Solve {l}{2x+3y=13}{2x-1/3+1+y/2=1}

$$\left. \begin{array} { l } { 2 x + 3 y = 13 } \\ { \frac { 2 x - 1 } { 3 } + \frac { 1 + y } { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-4$$

$$y=7$$

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.

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

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

$$2x+3y=13$$

Subtract 3y from both sides of the equation.

$$2x=-3y+13$$

Divide both sides by 2.

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

Multiply \frac{1}{2} times -3y+13.

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

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}.

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

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

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

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

$$-\frac{1}{2}y=-\frac{7}{2}$$

Multiply both sides by -2.

$$y=7$$

Substitute 7 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.

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

Multiply -\frac{3}{2} times 7.

$$x=\frac{-21+13}{2}$$

Add \frac{13}{2} to -\frac{21}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-4$$

The system is now solved.

$$x=-4,y=7$$

Steps Using Matrices

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

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

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.

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

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

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

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

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

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

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

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.

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

Do the arithmetic.

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

Multiply the matrices.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\\7\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-4,y=7$$