Finite Math Examples

$3 x \cdot 1 + 5 x \cdot 2 - x \cdot 3 = - 7$ , $x \cdot 1 + x \cdot 2 + x \cdot 3 = - 1$ , $2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1

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

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Step 1.1

Combine the opposite terms in $3 x \cdot 1 + 5 x \cdot 2 - x \cdot 3$ .

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Step 1.1.1

Reorder the factors in the terms $3 x \cdot 1$ and $- x \cdot 3$ .

$1 \cdot 3 x + 5 x \cdot 2 - 1 \cdot 3 x = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.1.2

Subtract $3 x$ from $1 \cdot 3 x$ .

$5 x \cdot 2 + 0 = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.1.3

Add $5 x \cdot 2$ and $0$ .

$5 x \cdot 2 = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

$5 x \cdot 2 = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.2

Multiply $2$ by $5$ .

$10 x = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $x$ by $1$ .

$10 x = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.3.2

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

$10 x = - 7$

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

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.3.3

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

$10 x = - 7$

$x + 2 x + 3 x = - 1$

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

$10 x = - 7$

$x + 2 x + 3 x = - 1$

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.4

Add $x$ and $2 x$ .

$10 x = - 7$

$3 x + 3 x = - 1$

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.5

Add $3 x$ and $3 x$ .

$10 x = - 7$

$6 x = - 1$

$2 x \cdot 1 + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.6

Simplify each term.

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Step 1.6.1

Multiply $2$ by $1$ .

$10 x = - 7$

$6 x = - 1$

$2 x + x \cdot 2 + 11 x \cdot 3 = 7$

Step 1.6.2

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

$10 x = - 7$

$6 x = - 1$

$2 x + 2 \cdot x + 11 x \cdot 3 = 7$

Step 1.6.3

Multiply $3$ by $11$ .

$10 x = - 7$

$6 x = - 1$

$2 x + 2 x + 33 x = 7$

$10 x = - 7$

$6 x = - 1$

$2 x + 2 x + 33 x = 7$

Step 1.7

Add $2 x$ and $2 x$ .

$10 x = - 7$

$6 x = - 1$

$4 x + 33 x = 7$

Step 1.8

Add $4 x$ and $33 x$ .

$10 x = - 7$

$6 x = - 1$

$37 x = 7$

$10 x = - 7$

$6 x = - 1$

$37 x = 7$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 10 & - 7 \\ 6 & - 1 \\ 37 & 7 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{10}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{10}{10} & \frac{- 7}{10} \\ 6 & - 1 \\ 37 & 7 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 6 & - 1 \\ 37 & 7 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 6 - 6 \cdot 1 & - 1 - 6 (- \frac{7}{10}) \\ 37 & 7 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 0 & \frac{16}{5} \\ 37 & 7 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} - 37 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} - 37 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 0 & \frac{16}{5} \\ 37 - 37 \cdot 1 & 7 - 37 (- \frac{7}{10}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 0 & \frac{16}{5} \\ 0 & \frac{329}{10} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{5}{16}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{5}{16}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ \frac{5}{16} \cdot 0 & \frac{5}{16} \cdot \frac{16}{5} \\ 0 & \frac{329}{10} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 0 & 1 \\ 0 & \frac{329}{10} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{329}{10} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{329}{10} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & - \frac{7}{10} \\ 0 & 1 \\ 0 - \frac{329}{10} \cdot 0 & \frac{329}{10} - \frac{329}{10} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

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

Step 3.6

Perform the row operation $R_{1} = R_{1} + \frac{7}{10} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} + \frac{7}{10} R_{2}$ to make the entry at $1, 2$ a $0$ .

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

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.

$(error = error)$