Solve {c}{x-11(y-3)=-2}{x+2y=7}

$$\left. \begin{array} { c } { x - 11 ( y - 3 ) = - 2 } \\ { x + 2 y = 7 } \end{array} \right.$$

Solve for x, y

$$x=\frac{7}{13}\approx 0.538461538$$

$$y = \frac{42}{13} = 3\frac{3}{13} \approx 3.230769231$$

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.

$$x-11\left(y-3\right)=-2,x+2y=7$$

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

$$x-11\left(y-3\right)=-2$$

Multiply -11 times y-3.

$$x-11y+33=-2$$

Subtract 33 from both sides of the equation.

$$x-11y=-35$$

Add 11y to both sides of the equation.

$$x=11y-35$$

Substitute 11y-35 for x in the other equation, x+2y=7.

$$11y-35+2y=7$$

Add 11y to 2y.

$$13y-35=7$$

Add 35 to both sides of the equation.

$$13y=42$$

Divide both sides by 13.

$$y=\frac{42}{13}$$

Substitute \frac{42}{13} for y in x=11y-35. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=11\times \frac{42}{13}-35$$

Multiply 11 times \frac{42}{13}.

$$x=\frac{462}{13}-35$$

Add -35 to \frac{462}{13}.

$$x=\frac{7}{13}$$

The system is now solved.

$$x=\frac{7}{13},y=\frac{42}{13}$$

Steps Using Matrices

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

$$x-11\left(y-3\right)=-2,x+2y=7$$

Simplify the first equation to put it in standard form.

$$x-11\left(y-3\right)=-2$$

Multiply -11 times y-3.

$$x-11y+33=-2$$

Subtract 33 from both sides of the equation.

$$x-11y=-35$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-11\\1&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-35\\7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-11\\1&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-11\\1&2\end{matrix}\right))\left(\begin{matrix}1&-11\\1&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-11\\1&2\end{matrix}\right))\left(\begin{matrix}-35\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-11\\1&2\end{matrix}\right))\left(\begin{matrix}-35\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-11\\1&2\end{matrix}\right))\left(\begin{matrix}-35\\7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2-\left(-11\right)}&-\frac{-11}{2-\left(-11\right)}\\-\frac{1}{2-\left(-11\right)}&\frac{1}{2-\left(-11\right)}\end{matrix}\right)\left(\begin{matrix}-35\\7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{13}&\frac{11}{13}\\-\frac{1}{13}&\frac{1}{13}\end{matrix}\right)\left(\begin{matrix}-35\\7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{13}\left(-35\right)+\frac{11}{13}\times 7\\-\frac{1}{13}\left(-35\right)+\frac{1}{13}\times 7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{13}\\\frac{42}{13}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7}{13},y=\frac{42}{13}$$