Finite Math Examples

$2 x \cdot 1 + x \cdot 2 + 3 x \cdot 3 = 6$ , $- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$ , $x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1

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

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $2$ by $1$ .

$2 x + x \cdot 2 + 3 x \cdot 3 = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.1.2

Move $2$ to the left of $x$ .

$2 x + 2 \cdot x + 3 x \cdot 3 = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.1.3

Multiply $3$ by $3$ .

$2 x + 2 x + 9 x = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

$2 x + 2 x + 9 x = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.2

Add $2 x$ and $2 x$ .

$4 x + 9 x = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.3

Add $4 x$ and $9 x$ .

$13 x = 6$

$- x \cdot 1 + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $- 1$ by $1$ .

$13 x = 6$

$- x + 2 x \cdot 2 + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.4.2

Multiply $2$ by $2$ .

$13 x = 6$

$- x + 4 x + x \cdot 3 = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.4.3

Move $3$ to the left of $x$ .

$13 x = 6$

$- x + 4 x + 3 x = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

$13 x = 6$

$- x + 4 x + 3 x = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.5

Add $- x$ and $4 x$ .

$13 x = 6$

$3 x + 3 x = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.6

Add $3 x$ and $3 x$ .

$13 x = 6$

$6 x = 5$

$x \cdot 1 + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.7

Simplify each term.

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Step 1.7.1

Multiply $x$ by $1$ .

$13 x = 6$

$6 x = 5$

$x + 3 x \cdot 2 + 4 x \cdot 3 = 11$

Step 1.7.2

Multiply $2$ by $3$ .

$13 x = 6$

$6 x = 5$

$x + 6 x + 4 x \cdot 3 = 11$

Step 1.7.3

Multiply $3$ by $4$ .

$13 x = 6$

$6 x = 5$

$x + 6 x + 12 x = 11$

$13 x = 6$

$6 x = 5$

$x + 6 x + 12 x = 11$

Step 1.8

Add $x$ and $6 x$ .

$13 x = 6$

$6 x = 5$

$7 x + 12 x = 11$

Step 1.9

Add $7 x$ and $12 x$ .

$13 x = 6$

$6 x = 5$

$19 x = 11$

$13 x = 6$

$6 x = 5$

$19 x = 11$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 13 & 6 \\ 6 & 5 \\ 19 & 11 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{13}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{13}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{13}{13} & \frac{6}{13} \\ 6 & 5 \\ 19 & 11 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 6 & 5 \\ 19 & 11 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 6 - 6 \cdot 1 & 5 - 6 (\frac{6}{13}) \\ 19 & 11 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & \frac{29}{13} \\ 19 & 11 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 19 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 19 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & \frac{29}{13} \\ 19 - 19 \cdot 1 & 11 - 19 (\frac{6}{13}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & \frac{29}{13} \\ 0 & \frac{29}{13} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{13}{29}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{13}{29}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ \frac{13}{29} \cdot 0 & \frac{13}{29} \cdot \frac{29}{13} \\ 0 & \frac{29}{13} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & 1 \\ 0 & \frac{29}{13} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{29}{13} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{29}{13} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & 1 \\ 0 - \frac{29}{13} \cdot 0 & \frac{29}{13} - \frac{29}{13} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{6}{13} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} - \frac{6}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} - \frac{6}{13} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{6}{13} \cdot 0 & \frac{6}{13} - \frac{6}{13} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 0$

$0 = 1$

$0 = 0$

Step 5

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

$(error = error)$