Finite Math Examples

$5 x \cdot 1 + 6 x \cdot 2 - 4 x \cdot 3 = 8$ , $x \cdot 1 - 2 x \cdot 4 = 4$ , $8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

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 $5$ by $1$ .

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

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.1.2

Multiply $2$ by $6$ .

$5 x + 12 x - 4 x \cdot 3 = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.1.3

Multiply $3$ by $- 4$ .

$5 x + 12 x - 12 x = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

$5 x + 12 x - 12 x = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.2

Combine the opposite terms in $5 x + 12 x - 12 x$ .

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Step 1.2.1

Subtract $12 x$ from $12 x$ .

$5 x + 0 = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.2.2

Add $5 x$ and $0$ .

$5 x = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

$5 x = 8$

$x \cdot 1 - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $x$ by $1$ .

$5 x = 8$

$x - 2 x \cdot 4 = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.3.2

Multiply $4$ by $- 2$ .

$5 x = 8$

$x - 8 x = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

$5 x = 8$

$x - 8 x = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.4

Subtract $8 x$ from $x$ .

$5 x = 8$

$- 7 x = 4$

$8 x \cdot 1 - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.5

Simplify each term.

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Step 1.5.1

Multiply $8$ by $1$ .

$5 x = 8$

$- 7 x = 4$

$8 x - 9 x \cdot 2 + x \cdot 3 = - 5$

Step 1.5.2

Multiply $2$ by $- 9$ .

$5 x = 8$

$- 7 x = 4$

$8 x - 18 x + x \cdot 3 = - 5$

Step 1.5.3

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

$5 x = 8$

$- 7 x = 4$

$8 x - 18 x + 3 x = - 5$

$5 x = 8$

$- 7 x = 4$

$8 x - 18 x + 3 x = - 5$

Step 1.6

Subtract $18 x$ from $8 x$ .

$5 x = 8$

$- 7 x = 4$

$- 10 x + 3 x = - 5$

Step 1.7

Add $- 10 x$ and $3 x$ .

$5 x = 8$

$- 7 x = 4$

$- 7 x = - 5$

$5 x = 8$

$- 7 x = 4$

$- 7 x = - 5$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 5 & 8 \\ - 7 & 4 \\ - 7 & - 5 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} \frac{5}{5} & \frac{8}{5} \\ - 7 & 4 \\ - 7 & - 5 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ - 7 & 4 \\ - 7 & - 5 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & \frac{8}{5} \\ - 7 + 7 \cdot 1 & 4 + 7 (\frac{8}{5}) \\ - 7 & - 5 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & \frac{76}{5} \\ - 7 & - 5 \end{array}]$

Step 3.3

Perform the row operation $R_{3} = R_{3} + 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.3.1

Perform the row operation $R_{3} = R_{3} + 7 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & \frac{76}{5} \\ - 7 + 7 \cdot 1 & - 5 + 7 (\frac{8}{5}) \end{array}]$

Step 3.3.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & \frac{76}{5} \\ 0 & \frac{31}{5} \end{array}]$

Step 3.4

Multiply each element of $R_{2}$ by $\frac{5}{76}$ to make the entry at $2, 2$ a $1$ .

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Step 3.4.1

Multiply each element of $R_{2}$ by $\frac{5}{76}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ \frac{5}{76} \cdot 0 & \frac{5}{76} \cdot \frac{76}{5} \\ 0 & \frac{31}{5} \end{array}]$

Step 3.4.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & 1 \\ 0 & \frac{31}{5} \end{array}]$

Step 3.5

Perform the row operation $R_{3} = R_{3} - \frac{31}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.5.1

Perform the row operation $R_{3} = R_{3} - \frac{31}{5} R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & 1 \\ 0 - \frac{31}{5} \cdot 0 & \frac{31}{5} - \frac{31}{5} \cdot 1 \end{array}]$

Step 3.5.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & \frac{8}{5} \\ 0 & 1 \\ 0 & 0 \end{array}]$

Step 3.6

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.6.1

Perform the row operation $R_{1} = R_{1} - \frac{8}{5} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{8}{5} \cdot 0 & \frac{8}{5} - \frac{8}{5} \cdot 1 \\ 0 & 1 \\ 0 & 0 \end{array}]$

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.

$(error = error)$