Solve {l}{5m+6n=-32}{10m+4n=-48}

$$\left. \begin{array} { l } { 5 m + 6 n = - 32 } \\ { 10 m + 4 n = - 48 } \end{array} \right.$$

Solve for m, n

$$m=-4$$

$$n=-2$$

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.

$$5m+6n=-32,10m+4n=-48$$

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

$$5m+6n=-32$$

Subtract 6n from both sides of the equation.

$$5m=-6n-32$$

Divide both sides by 5.

$$m=\frac{1}{5}\left(-6n-32\right)$$

Multiply \frac{1}{5} times -6n-32.

$$m=-\frac{6}{5}n-\frac{32}{5}$$

Substitute \frac{-6n-32}{5} for m in the other equation, 10m+4n=-48.

$$10\left(-\frac{6}{5}n-\frac{32}{5}\right)+4n=-48$$

Multiply 10 times \frac{-6n-32}{5}.

$$-12n-64+4n=-48$$

Add -12n to 4n.

$$-8n-64=-48$$

Add 64 to both sides of the equation.

$$-8n=16$$

Divide both sides by -8.

$$n=-2$$

Substitute -2 for n in m=-\frac{6}{5}n-\frac{32}{5}. Because the resulting equation contains only one variable, you can solve for m directly.

$$m=-\frac{6}{5}\left(-2\right)-\frac{32}{5}$$

Multiply -\frac{6}{5} times -2.

$$m=\frac{12-32}{5}$$

Add -\frac{32}{5} to \frac{12}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m=-4$$

The system is now solved.

$$m=-4,n=-2$$

Steps Using Matrices

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

$$5m+6n=-32,10m+4n=-48$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&6\\10&4\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&6\\10&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&6\\10&4\end{matrix}\right))\left(\begin{matrix}5&6\\10&4\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\10&4\end{matrix}\right))\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\10&4\end{matrix}\right))\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

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

$$\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}5&6\\10&4\end{matrix}\right))\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

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.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5\times 4-6\times 10}&-\frac{6}{5\times 4-6\times 10}\\-\frac{10}{5\times 4-6\times 10}&\frac{5}{5\times 4-6\times 10}\end{matrix}\right)\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}&\frac{3}{20}\\\frac{1}{4}&-\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}-32\\-48\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}\left(-32\right)+\frac{3}{20}\left(-48\right)\\\frac{1}{4}\left(-32\right)-\frac{1}{8}\left(-48\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-4\\-2\end{matrix}\right)$$

Extract the matrix elements m and n.

$$m=-4,n=-2$$

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.

$$5m+6n=-32,10m+4n=-48$$

To make 5m and 10m equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 5.

$$10\times 5m+10\times 6n=10\left(-32\right),5\times 10m+5\times 4n=5\left(-48\right)$$

Simplify.

$$50m+60n=-320,50m+20n=-240$$

Subtract 50m+20n=-240 from 50m+60n=-320 by subtracting like terms on each side of the equal sign.

$$50m-50m+60n-20n=-320+240$$

Add 50m to -50m. Terms 50m and -50m cancel out, leaving an equation with only one variable that can be solved.

$$60n-20n=-320+240$$

Add 60n to -20n.

$$40n=-320+240$$

Add -320 to 240.

$$40n=-80$$

Divide both sides by 40.

$$n=-2$$

Substitute -2 for n in 10m+4n=-48. Because the resulting equation contains only one variable, you can solve for m directly.

$$10m+4\left(-2\right)=-48$$

Multiply 4 times -2.

$$10m-8=-48$$

Add 8 to both sides of the equation.

$$10m=-40$$

Divide both sides by 10.

$$m=-4$$

The system is now solved.

$$m=-4,n=-2$$