Finite Math Examples

$4 a + 4 y - 4 Z = 24$ , $2 a - y + Z = - 9$ , $a - 2 y + 3 Z = 1$

Step 1

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

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Step 1.1

Move $4 y$ .

$4 a - 4 Z + 4 y = 24$

$2 a - y + Z = - 9$

$a - 2 y + 3 Z = 1$

Step 1.2

Reorder $4 a$ and $- 4 Z$ .

$- 4 Z + 4 a + 4 y = 24$

$2 a - y + Z = - 9$

$a - 2 y + 3 Z = 1$

Step 1.3

Move $- y$ .

$- 4 Z + 4 a + 4 y = 24$

$2 a + Z - y = - 9$

$a - 2 y + 3 Z = 1$

Step 1.4

Reorder $2 a$ and $Z$ .

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$a - 2 y + 3 Z = 1$

Step 1.5

Move $- 2 y$ .

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$a + 3 Z - 2 y = 1$

Step 1.6

Reorder $a$ and $3 Z$ .

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$3 Z + a - 2 y = 1$

$- 4 Z + 4 a + 4 y = 24$

$Z + 2 a - y = - 9$

$3 Z + a - 2 y = 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 4 & 4 & 4 & 24 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{4}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{4} \cdot - 4 & - \frac{1}{4} \cdot 4 & - \frac{1}{4} \cdot 4 & - \frac{1}{4} \cdot 24 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 1 & 2 & - 1 & - 9 \\ 3 & 1 & - 2 & 1 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 1 - 1 & 2 + 1 & - 1 + 1 & - 9 + 6 \\ 3 & 1 & - 2 & 1 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 3 & 1 & - 2 & 1 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 3 - 3 \cdot 1 & 1 - 3 \cdot - 1 & - 2 - 3 \cdot - 1 & 1 - 3 \cdot - 6 \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 3 & 0 & - 3 \\ 0 & 4 & 1 & 19 \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{1}{3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ \frac{0}{3} & \frac{3}{3} & \frac{0}{3} & \frac{- 3}{3} \\ 0 & 4 & 1 & 19 \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 4 & 1 & 19 \end{array}]$

Step 3.5

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

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Step 3.5.1

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

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 - 4 \cdot 0 & 4 - 4 \cdot 1 & 1 - 4 \cdot 0 & 19 - 4 \cdot - 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - 1 & - 1 & - 6 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

Step 3.6

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

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Step 3.6.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 0 & - 1 + 1 \cdot 1 & - 6 + 1 \cdot 23 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 17 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

Step 3.7

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

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Step 3.7.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & 0 + 0 & 17 - 1 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

Step 3.7.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0 & 16 \\ 0 & 1 & 0 & - 1 \\ 0 & 0 & 1 & 23 \end{array}]$

Step 4

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

$Z = 16$

$a = - 1$

$y = 23$

Step 5

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

$(16, - 1, 23)$