Solve {l}{20n+50m=11}{70m+30n=16}

$$\left. \begin{array} { l } { 20 n + 50 m = 11 } \\ { 70 m + 30 n = 16 } \end{array} \right.$$

Solve for n, m

$$n=\frac{3}{10}=0.3$$

$$m=\frac{1}{10}=0.1$$

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.

$$20n+50m=11,30n+70m=16$$

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

$$20n+50m=11$$

Subtract 50m from both sides of the equation.

$$20n=-50m+11$$

Divide both sides by 20.

$$n=\frac{1}{20}\left(-50m+11\right)$$

Multiply \frac{1}{20} times -50m+11.

$$n=-\frac{5}{2}m+\frac{11}{20}$$

Substitute -\frac{5m}{2}+\frac{11}{20} for n in the other equation, 30n+70m=16.

$$30\left(-\frac{5}{2}m+\frac{11}{20}\right)+70m=16$$

Multiply 30 times -\frac{5m}{2}+\frac{11}{20}.

$$-75m+\frac{33}{2}+70m=16$$

Add -75m to 70m.

$$-5m+\frac{33}{2}=16$$

Subtract \frac{33}{2} from both sides of the equation.

$$-5m=-\frac{1}{2}$$

Divide both sides by -5.

$$m=\frac{1}{10}$$

Substitute \frac{1}{10} for m in n=-\frac{5}{2}m+\frac{11}{20}. Because the resulting equation contains only one variable, you can solve for n directly.

$$n=-\frac{5}{2}\times \frac{1}{10}+\frac{11}{20}$$

Multiply -\frac{5}{2} times \frac{1}{10} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$n=-\frac{1}{4}+\frac{11}{20}$$

Add \frac{11}{20} to -\frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n=\frac{3}{10}$$

The system is now solved.

$$n=\frac{3}{10},m=\frac{1}{10}$$

Steps Using Matrices

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

$$20n+50m=11,30n+70m=16$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&50\\30&70\end{matrix}\right)\left(\begin{matrix}n\\m\end{matrix}\right)=\left(\begin{matrix}11\\16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&50\\30&70\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&50\\30&70\end{matrix}\right))\left(\begin{matrix}20&50\\30&70\end{matrix}\right)\left(\begin{matrix}n\\m\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\30&70\end{matrix}\right))\left(\begin{matrix}11\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}n\\m\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\30&70\end{matrix}\right))\left(\begin{matrix}11\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}n\\m\end{matrix}\right)=inverse(\left(\begin{matrix}20&50\\30&70\end{matrix}\right))\left(\begin{matrix}11\\16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}n\\m\end{matrix}\right)=\left(\begin{matrix}\frac{70}{20\times 70-50\times 30}&-\frac{50}{20\times 70-50\times 30}\\-\frac{30}{20\times 70-50\times 30}&\frac{20}{20\times 70-50\times 30}\end{matrix}\right)\left(\begin{matrix}11\\16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}n\\m\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{10}&\frac{1}{2}\\\frac{3}{10}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}11\\16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}n\\m\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{10}\times 11+\frac{1}{2}\times 16\\\frac{3}{10}\times 11-\frac{1}{5}\times 16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}n\\m\end{matrix}\right)=\left(\begin{matrix}\frac{3}{10}\\\frac{1}{10}\end{matrix}\right)$$

Extract the matrix elements n and m.

$$n=\frac{3}{10},m=\frac{1}{10}$$

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.

$$20n+50m=11,30n+70m=16$$

To make 20n and 30n equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 20.

$$30\times 20n+30\times 50m=30\times 11,20\times 30n+20\times 70m=20\times 16$$

Simplify.

$$600n+1500m=330,600n+1400m=320$$

Subtract 600n+1400m=320 from 600n+1500m=330 by subtracting like terms on each side of the equal sign.

$$600n-600n+1500m-1400m=330-320$$

Add 600n to -600n. Terms 600n and -600n cancel out, leaving an equation with only one variable that can be solved.

$$1500m-1400m=330-320$$

Add 1500m to -1400m.

$$100m=330-320$$

Add 330 to -320.

$$100m=10$$

Divide both sides by 100.

$$m=\frac{1}{10}$$

Substitute \frac{1}{10} for m in 30n+70m=16. Because the resulting equation contains only one variable, you can solve for n directly.

$$30n+70\times \frac{1}{10}=16$$

Multiply 70 times \frac{1}{10}.

$$30n+7=16$$

Subtract 7 from both sides of the equation.

$$30n=9$$

Divide both sides by 30.

$$n=\frac{3}{10}$$

The system is now solved.

$$n=\frac{3}{10},m=\frac{1}{10}$$