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

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

Solve for x, y

$$x=3$$

$$y = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

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.

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

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

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

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

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

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

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

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.

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

Add \frac{1}{6} to both sides of the equation.

$$\frac{1}{3}x+\frac{1}{2}y=\frac{13}{6}$$

Subtract \frac{y}{2} from both sides of the equation.

$$\frac{1}{3}x=-\frac{1}{2}y+\frac{13}{6}$$

Multiply both sides by 3.

$$x=3\left(-\frac{1}{2}y+\frac{13}{6}\right)$$

Multiply 3 times -\frac{y}{2}+\frac{13}{6}.

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

Substitute \frac{-3y+13}{2} for x in the other equation, \frac{1}{3}\left(2x-1\right)+\frac{1}{2}\left(-y+1\right)=1.

$$\frac{1}{3}\left(2\left(-\frac{3}{2}y+\frac{13}{2}\right)-1\right)+\frac{1}{2}\left(-y+1\right)=1$$

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

$$\frac{1}{3}\left(-3y+13-1\right)+\frac{1}{2}\left(-y+1\right)=1$$

Add 13 to -1.

$$\frac{1}{3}\left(-3y+12\right)+\frac{1}{2}\left(-y+1\right)=1$$

Multiply \frac{1}{3} times -3y+12.

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

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

$$-y+4-\frac{1}{2}y+\frac{1}{2}=1$$

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

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

Add 4 to \frac{1}{2}.

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

Subtract \frac{9}{2} from both sides of the equation.

$$-\frac{3}{2}y=-\frac{7}{2}$$

Divide both sides of the equation by -\frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{7}{3}$$

Substitute \frac{7}{3} for y in x=-\frac{3}{2}y+\frac{13}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times \frac{7}{3}+\frac{13}{2}$$

Multiply -\frac{3}{2} times \frac{7}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-7+13}{2}$$

Add \frac{13}{2} to -\frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=3$$

The system is now solved.

$$x=3,y=\frac{7}{3}$$

Steps Using Matrices

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

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

Simplify the first equation to put it in standard form.

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

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

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

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

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

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.

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

Add \frac{1}{6} to both sides of the equation.

$$\frac{1}{3}x+\frac{1}{2}y=\frac{13}{6}$$

Simplify the second equation to put it in standard form.

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

Multiply \frac{1}{3} times 2x-1.

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

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

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

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.

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

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

$$\frac{2}{3}x-\frac{1}{2}y=\frac{5}{6}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{2}{3}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{2}{3}}&-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{2}{3}}\\-\frac{\frac{2}{3}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{2}{3}}&\frac{\frac{1}{3}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{2}{3}}\end{matrix}\right)\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1&1\\\frac{4}{3}&-\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}\frac{13}{6}\\\frac{5}{6}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13+5}{6}\\\frac{4}{3}\times \frac{13}{6}-\frac{2}{3}\times \frac{5}{6}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\\frac{7}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=\frac{7}{3}$$