Solve {c}{5m+6n=64}{15m+5n=140}

$$\left. \begin{array} { c } { 5 m + 6 n = 64 } \\ { 15 m + 5 n = 140 } \end{array} \right.$$

Solve for m, n

$$m=8$$

$$n=4$$

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.

$$5m+6n=64,15m+5n=140$$

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

$$5m+6n=64$$

Subtract 6n from both sides of the equation.

$$5m=-6n+64$$

Divide both sides by 5.

$$m=\frac{1}{5}\left(-6n+64\right)$$

Multiply \frac{1}{5} times -6n+64.

$$m=-\frac{6}{5}n+\frac{64}{5}$$

Substitute \frac{-6n+64}{5} for m in the other equation, 15m+5n=140.

$$15\left(-\frac{6}{5}n+\frac{64}{5}\right)+5n=140$$

Multiply 15 times \frac{-6n+64}{5}.

$$-18n+192+5n=140$$

Add -18n to 5n.

$$-13n+192=140$$

Subtract 192 from both sides of the equation.

$$-13n=-52$$

Divide both sides by -13.

$$n=4$$

Substitute 4 for n in m=-\frac{6}{5}n+\frac{64}{5}. Because the resulting equation contains only one variable, you can solve for m directly.

$$m=-\frac{6}{5}\times 4+\frac{64}{5}$$

Multiply -\frac{6}{5} times 4.

$$m=\frac{-24+64}{5}$$

Add \frac{64}{5} to -\frac{24}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m=8$$

The system is now solved.

$$m=8,n=4$$

Steps Using Matrices

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

$$5m+6n=64,15m+5n=140$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&6\\15&5\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}64\\140\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&6\\15&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&6\\15&5\end{matrix}\right))\left(\begin{matrix}5&6\\15&5\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\15&5\end{matrix}\right))\left(\begin{matrix}64\\140\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\15&5\end{matrix}\right))\left(\begin{matrix}64\\140\end{matrix}\right)$$

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

$$\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\15&5\end{matrix}\right))\left(\begin{matrix}64\\140\end{matrix}\right)$$

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.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5\times 5-6\times 15}&-\frac{6}{5\times 5-6\times 15}\\-\frac{15}{5\times 5-6\times 15}&\frac{5}{5\times 5-6\times 15}\end{matrix}\right)\left(\begin{matrix}64\\140\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{13}&\frac{6}{65}\\\frac{3}{13}&-\frac{1}{13}\end{matrix}\right)\left(\begin{matrix}64\\140\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{13}\times 64+\frac{6}{65}\times 140\\\frac{3}{13}\times 64-\frac{1}{13}\times 140\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}8\\4\end{matrix}\right)$$

Extract the matrix elements m and n.

$$m=8,n=4$$

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.

$$5m+6n=64,15m+5n=140$$

To make 5m and 15m equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 5.

$$15\times 5m+15\times 6n=15\times 64,5\times 15m+5\times 5n=5\times 140$$

Simplify.

$$75m+90n=960,75m+25n=700$$

Subtract 75m+25n=700 from 75m+90n=960 by subtracting like terms on each side of the equal sign.

$$75m-75m+90n-25n=960-700$$

Add 75m to -75m. Terms 75m and -75m cancel out, leaving an equation with only one variable that can be solved.

$$90n-25n=960-700$$

Add 90n to -25n.

$$65n=960-700$$

Add 960 to -700.

$$65n=260$$

Divide both sides by 65.

$$n=4$$

Substitute 4 for n in 15m+5n=140. Because the resulting equation contains only one variable, you can solve for m directly.

$$15m+5\times 4=140$$

Multiply 5 times 4.

$$15m+20=140$$

Subtract 20 from both sides of the equation.

$$15m=120$$

Divide both sides by 15.

$$m=8$$

The system is now solved.

$$m=8,n=4$$