Finite Math Examples

$5 x \cdot 1 - 4 x \cdot 2 = 10$ , $8 x \cdot 1 - 6 x \cdot 2 = 20$

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 $1$ .

$5 x - 4 x \cdot 2 = 10$

$8 x \cdot 1 - 6 x \cdot 2 = 20$

Step 1.1.2

Multiply $2$ by $- 4$ .

$5 x - 8 x = 10$

$8 x \cdot 1 - 6 x \cdot 2 = 20$

$5 x - 8 x = 10$

$8 x \cdot 1 - 6 x \cdot 2 = 20$

Step 1.2

Subtract $8 x$ from $5 x$ .

$- 3 x = 10$

$8 x \cdot 1 - 6 x \cdot 2 = 20$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $8$ by $1$ .

$- 3 x = 10$

$8 x - 6 x \cdot 2 = 20$

Step 1.3.2

Multiply $2$ by $- 6$ .

$- 3 x = 10$

$8 x - 12 x = 20$

$- 3 x = 10$

$8 x - 12 x = 20$

Step 1.4

Subtract $12 x$ from $8 x$ .

$- 3 x = 10$

$- 4 x = 20$

$- 3 x = 10$

$- 4 x = 20$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & 10 \\ - 4 & 20 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 10 \\ - 4 & 20 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{10}{3} \\ - 4 & 20 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{10}{3} \\ - 4 + 4 \cdot 1 & 20 + 4 (- \frac{10}{3}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{10}{3} \\ 0 & \frac{20}{3} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{3}{20}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{3}{20}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - \frac{10}{3} \\ \frac{3}{20} \cdot 0 & \frac{3}{20} \cdot \frac{20}{3} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - \frac{10}{3} \\ 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + \frac{10}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + \frac{10}{3} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + \frac{10}{3} \cdot 0 & - \frac{10}{3} + \frac{10}{3} \cdot 1 \\ 0 & 1 \end{array}]$

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.

$No solution$