Solve {l}{k/L=100}{5k+50L=110}

$$\left. \begin{array} { l } { \frac { k } { L } = 100 } \\ { 5 k + 50 L = 110 } \end{array} \right.$$

Solve for k, L

$$k=20$$

$$L=\frac{1}{5}=0.2$$

Steps Using Substitution

Consider the first equation. Variable L cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by L.

$$k=100L$$

Substitute 100L for k in the other equation, 5k+50L=110.

$$5\times 100L+50L=110$$

Multiply 5 times 100L.

$$500L+50L=110$$

Add 500L to 50L.

$$550L=110$$

Divide both sides by 550.

$$L=\frac{1}{5}$$

Substitute \frac{1}{5} for L in k=100L. Because the resulting equation contains only one variable, you can solve for k directly.

$$k=100\times \frac{1}{5}$$

Multiply 100 times \frac{1}{5}.

$$k=20$$

The system is now solved.

$$k=20,L=\frac{1}{5}$$

Steps Using Matrices

Consider the first equation. Variable L cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by L.

$$k=100L$$

Subtract 100L from both sides.

$$k-100L=0$$

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

$$k-100L=0,5k+50L=110$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-100\\5&50\end{matrix}\right)\left(\begin{matrix}k\\L\end{matrix}\right)=\left(\begin{matrix}0\\110\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-100\\5&50\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-100\\5&50\end{matrix}\right))\left(\begin{matrix}1&-100\\5&50\end{matrix}\right)\left(\begin{matrix}k\\L\end{matrix}\right)=inverse(\left(\begin{matrix}1&-100\\5&50\end{matrix}\right))\left(\begin{matrix}0\\110\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k\\L\end{matrix}\right)=inverse(\left(\begin{matrix}1&-100\\5&50\end{matrix}\right))\left(\begin{matrix}0\\110\end{matrix}\right)$$

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

$$\left(\begin{matrix}k\\L\end{matrix}\right)=inverse(\left(\begin{matrix}1&-100\\5&50\end{matrix}\right))\left(\begin{matrix}0\\110\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k\\L\end{matrix}\right)=\left(\begin{matrix}\frac{50}{50-\left(-100\times 5\right)}&-\frac{-100}{50-\left(-100\times 5\right)}\\-\frac{5}{50-\left(-100\times 5\right)}&\frac{1}{50-\left(-100\times 5\right)}\end{matrix}\right)\left(\begin{matrix}0\\110\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k\\L\end{matrix}\right)=\left(\begin{matrix}\frac{1}{11}&\frac{2}{11}\\-\frac{1}{110}&\frac{1}{550}\end{matrix}\right)\left(\begin{matrix}0\\110\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k\\L\end{matrix}\right)=\left(\begin{matrix}\frac{2}{11}\times 110\\\frac{1}{550}\times 110\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k\\L\end{matrix}\right)=\left(\begin{matrix}20\\\frac{1}{5}\end{matrix}\right)$$

Extract the matrix elements k and L.

$$k=20,L=\frac{1}{5}$$

Steps Using Elimination

Consider the first equation. Variable L cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by L.

$$k=100L$$

Subtract 100L from both sides.

$$k-100L=0$$

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.

$$k-100L=0,5k+50L=110$$

To make k and 5k equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 1.

$$5k+5\left(-100\right)L=0,5k+50L=110$$

Simplify.

$$5k-500L=0,5k+50L=110$$

Subtract 5k+50L=110 from 5k-500L=0 by subtracting like terms on each side of the equal sign.

$$5k-5k-500L-50L=-110$$

Add 5k to -5k. Terms 5k and -5k cancel out, leaving an equation with only one variable that can be solved.

$$-500L-50L=-110$$

Add -500L to -50L.

$$-550L=-110$$

Divide both sides by -550.

$$L=\frac{1}{5}$$

Substitute \frac{1}{5} for L in 5k+50L=110. Because the resulting equation contains only one variable, you can solve for k directly.

$$5k+50\times \frac{1}{5}=110$$

Multiply 50 times \frac{1}{5}.

$$5k+10=110$$

Subtract 10 from both sides of the equation.

$$5k=100$$

Divide both sides by 5.

$$k=20$$

The system is now solved.

$$k=20,L=\frac{1}{5}$$