Solve {l}{x-2(3y-1)=-4}{2/3y-(-x-7)=1}

$$\left\{ \begin{array} { l } { x - 2 ( 3 y - 1 ) = - 4 } \\ { \frac { 2 } { 3 } y - ( - x - 7 ) = 1 } \end{array} \right.$$

Solve for x, y

$$x=-6$$

$$y=0$$

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.

$$x-2\left(3y-1\right)=-4,-\left(-x-7\right)+\frac{2}{3}y=1$$

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

$$x-2\left(3y-1\right)=-4$$

Multiply -2 times 3y-1.

$$x-6y+2=-4$$

Subtract 2 from both sides of the equation.

$$x-6y=-6$$

Add 6y to both sides of the equation.

$$x=6y-6$$

Substitute -6+6y for x in the other equation, -\left(-x-7\right)+\frac{2}{3}y=1.

$$-\left(-\left(6y-6\right)-7\right)+\frac{2}{3}y=1$$

Multiply -1 times -6+6y.

$$-\left(-6y+6-7\right)+\frac{2}{3}y=1$$

Add 6 to -7.

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

Multiply -1 times -6y-1.

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

Add 6y to \frac{2y}{3}.

$$\frac{20}{3}y+1=1$$

Subtract 1 from both sides of the equation.

$$\frac{20}{3}y=0$$

Divide both sides of the equation by \frac{20}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=0$$

Substitute 0 for y in x=6y-6. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-6$$

The system is now solved.

$$x=-6,y=0$$

Steps Using Matrices

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

$$x-2\left(3y-1\right)=-4,-\left(-x-7\right)+\frac{2}{3}y=1$$

Simplify the first equation to put it in standard form.

$$x-2\left(3y-1\right)=-4$$

Multiply -2 times 3y-1.

$$x-6y+2=-4$$

Subtract 2 from both sides of the equation.

$$x-6y=-6$$

Simplify the second equation to put it in standard form.

$$-\left(-x-7\right)+\frac{2}{3}y=1$$

Multiply -1 times -x-7.

$$x+7+\frac{2}{3}y=1$$

Subtract 7 from both sides of the equation.

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

Write the equations in matrix form.

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

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

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

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

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

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

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

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.

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

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}&\frac{9}{10}\\-\frac{3}{20}&\frac{3}{20}\end{matrix}\right)\left(\begin{matrix}-6\\-6\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}\left(-6\right)+\frac{9}{10}\left(-6\right)\\-\frac{3}{20}\left(-6\right)+\frac{3}{20}\left(-6\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-6\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-6,y=0$$