Finite Math Examples

$- 2 x \cdot 3 + 7 x \cdot 5 = 12$ , $2 x \cdot 1 + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$ , $2 x \cdot 1 + 4 x \cdot 2 - 5 x \cdot 3 + 6 x \cdot 4 - 5 x \cdot 5 = - 1$

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 $3$ by $- 2$ .

$- 6 x + 7 x \cdot 5 = 12$

$2 x \cdot 1 + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.1.2

Multiply $5$ by $7$ .

$- 6 x + 35 x = 12$

$2 x \cdot 1 + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

$- 6 x + 35 x = 12$

$2 x \cdot 1 + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.2

Add $- 6 x$ and $35 x$ .

$29 x = 12$

$2 x \cdot 1 + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $2$ by $1$ .

$29 x = 12$

$2 x + 4 x \cdot 2 + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.3.2

Multiply $2$ by $4$ .

$29 x = 12$

$2 x + 8 x + 10 x \cdot 3 + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.3.3

Multiply $3$ by $10$ .

$29 x = 12$

$2 x + 8 x + 30 x + 6 x \cdot 4 + 12 x \cdot 5 = 28$

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

Step 1.3.4

Multiply $4$ by $6$ .

$29 x = 12$

$2 x + 8 x + 30 x + 24 x + 12 x \cdot 5 = 28$

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

Step 1.3.5

Multiply $5$ by $12$ .

$29 x = 12$

$2 x + 8 x + 30 x + 24 x + 60 x = 28$

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

$29 x = 12$

$2 x + 8 x + 30 x + 24 x + 60 x = 28$

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

Step 1.4

Add $2 x$ and $8 x$ .

$29 x = 12$

$10 x + 30 x + 24 x + 60 x = 28$

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

Step 1.5

Add $10 x$ and $30 x$ .

$29 x = 12$

$40 x + 24 x + 60 x = 28$

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

Step 1.6

Add $40 x$ and $24 x$ .

$29 x = 12$

$64 x + 60 x = 28$

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

Step 1.7

Add $64 x$ and $60 x$ .

$29 x = 12$

$124 x = 28$

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

Step 1.8

Simplify each term.

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Step 1.8.1

Multiply $2$ by $1$ .

$29 x = 12$

$124 x = 28$

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

Step 1.8.2

Multiply $2$ by $4$ .

$29 x = 12$

$124 x = 28$

$2 x + 8 x - 5 x \cdot 3 + 6 x \cdot 4 - 5 x \cdot 5 = - 1$

Step 1.8.3

Multiply $3$ by $- 5$ .

$29 x = 12$

$124 x = 28$

$2 x + 8 x - 15 x + 6 x \cdot 4 - 5 x \cdot 5 = - 1$

Step 1.8.4

Multiply $4$ by $6$ .

$29 x = 12$

$124 x = 28$

$2 x + 8 x - 15 x + 24 x - 5 x \cdot 5 = - 1$

Step 1.8.5

Multiply $5$ by $- 5$ .

$29 x = 12$

$124 x = 28$

$2 x + 8 x - 15 x + 24 x - 25 x = - 1$

$29 x = 12$

$124 x = 28$

$2 x + 8 x - 15 x + 24 x - 25 x = - 1$

Step 1.9

Add $2 x$ and $8 x$ .

$29 x = 12$

$124 x = 28$

$10 x - 15 x + 24 x - 25 x = - 1$

Step 1.10

Subtract $15 x$ from $10 x$ .

$29 x = 12$

$124 x = 28$

$- 5 x + 24 x - 25 x = - 1$

Step 1.11

Add $- 5 x$ and $24 x$ .

$29 x = 12$

$124 x = 28$

$19 x - 25 x = - 1$

Step 1.12

Subtract $25 x$ from $19 x$ .

$29 x = 12$

$124 x = 28$

$- 6 x = - 1$

$29 x = 12$

$124 x = 28$

$- 6 x = - 1$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 29 & 12 \\ 124 & 28 \\ - 6 & - 1 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{29}{29} & \frac{12}{29} \\ 124 & 28 \\ - 6 & - 1 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 124 & 28 \\ - 6 & - 1 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} - 124 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} - 124 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 124 - 124 \cdot 1 & 28 - 124 (\frac{12}{29}) \\ - 6 & - 1 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & - \frac{676}{29} \\ - 6 & - 1 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 6 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & - \frac{676}{29} \\ - 6 + 6 \cdot 1 & - 1 + 6 (\frac{12}{29}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & - \frac{676}{29} \\ 0 & \frac{43}{29} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $- \frac{29}{676}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $- \frac{29}{676}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ - \frac{29}{676} \cdot 0 & - \frac{29}{676} (- \frac{676}{29}) \\ 0 & \frac{43}{29} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & 1 \\ 0 & \frac{43}{29} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{43}{29} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{43}{29} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & 1 \\ 0 - \frac{43}{29} \cdot 0 & \frac{43}{29} - \frac{43}{29} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{12}{29} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} - \frac{12}{29} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} - \frac{12}{29} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{12}{29} \cdot 0 & \frac{12}{29} - \frac{12}{29} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

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.

$(error = error)$