Solve {l}{26u_{2}-15u_{3}=288}{-6u_{2}+13u_{3}=68}

$$\left. \begin{array} { l } { 26 u _ { 2 } - 15 u _ { 3 } = 288 } \\ { - 6 u _ { 2 } + 13 u _ { 3 } = 68 } \end{array} \right.$$

Solve for u_2, u_3

$$u_{2} = \frac{1191}{62} = 19\frac{13}{62} \approx 19.209677419$$

$$u_{3} = \frac{437}{31} = 14\frac{3}{31} \approx 14.096774194$$

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.

$$26u_{2}-15u_{3}=288,-6u_{2}+13u_{3}=68$$

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

$$26u_{2}-15u_{3}=288$$

Add 15u_{3} to both sides of the equation.

$$26u_{2}=15u_{3}+288$$

Divide both sides by 26.

$$u_{2}=\frac{1}{26}\left(15u_{3}+288\right)$$

Multiply \frac{1}{26} times 15u_{3}+288.

$$u_{2}=\frac{15}{26}u_{3}+\frac{144}{13}$$

Substitute \frac{15u_{3}}{26}+\frac{144}{13} for u_{2} in the other equation, -6u_{2}+13u_{3}=68.

$$-6\left(\frac{15}{26}u_{3}+\frac{144}{13}\right)+13u_{3}=68$$

Multiply -6 times \frac{15u_{3}}{26}+\frac{144}{13}.

$$-\frac{45}{13}u_{3}-\frac{864}{13}+13u_{3}=68$$

Add -\frac{45u_{3}}{13} to 13u_{3}.

$$\frac{124}{13}u_{3}-\frac{864}{13}=68$$

Add \frac{864}{13} to both sides of the equation.

$$\frac{124}{13}u_{3}=\frac{1748}{13}$$

Divide both sides of the equation by \frac{124}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$u_{3}=\frac{437}{31}$$

Substitute \frac{437}{31} for u_{3} in u_{2}=\frac{15}{26}u_{3}+\frac{144}{13}. Because the resulting equation contains only one variable, you can solve for u_{2} directly.

$$u_{2}=\frac{15}{26}\times \frac{437}{31}+\frac{144}{13}$$

Multiply \frac{15}{26} times \frac{437}{31} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$u_{2}=\frac{6555}{806}+\frac{144}{13}$$

Add \frac{144}{13} to \frac{6555}{806} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$u_{2}=\frac{1191}{62}$$

The system is now solved.

$$u_{2}=\frac{1191}{62},u_{3}=\frac{437}{31}$$

Steps Using Matrices

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

$$26u_{2}-15u_{3}=288,-6u_{2}+13u_{3}=68$$

Write the equations in matrix form.

$$\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right)\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=\left(\begin{matrix}288\\68\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}26&-15\\-6&13\end{matrix}\right).

$$inverse(\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right))\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right)\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right))\left(\begin{matrix}288\\68\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right))\left(\begin{matrix}288\\68\end{matrix}\right)$$

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

$$\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=inverse(\left(\begin{matrix}26&-15\\-6&13\end{matrix}\right))\left(\begin{matrix}288\\68\end{matrix}\right)$$

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.

$$\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{13}{26\times 13-\left(-15\left(-6\right)\right)}&-\frac{-15}{26\times 13-\left(-15\left(-6\right)\right)}\\-\frac{-6}{26\times 13-\left(-15\left(-6\right)\right)}&\frac{26}{26\times 13-\left(-15\left(-6\right)\right)}\end{matrix}\right)\left(\begin{matrix}288\\68\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{13}{248}&\frac{15}{248}\\\frac{3}{124}&\frac{13}{124}\end{matrix}\right)\left(\begin{matrix}288\\68\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{13}{248}\times 288+\frac{15}{248}\times 68\\\frac{3}{124}\times 288+\frac{13}{124}\times 68\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}u_{2}\\u_{3}\end{matrix}\right)=\left(\begin{matrix}\frac{1191}{62}\\\frac{437}{31}\end{matrix}\right)$$

Extract the matrix elements u_{2} and u_{3}.

$$u_{2}=\frac{1191}{62},u_{3}=\frac{437}{31}$$

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.

$$26u_{2}-15u_{3}=288,-6u_{2}+13u_{3}=68$$

To make 26u_{2} and -6u_{2} equal, multiply all terms on each side of the first equation by -6 and all terms on each side of the second by 26.

$$-6\times 26u_{2}-6\left(-15\right)u_{3}=-6\times 288,26\left(-6\right)u_{2}+26\times 13u_{3}=26\times 68$$

Simplify.

$$-156u_{2}+90u_{3}=-1728,-156u_{2}+338u_{3}=1768$$

Subtract -156u_{2}+338u_{3}=1768 from -156u_{2}+90u_{3}=-1728 by subtracting like terms on each side of the equal sign.

$$-156u_{2}+156u_{2}+90u_{3}-338u_{3}=-1728-1768$$

Add -156u_{2} to 156u_{2}. Terms -156u_{2} and 156u_{2} cancel out, leaving an equation with only one variable that can be solved.

$$90u_{3}-338u_{3}=-1728-1768$$

Add 90u_{3} to -338u_{3}.

$$-248u_{3}=-1728-1768$$

Add -1728 to -1768.

$$-248u_{3}=-3496$$

Divide both sides by -248.

$$u_{3}=\frac{437}{31}$$

Substitute \frac{437}{31} for u_{3} in -6u_{2}+13u_{3}=68. Because the resulting equation contains only one variable, you can solve for u_{2} directly.

$$-6u_{2}+13\times \frac{437}{31}=68$$

Multiply 13 times \frac{437}{31}.

$$-6u_{2}+\frac{5681}{31}=68$$

Subtract \frac{5681}{31} from both sides of the equation.

$$-6u_{2}=-\frac{3573}{31}$$

Divide both sides by -6.

$$u_{2}=\frac{1191}{62}$$

The system is now solved.

$$u_{2}=\frac{1191}{62},u_{3}=\frac{437}{31}$$