Solve {l}{F_{1}=100+F_{2}}{84F_{1}=19F_{2}}

$$\left. \begin{array} { l } { F _ { 1 } = 100 + F _ { 2 } } \\ { 84 F _ { 1 } = 19 F _ { 2 } } \end{array} \right.$$

Solve for F_1, F_2

$$F_{1} = -\frac{380}{13} = -29\frac{3}{13} \approx -29.230769231$$

$$F_{2} = -\frac{1680}{13} = -129\frac{3}{13} \approx -129.230769231$$

Steps Using Substitution

Consider the first equation. Subtract F_{2} from both sides.

$$F_{1}-F_{2}=100$$

Consider the second equation. Subtract 19F_{2} from both sides.

$$84F_{1}-19F_{2}=0$$

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.

$$F_{1}-F_{2}=100,84F_{1}-19F_{2}=0$$

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

$$F_{1}-F_{2}=100$$

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

$$F_{1}=F_{2}+100$$

Substitute F_{2}+100 for F_{1} in the other equation, 84F_{1}-19F_{2}=0.

$$84\left(F_{2}+100\right)-19F_{2}=0$$

Multiply 84 times F_{2}+100.

$$84F_{2}+8400-19F_{2}=0$$

Add 84F_{2} to -19F_{2}.

$$65F_{2}+8400=0$$

Subtract 8400 from both sides of the equation.

$$65F_{2}=-8400$$

Divide both sides by 65.

$$F_{2}=-\frac{1680}{13}$$

Substitute -\frac{1680}{13} for F_{2} in F_{1}=F_{2}+100. Because the resulting equation contains only one variable, you can solve for F_{1} directly.

$$F_{1}=-\frac{1680}{13}+100$$

Add 100 to -\frac{1680}{13}.

$$F_{1}=-\frac{380}{13}$$

The system is now solved.

$$F_{1}=-\frac{380}{13},F_{2}=-\frac{1680}{13}$$

Steps Using Matrices

Consider the first equation. Subtract F_{2} from both sides.

$$F_{1}-F_{2}=100$$

Consider the second equation. Subtract 19F_{2} from both sides.

$$84F_{1}-19F_{2}=0$$

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

$$F_{1}-F_{2}=100,84F_{1}-19F_{2}=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right)\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=\left(\begin{matrix}100\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\84&-19\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right))\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right)\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\84&-19\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{-19-\left(-84\right)}&-\frac{-1}{-19-\left(-84\right)}\\-\frac{84}{-19-\left(-84\right)}&\frac{1}{-19-\left(-84\right)}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{65}&\frac{1}{65}\\-\frac{84}{65}&\frac{1}{65}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{65}\times 100\\-\frac{84}{65}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}F_{1}\\F_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{380}{13}\\-\frac{1680}{13}\end{matrix}\right)$$

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

$$F_{1}=-\frac{380}{13},F_{2}=-\frac{1680}{13}$$

Steps Using Elimination

Consider the first equation. Subtract F_{2} from both sides.

$$F_{1}-F_{2}=100$$

Consider the second equation. Subtract 19F_{2} from both sides.

$$84F_{1}-19F_{2}=0$$

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.

$$F_{1}-F_{2}=100,84F_{1}-19F_{2}=0$$

To make F_{1} and 84F_{1} equal, multiply all terms on each side of the first equation by 84 and all terms on each side of the second by 1.

$$84F_{1}+84\left(-1\right)F_{2}=84\times 100,84F_{1}-19F_{2}=0$$

Simplify.

$$84F_{1}-84F_{2}=8400,84F_{1}-19F_{2}=0$$

Subtract 84F_{1}-19F_{2}=0 from 84F_{1}-84F_{2}=8400 by subtracting like terms on each side of the equal sign.

$$84F_{1}-84F_{1}-84F_{2}+19F_{2}=8400$$

Add 84F_{1} to -84F_{1}. Terms 84F_{1} and -84F_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-84F_{2}+19F_{2}=8400$$

Add -84F_{2} to 19F_{2}.

$$-65F_{2}=8400$$

Divide both sides by -65.

$$F_{2}=-\frac{1680}{13}$$

Substitute -\frac{1680}{13} for F_{2} in 84F_{1}-19F_{2}=0. Because the resulting equation contains only one variable, you can solve for F_{1} directly.

$$84F_{1}-19\left(-\frac{1680}{13}\right)=0$$

Multiply -19 times -\frac{1680}{13}.

$$84F_{1}+\frac{31920}{13}=0$$

Subtract \frac{31920}{13} from both sides of the equation.

$$84F_{1}=-\frac{31920}{13}$$

Divide both sides by 84.

$$F_{1}=-\frac{380}{13}$$

The system is now solved.

$$F_{1}=-\frac{380}{13},F_{2}=-\frac{1680}{13}$$