Finite Math Examples

$4 x \cdot 1 - 14 x \cdot 2 - 2 x \cdot 3 = 68$ , $x \cdot 1 - 4 x \cdot 2 = 19$

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 $4$ by $1$ .

$4 x - 14 x \cdot 2 - 2 x \cdot 3 = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

Step 1.1.2

Multiply $2$ by $- 14$ .

$4 x - 28 x - 2 x \cdot 3 = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

Step 1.1.3

Multiply $3$ by $- 2$ .

$4 x - 28 x - 6 x = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

$4 x - 28 x - 6 x = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

Step 1.2

Subtract $28 x$ from $4 x$ .

$- 24 x - 6 x = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

Step 1.3

Subtract $6 x$ from $- 24 x$ .

$- 30 x = 68$

$x \cdot 1 - 4 x \cdot 2 = 19$

Step 1.4

Simplify each term.

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Step 1.4.1

Multiply $x$ by $1$ .

$- 30 x = 68$

$x - 4 x \cdot 2 = 19$

Step 1.4.2

Multiply $2$ by $- 4$ .

$- 30 x = 68$

$x - 8 x = 19$

$- 30 x = 68$

$x - 8 x = 19$

Step 1.5

Subtract $8 x$ from $x$ .

$- 30 x = 68$

$- 7 x = 19$

$- 30 x = 68$

$- 7 x = 19$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 30 & 68 \\ - 7 & 19 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{30}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{30}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{30} \cdot - 30 & - \frac{1}{30} \cdot 68 \\ - 7 & 19 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{34}{15} \\ - 7 & 19 \end{array}]$

$[\begin{array}{c} 1 & - \frac{34}{15} \\ - 7 & 19 \end{array}]$

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 $0$ .

$[\begin{array}{c} 1 & - \frac{34}{15} \\ - 7 + 7 \cdot 1 & 19 + 7 (- \frac{34}{15}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{34}{15} \\ 0 & \frac{47}{15} \end{array}]$

$[\begin{array}{c} 1 & - \frac{34}{15} \\ 0 & \frac{47}{15} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{15}{47}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{15}{47}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{34}{15} \\ \frac{15}{47} \cdot 0 & \frac{15}{47} \cdot \frac{47}{15} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{34}{15} \\ 0 & 1 \end{array}]$

$[\begin{array}{c} 1 & - \frac{34}{15} \\ 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{34}{15} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{34}{15} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{34}{15} \cdot 0 & - \frac{34}{15} + \frac{34}{15} \cdot 1 \\ 0 & 1 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

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

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

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

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.

$No solution$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$