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

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

Solve for x, y

$$x = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

$$y = \frac{13}{5} = 2\frac{3}{5} = 2.6$$

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.

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

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

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

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

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

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

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

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

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

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

$$\frac{1}{3}x+\frac{1}{2}y=\frac{11}{6}$$

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

$$\frac{1}{3}x=-\frac{1}{2}y+\frac{11}{6}$$

Multiply both sides by 3.

$$x=3\left(-\frac{1}{2}y+\frac{11}{6}\right)$$

Multiply 3 times -\frac{y}{2}+\frac{11}{6}.

$$x=-\frac{3}{2}y+\frac{11}{2}$$

Substitute \frac{-3y+11}{2} for x in the other equation, \frac{1}{2}\left(x+2\right)+\frac{1}{2}\left(-y+1\right)=1.

$$\frac{1}{2}\left(-\frac{3}{2}y+\frac{11}{2}+2\right)+\frac{1}{2}\left(-y+1\right)=1$$

Add \frac{11}{2} to 2.

$$\frac{1}{2}\left(-\frac{3}{2}y+\frac{15}{2}\right)+\frac{1}{2}\left(-y+1\right)=1$$

Multiply \frac{1}{2} times \frac{-3y+15}{2}.

$$-\frac{3}{4}y+\frac{15}{4}+\frac{1}{2}\left(-y+1\right)=1$$

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

$$-\frac{3}{4}y+\frac{15}{4}-\frac{1}{2}y+\frac{1}{2}=1$$

Add -\frac{3y}{4} to -\frac{y}{2}.

$$-\frac{5}{4}y+\frac{15}{4}+\frac{1}{2}=1$$

Add \frac{15}{4} to \frac{1}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-\frac{5}{4}y+\frac{17}{4}=1$$

Subtract \frac{17}{4} from both sides of the equation.

$$-\frac{5}{4}y=-\frac{13}{4}$$

Divide both sides of the equation by -\frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{13}{5}$$

Substitute \frac{13}{5} for y in x=-\frac{3}{2}y+\frac{11}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times \frac{13}{5}+\frac{11}{2}$$

Multiply -\frac{3}{2} times \frac{13}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{39}{10}+\frac{11}{2}$$

Add \frac{11}{2} to -\frac{39}{10} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{8}{5}$$

The system is now solved.

$$x=\frac{8}{5},y=\frac{13}{5}$$

Steps Using Matrices

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

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

Simplify the first equation to put it in standard form.

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

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

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

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

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

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

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

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

$$\frac{1}{3}x+\frac{1}{2}y=\frac{11}{6}$$

Simplify the second equation to put it in standard form.

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

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

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

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

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

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

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

Subtract \frac{3}{2} from both sides of the equation.

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

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{1}{2}}&-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{1}{2}}\\-\frac{\frac{1}{2}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{1}{2}}&\frac{\frac{1}{3}}{\frac{1}{3}\left(-\frac{1}{2}\right)-\frac{1}{2}\times \frac{1}{2}}\end{matrix}\right)\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}&\frac{6}{5}\\\frac{6}{5}&-\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}\frac{11}{6}\\-\frac{1}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{5}\times \frac{11}{6}+\frac{6}{5}\left(-\frac{1}{2}\right)\\\frac{6}{5}\times \frac{11}{6}-\frac{4}{5}\left(-\frac{1}{2}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{5}\\\frac{13}{5}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{8}{5},y=\frac{13}{5}$$