Solve {l}{64+11x_{2}-18x_{1}=0}{6-18x_{2}+11x_{1}=0}

$$\left\{ \begin{array} { l } { 64 + 11 x _ { 2 } - 18 x _ { 1 } = 0 } \\ { 6 - 18 x _ { 2 } + 11 x _ { 1 } = 0 } \end{array} \right.$$

Solve for x_2, x_1

$$x_{2}=4$$

$$x_{1}=6$$

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.

$$11x_{2}-18x_{1}+64=0,-18x_{2}+11x_{1}+6=0$$

Choose one of the equations and solve it for x_{2} by isolating x_{2} on the left hand side of the equal sign.

$$11x_{2}-18x_{1}+64=0$$

Subtract 64 from both sides of the equation.

$$11x_{2}-18x_{1}=-64$$

Add 18x_{1} to both sides of the equation.

$$11x_{2}=18x_{1}-64$$

Divide both sides by 11.

$$x_{2}=\frac{1}{11}\left(18x_{1}-64\right)$$

Multiply \frac{1}{11} times 18x_{1}-64.

$$x_{2}=\frac{18}{11}x_{1}-\frac{64}{11}$$

Substitute \frac{18x_{1}-64}{11} for x_{2} in the other equation, -18x_{2}+11x_{1}+6=0.

$$-18\left(\frac{18}{11}x_{1}-\frac{64}{11}\right)+11x_{1}+6=0$$

Multiply -18 times \frac{18x_{1}-64}{11}.

$$-\frac{324}{11}x_{1}+\frac{1152}{11}+11x_{1}+6=0$$

Add -\frac{324x_{1}}{11} to 11x_{1}.

$$-\frac{203}{11}x_{1}+\frac{1152}{11}+6=0$$

Add \frac{1152}{11} to 6.

$$-\frac{203}{11}x_{1}+\frac{1218}{11}=0$$

Subtract \frac{1218}{11} from both sides of the equation.

$$-\frac{203}{11}x_{1}=-\frac{1218}{11}$$

Divide both sides of the equation by -\frac{203}{11}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{1}=6$$

Substitute 6 for x_{1} in x_{2}=\frac{18}{11}x_{1}-\frac{64}{11}. Because the resulting equation contains only one variable, you can solve for x_{2} directly.

$$x_{2}=\frac{18}{11}\times 6-\frac{64}{11}$$

Multiply \frac{18}{11} times 6.

$$x_{2}=\frac{108-64}{11}$$

Add -\frac{64}{11} to \frac{108}{11} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x_{2}=4$$

The system is now solved.

$$x_{2}=4,x_{1}=6$$

Steps Using Matrices

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

$$11x_{2}-18x_{1}+64=0,-18x_{2}+11x_{1}+6=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right)\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}11&-18\\-18&11\end{matrix}\right).

$$inverse(\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right))\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right)\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right))\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right))\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

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

$$\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=inverse(\left(\begin{matrix}11&-18\\-18&11\end{matrix}\right))\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=\left(\begin{matrix}\frac{11}{11\times 11-\left(-18\left(-18\right)\right)}&-\frac{-18}{11\times 11-\left(-18\left(-18\right)\right)}\\-\frac{-18}{11\times 11-\left(-18\left(-18\right)\right)}&\frac{11}{11\times 11-\left(-18\left(-18\right)\right)}\end{matrix}\right)\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{203}&-\frac{18}{203}\\-\frac{18}{203}&-\frac{11}{203}\end{matrix}\right)\left(\begin{matrix}-64\\-6\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{203}\left(-64\right)-\frac{18}{203}\left(-6\right)\\-\frac{18}{203}\left(-64\right)-\frac{11}{203}\left(-6\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{2}\\x_{1}\end{matrix}\right)=\left(\begin{matrix}4\\6\end{matrix}\right)$$

Extract the matrix elements x_{2} and x_{1}.

$$x_{2}=4,x_{1}=6$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$11x_{2}-18x_{1}+64=0,-18x_{2}+11x_{1}+6=0$$

To make 11x_{2} and -18x_{2} equal, multiply all terms on each side of the first equation by -18 and all terms on each side of the second by 11.

$$-18\times 11x_{2}-18\left(-18\right)x_{1}-18\times 64=0,11\left(-18\right)x_{2}+11\times 11x_{1}+11\times 6=0$$

Simplify.

$$-198x_{2}+324x_{1}-1152=0,-198x_{2}+121x_{1}+66=0$$

Subtract -198x_{2}+121x_{1}+66=0 from -198x_{2}+324x_{1}-1152=0 by subtracting like terms on each side of the equal sign.

$$-198x_{2}+198x_{2}+324x_{1}-121x_{1}-1152-66=0$$

Add -198x_{2} to 198x_{2}. Terms -198x_{2} and 198x_{2} cancel out, leaving an equation with only one variable that can be solved.

$$324x_{1}-121x_{1}-1152-66=0$$

Add 324x_{1} to -121x_{1}.

$$203x_{1}-1152-66=0$$

Add -1152 to -66.

$$203x_{1}-1218=0$$

Add 1218 to both sides of the equation.

$$203x_{1}=1218$$

Divide both sides by 203.

$$x_{1}=6$$

Substitute 6 for x_{1} in -18x_{2}+11x_{1}+6=0. Because the resulting equation contains only one variable, you can solve for x_{2} directly.

$$-18x_{2}+11\times 6+6=0$$

Multiply 11 times 6.

$$-18x_{2}+66+6=0$$

Add 66 to 6.

$$-18x_{2}+72=0$$

Subtract 72 from both sides of the equation.

$$-18x_{2}=-72$$

Divide both sides by -18.

$$x_{2}=4$$

The system is now solved.

$$x_{2}=4,x_{1}=6$$