Finite Math Examples

$y = - 1 + x$ , $y = 3 + z$ , $z = - 2 x$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Subtract $x$ from both sides of the equation.

$y - x = - 1$

$y = 3 + z$

$z = - 2 x$

Step 1.2

Reorder $y$ and $- x$ .

$- x + y = - 1$

$y = 3 + z$

$z = - 2 x$

Step 1.3

Subtract $z$ from both sides of the equation.

$- x + y = - 1$

$y - z = 3$

$z = - 2 x$

Step 1.4

Add $2 x$ to both sides of the equation.

$- x + y = - 1$

$y - z = 3$

$z + 2 x = 0$

Step 1.5

Reorder $z$ and $2 x$ .

$- x + y = - 1$

$y - z = 3$

$2 x + z = 0$

$- x + y = - 1$

$y - z = 3$

$2 x + z = 0$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 1 & 1 & 0 & - 1 \\ 0 & 1 & - 1 & 3 \\ 2 & 0 & 1 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- 1$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - - 1 & - 1 \cdot 1 & - 0 & - - 1 \\ 0 & 1 & - 1 & 3 \\ 2 & 0 & 1 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 1 & 3 \\ 2 & 0 & 1 & 0 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 1 & 3 \\ 2 - 2 \cdot 1 & 0 - 2 \cdot - 1 & 1 - 2 \cdot 0 & 0 - 2 \cdot 1 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 1 & 3 \\ 0 & 2 & 1 & - 2 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 & 1 & - 1 & 3 \\ 0 - 2 \cdot 0 & 2 - 2 \cdot 1 & 1 - 2 \cdot - 1 & - 2 - 2 \cdot 3 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

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

Step 3.4

Multiply each element of $R_{3}$ by $\frac{1}{3}$ to make the entry at $3, 3$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{3}$ by $\frac{1}{3}$ to make the entry at $3, 3$ a $1$ .

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

Step 3.4.2

Simplify $R_{3}$ .

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

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & - 1 & 0 & 1 \\ 0 + 0 & 1 + 0 & - 1 + 1 \cdot 1 & 3 - \frac{8}{3} \\ 0 & 0 & 1 & - \frac{8}{3} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

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

Step 3.6

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

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Step 3.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 1 + \frac{1}{3} \\ 0 & 1 & 0 & \frac{1}{3} \\ 0 & 0 & 1 & - \frac{8}{3} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

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

Step 4

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

$x = \frac{4}{3}$

$y = \frac{1}{3}$

$z = - \frac{8}{3}$

Step 5

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

$(\frac{4}{3}, \frac{1}{3}, - \frac{8}{3})$