(y-1\right)=8$$

Multiply \frac{1}{3} times y-1.

$$\frac{1}{2}x+\frac{1}{2}+\frac{1}{3}y-\frac{1}{3}=8$$

Add \frac{1}{2} to -\frac{1}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{2}x+\frac{1}{3}y+\frac{1}{6}=8$$

Subtract \frac{1}{6} from both sides of the equation.

$$\frac{1}{2}x+\frac{1}{3}y=\frac{47}{6}$$

Simplify the second equation to put it in standard form.

$$\frac{1}{3}\left(x-1\right)+\frac{1}{2}\left(y+1\right)=9$$

Multiply \frac{1}{3} times x-1.

$$\frac{1}{3}x-\frac{1}{3}+\frac{1}{2}\left(y+1\right)=9$$

Multiply \frac{1}{2} times y+1.

$$\frac{1}{3}x-\frac{1}{3}+\frac{1}{2}y+\frac{1}{2}=9$$

Add -\frac{1}{3} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$\frac{1}{3}x+\frac{1}{2}y+\frac{1}{6}=9$$

Subtract \frac{1}{6} from both sides of the equation.

$$\frac{1}{3}x+\frac{1}{2}y=\frac{53}{6}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{2}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}-\frac{1}{3}\times \frac{1}{3}}&-\frac{\frac{1}{3}}{\frac{1}{2}\times \frac{1}{2}-\frac{1}{3}\times \frac{1}{3}}\\-\frac{\frac{1}{3}}{\frac{1}{2}\times \frac{1}{2}-\frac{1}{3}\times \frac{1}{3}}&\frac{\frac{1}{2}}{\frac{1}{2}\times \frac{1}{2}-\frac{1}{3}\times \frac{1}{3}}\end{matrix}\right)\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{18}{5}&-\frac{12}{5}\\-\frac{12}{5}&\frac{18}{5}\end{matrix}\right)\left(\begin{matrix}\frac{47}{6}\\\frac{53}{6}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{18}{5}\times \frac{47}{6}-\frac{12}{5}\times \frac{53}{6}\\-\frac{12}{5}\times \frac{47}{6}+\frac{18}{5}\times \frac{53}{6}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\13\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=7,y=13$$