Solve {l}{x+2/3-2(y-1)=0}{x+y=-1}

$$\left\{ \begin{array} { l } { \frac { x + 2 } { 3 } - 2 ( y - 1 ) = 0 } \\ { x + y = - 1 } \end{array} \right.$$

Solve for x, y

$$x=-2$$

$$y=1$$

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.

$$\frac{1}{3}\left(x+2\right)-2\left(y-1\right)=0,x+y=-1$$

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

$$\frac{1}{3}\left(x+2\right)-2\left(y-1\right)=0$$

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

$$\frac{1}{3}x+\frac{2}{3}-2\left(y-1\right)=0$$

Multiply -2 times y-1.

$$\frac{1}{3}x+\frac{2}{3}-2y+2=0$$

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

$$\frac{1}{3}x-2y+\frac{8}{3}=0$$

Subtract \frac{8}{3} from both sides of the equation.

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

Add 2y to both sides of the equation.

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

Multiply both sides by 3.

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

Multiply 3 times 2y-\frac{8}{3}.

$$x=6y-8$$

Substitute 6y-8 for x in the other equation, x+y=-1.

$$6y-8+y=-1$$

Add 6y to y.

$$7y-8=-1$$

Add 8 to both sides of the equation.

$$7y=7$$

Divide both sides by 7.

$$y=1$$

Substitute 1 for y in x=6y-8. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=6-8$$

Add -8 to 6.

$$x=-2$$

The system is now solved.

$$x=-2,y=1$$

Steps Using Matrices

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

$$\frac{1}{3}\left(x+2\right)-2\left(y-1\right)=0,x+y=-1$$

Simplify the first equation to put it in standard form.

$$\frac{1}{3}\left(x+2\right)-2\left(y-1\right)=0$$

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

$$\frac{1}{3}x+\frac{2}{3}-2\left(y-1\right)=0$$

Multiply -2 times y-1.

$$\frac{1}{3}x+\frac{2}{3}-2y+2=0$$

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

$$\frac{1}{3}x-2y+\frac{8}{3}=0$$

Subtract \frac{8}{3} from both sides of the equation.

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

Write the equations in matrix form.

$$\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right))\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right))\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right))\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{1}{3}&-2\\1&1\end{matrix}\right))\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{\frac{1}{3}-\left(-2\right)}&-\frac{-2}{\frac{1}{3}-\left(-2\right)}\\-\frac{1}{\frac{1}{3}-\left(-2\right)}&\frac{\frac{1}{3}}{\frac{1}{3}-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7}&\frac{6}{7}\\-\frac{3}{7}&\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}-\frac{8}{3}\\-1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{7}\left(-\frac{8}{3}\right)+\frac{6}{7}\left(-1\right)\\-\frac{3}{7}\left(-\frac{8}{3}\right)+\frac{1}{7}\left(-1\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2,y=1$$