Solve {c}{k_{1}-k_{2}=2}{200k_{1}+1400=300k_{2}+6000}

$$\left. \begin{array} { c } { k _ { 1 } - k _ { 2 } = 2 } \\ { 200 k _ { 1 } + 1400 = 300 k _ { 2 } + 6000 } \end{array} \right.$$

Solve for k_1, k_2

$$k_{1}=-40$$

$$k_{2}=-42$$

Steps Using Substitution

Consider the second equation. Subtract 300k_{2} from both sides.

$$200k_{1}+1400-300k_{2}=6000$$

Subtract 1400 from both sides.

$$200k_{1}-300k_{2}=6000-1400$$

Subtract 1400 from 6000 to get 4600.

$$200k_{1}-300k_{2}=4600$$

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.

$$k_{1}-k_{2}=2,200k_{1}-300k_{2}=4600$$

Choose one of the equations and solve it for k_{1} by isolating k_{1} on the left hand side of the equal sign.

$$k_{1}-k_{2}=2$$

Add k_{2} to both sides of the equation.

$$k_{1}=k_{2}+2$$

Substitute k_{2}+2 for k_{1} in the other equation, 200k_{1}-300k_{2}=4600.

$$200\left(k_{2}+2\right)-300k_{2}=4600$$

Multiply 200 times k_{2}+2.

$$200k_{2}+400-300k_{2}=4600$$

Add 200k_{2} to -300k_{2}.

$$-100k_{2}+400=4600$$

Subtract 400 from both sides of the equation.

$$-100k_{2}=4200$$

Divide both sides by -100.

$$k_{2}=-42$$

Substitute -42 for k_{2} in k_{1}=k_{2}+2. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$k_{1}=-42+2$$

Add 2 to -42.

$$k_{1}=-40$$

The system is now solved.

$$k_{1}=-40,k_{2}=-42$$

Steps Using Matrices

Consider the second equation. Subtract 300k_{2} from both sides.

$$200k_{1}+1400-300k_{2}=6000$$

Subtract 1400 from both sides.

$$200k_{1}-300k_{2}=6000-1400$$

Subtract 1400 from 6000 to get 4600.

$$200k_{1}-300k_{2}=4600$$

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

$$k_{1}-k_{2}=2,200k_{1}-300k_{2}=4600$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}2\\4600\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\200&-300\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right))\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right))\left(\begin{matrix}2\\4600\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right))\left(\begin{matrix}2\\4600\end{matrix}\right)$$

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

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\200&-300\end{matrix}\right))\left(\begin{matrix}2\\4600\end{matrix}\right)$$

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.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{300}{-300-\left(-200\right)}&-\frac{-1}{-300-\left(-200\right)}\\-\frac{200}{-300-\left(-200\right)}&\frac{1}{-300-\left(-200\right)}\end{matrix}\right)\left(\begin{matrix}2\\4600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}3&-\frac{1}{100}\\2&-\frac{1}{100}\end{matrix}\right)\left(\begin{matrix}2\\4600\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}3\times 2-\frac{1}{100}\times 4600\\2\times 2-\frac{1}{100}\times 4600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}k_{1}\\k_{2}\end{matrix}\right)=\left(\begin{matrix}-40\\-42\end{matrix}\right)$$

Extract the matrix elements k_{1} and k_{2}.

$$k_{1}=-40,k_{2}=-42$$

Steps Using Elimination

Consider the second equation. Subtract 300k_{2} from both sides.

$$200k_{1}+1400-300k_{2}=6000$$

Subtract 1400 from both sides.

$$200k_{1}-300k_{2}=6000-1400$$

Subtract 1400 from 6000 to get 4600.

$$200k_{1}-300k_{2}=4600$$

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.

$$k_{1}-k_{2}=2,200k_{1}-300k_{2}=4600$$

To make k_{1} and 200k_{1} equal, multiply all terms on each side of the first equation by 200 and all terms on each side of the second by 1.

$$200k_{1}+200\left(-1\right)k_{2}=200\times 2,200k_{1}-300k_{2}=4600$$

Simplify.

$$200k_{1}-200k_{2}=400,200k_{1}-300k_{2}=4600$$

Subtract 200k_{1}-300k_{2}=4600 from 200k_{1}-200k_{2}=400 by subtracting like terms on each side of the equal sign.

$$200k_{1}-200k_{1}-200k_{2}+300k_{2}=400-4600$$

Add 200k_{1} to -200k_{1}. Terms 200k_{1} and -200k_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-200k_{2}+300k_{2}=400-4600$$

Add -200k_{2} to 300k_{2}.

$$100k_{2}=400-4600$$

Add 400 to -4600.

$$100k_{2}=-4200$$

Divide both sides by 100.

$$k_{2}=-42$$

Substitute -42 for k_{2} in 200k_{1}-300k_{2}=4600. Because the resulting equation contains only one variable, you can solve for k_{1} directly.

$$200k_{1}-300\left(-42\right)=4600$$

Multiply -300 times -42.

$$200k_{1}+12600=4600$$

Subtract 12600 from both sides of the equation.

$$200k_{1}=-8000$$

Divide both sides by 200.

$$k_{1}=-40$$

The system is now solved.

$$k_{1}=-40,k_{2}=-42$$