Finite Math Examples

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

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 $2 x \cdot 1 - x \cdot 2 - 5 x \cdot 3$ .

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Step 1.1.1

Reorder the factors in the terms $2 x \cdot 1$ and $- x \cdot 2$ .

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

$x \cdot 1 - 3 x \cdot 2 = 9$

Step 1.1.2

Subtract $2 x$ from $1 \cdot 2 x$ .

$0 - 5 x \cdot 3 = 8$

$x \cdot 1 - 3 x \cdot 2 = 9$

$0 - 5 x \cdot 3 = 8$

$x \cdot 1 - 3 x \cdot 2 = 9$

Step 1.2

Subtract $5 x \cdot 3$ from $0$ .

$- 5 x \cdot 3 = 8$

$x \cdot 1 - 3 x \cdot 2 = 9$

Step 1.3

Multiply $3$ by $- 5$ .

$- 15 x = 8$

$x \cdot 1 - 3 x \cdot 2 = 9$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $x$ by $1$ .

$- 15 x = 8$

$x - 3 x \cdot 2 = 9$

Step 1.4.2

Multiply $2$ by $- 3$ .

$- 15 x = 8$

$x - 6 x = 9$

$- 15 x = 8$

$x - 6 x = 9$

Step 1.5

Subtract $6 x$ from $x$ .

$- 15 x = 8$

$- 5 x = 9$

$- 15 x = 8$

$- 5 x = 9$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 15 & 8 \\ - 5 & 9 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{15}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{15}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{15} \cdot - 15 & - \frac{1}{15} \cdot 8 \\ - 5 & 9 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{8}{15} \\ - 5 & 9 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - \frac{8}{15} \\ - 5 + 5 \cdot 1 & 9 + 5 (- \frac{8}{15}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{8}{15} \\ 0 & \frac{19}{3} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{3}{19}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{19}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{8}{15} \\ \frac{3}{19} \cdot 0 & \frac{3}{19} \cdot \frac{19}{3} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{8}{15} \\ 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{8}{15} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{8}{15} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{8}{15} \cdot 0 & - \frac{8}{15} + \frac{8}{15} \cdot 1 \\ 0 & 1 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 0$

$0 = 1$

Step 5

The system is inconsistent so there is no solution.

$No solution$