Finite Math Examples

$y + 3 z = - 4$ , $x + 2 y + 2 = 7$ , $x - 2 y = 1$

Step 1

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

Tap for more steps...

Step 1.1

Subtract $2$ from both sides of the equation.

$y + 3 z = - 4$

$x + 2 y = 7 - 2$

$x - 2 y = 1$

Step 1.2

Subtract $2$ from $7$ .

$y + 3 z = - 4$

$x + 2 y = 5$

$x - 2 y = 1$

$y + 3 z = - 4$

$x + 2 y = 5$

$x - 2 y = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 0 & 1 & 3 & - 4 \\ 1 & 2 & 0 & 5 \\ 1 & - 2 & 0 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

Tap for more steps...

Step 3.1

Swap $R_{2}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 1 & - 2 & 0 & 1 \end{array}]$

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 1 - 1 & - 2 - 2 & 0 - 0 & 1 - 5 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 0 & - 4 & 0 & - 4 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

Tap for more steps...

Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 4 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 0 + 4 \cdot 0 & - 4 + 4 \cdot 1 & 0 + 4 \cdot 3 & - 4 + 4 \cdot - 4 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 0 & 0 & 12 & - 20 \end{array}]$

Step 3.4

Multiply each element of $R_{3}$ by $\frac{1}{12}$ to make the entry at $3, 3$ a $1$ .

Tap for more steps...

Step 3.4.1

Multiply each element of $R_{3}$ by $\frac{1}{12}$ to make the entry at $3, 3$ a $1$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ \frac{0}{12} & \frac{0}{12} & \frac{12}{12} & \frac{- 20}{12} \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 3 & - 4 \\ 0 & 0 & 1 & - \frac{5}{3} \end{array}]$

Step 3.5

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

Tap for more steps...

Step 3.5.1

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

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 - 3 \cdot 0 & 1 - 3 \cdot 0 & 3 - 3 \cdot 1 & - 4 - 3 (- \frac{5}{3}) \\ 0 & 0 & 1 & - \frac{5}{3} \end{array}]$

Step 3.5.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 0 & 5 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & - \frac{5}{3} \end{array}]$

Step 3.6

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

Tap for more steps...

Step 3.6.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 0 - 2 \cdot 0 & 5 - 2 \cdot 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & - \frac{5}{3} \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 3$

$y = 1$

$z = - \frac{5}{3}$

Step 5

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

$(3, 1, - \frac{5}{3})$