Finite Math Examples

$x + 3 y + 4 z = 10$ , $- 3 x + 7 y + 42 = 2$

Step 1

Move all terms not containing a variable to the right side of the equation.

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Step 1.1

Subtract $42$ from both sides of the equation.

$x + 3 y + 4 z = 10$

$- 3 x + 7 y = 2 - 42$

Step 1.2

Subtract $42$ from $2$ .

$x + 3 y + 4 z = 10$

$- 3 x + 7 y = - 40$

$x + 3 y + 4 z = 10$

$- 3 x + 7 y = - 40$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 3 & 4 & 10 \\ - 3 & 7 & 0 & - 40 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 3 & 4 & 10 \\ - 3 + 3 \cdot 1 & 7 + 3 \cdot 3 & 0 + 3 \cdot 4 & - 40 + 3 \cdot 10 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 4 & 10 \\ 0 & 16 & 12 & - 10 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $\frac{1}{16}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{16}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 3 & 4 & 10 \\ \frac{0}{16} & \frac{16}{16} & \frac{12}{16} & \frac{- 10}{16} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 4 & 10 \\ 0 & 1 & \frac{3}{4} & - \frac{5}{8} \end{array}]$

Step 3.3

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{1} = R_{1} - 3 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & 4 - 3 (\frac{3}{4}) & 10 - 3 (- \frac{5}{8}) \\ 0 & 1 & \frac{3}{4} & - \frac{5}{8} \end{array}]$

Step 3.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & \frac{7}{4} & \frac{95}{8} \\ 0 & 1 & \frac{3}{4} & - \frac{5}{8} \end{array}]$

Step 4

Use the result matrix to declare the final solution to the system of equations.

$x + \frac{7}{4} z = \frac{95}{8}$

$y + \frac{3}{4} z = - \frac{5}{8}$

Step 5

The solution is the set of ordered pairs that make the system true.

$(\frac{95}{8} - \frac{7 z}{4}, - \frac{5}{8} - \frac{3 z}{4}, z)$