Solve {l}{2/3x+y=-1}{x+1/3+y-1/6=-1}

$$\left\{ \begin{array} { l } { \frac { 2 } { 3 } x + y = - 1 } \\ { \frac { x + 1 } { 3 } + \frac { y - 1 } { 6 } = - 1 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

$$y=2$$

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.

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

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

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

Subtract y from both sides of the equation.

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

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

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

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

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

Substitute \frac{-3y-3}{2} for x in the other equation, \frac{1}{3}\left(x+1\right)+\frac{1}{6}\left(y-1\right)=-1.

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

Add -\frac{3}{2} to 1.

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

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

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

Multiply \frac{1}{6} times y-1.

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

Add -\frac{y}{2} to \frac{y}{6}.

$$-\frac{1}{3}y-\frac{1}{6}-\frac{1}{6}=-1$$

Add -\frac{1}{6} to -\frac{1}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-\frac{1}{3}y-\frac{1}{3}=-1$$

Add \frac{1}{3} to both sides of the equation.

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

Multiply both sides by -3.

$$y=2$$

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

$$x=-\frac{3}{2}\times 2-\frac{3}{2}$$

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

$$x=-3-\frac{3}{2}$$

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

$$x=-\frac{9}{2}$$

The system is now solved.

$$x=-\frac{9}{2},y=2$$

Steps Using Matrices

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

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

Simplify the second equation to put it in standard form.

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

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

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

Multiply \frac{1}{6} times y-1.

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

Add \frac{1}{3} to -\frac{1}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

$$\frac{1}{3}x+\frac{1}{6}y=-\frac{7}{6}$$

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right).

$$inverse(\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{2}{3}&1\\\frac{1}{3}&\frac{1}{6}\end{matrix}\right))\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{\frac{1}{6}}{\frac{2}{3}\times \frac{1}{6}-\frac{1}{3}}&-\frac{1}{\frac{2}{3}\times \frac{1}{6}-\frac{1}{3}}\\-\frac{\frac{1}{3}}{\frac{2}{3}\times \frac{1}{6}-\frac{1}{3}}&\frac{\frac{2}{3}}{\frac{2}{3}\times \frac{1}{6}-\frac{1}{3}}\end{matrix}\right)\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{4}&\frac{9}{2}\\\frac{3}{2}&-3\end{matrix}\right)\left(\begin{matrix}-1\\-\frac{7}{6}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{4}\left(-1\right)+\frac{9}{2}\left(-\frac{7}{6}\right)\\\frac{3}{2}\left(-1\right)-3\left(-\frac{7}{6}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{2}\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{9}{2},y=2$$