Finite Math Examples

$x \cdot 1 - 2 x \cdot 2 = k \cdot 1$ , $x \cdot 1 - x \cdot 2 = k \cdot 2$

Step 1

Move variables to the left and constant terms to the right.

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Step 1.1

Subtract $k \cdot 1$ from both sides of the equation.

$x \cdot 1 - 2 x \cdot 2 - k \cdot 1 = 0$

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

Step 1.2

Simplify each term.

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Step 1.2.1

Multiply $x$ by $1$ .

$x - 2 x \cdot 2 - k \cdot 1 = 0$

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

Step 1.2.2

Multiply $2$ by $- 2$ .

$x - 4 x - k \cdot 1 = 0$

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

Step 1.2.3

Multiply $- 1$ by $1$ .

$x - 4 x - k = 0$

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

$x - 4 x - k = 0$

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

Step 1.3

Subtract $4 x$ from $x$ .

$- 3 x - k = 0$

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

Step 1.4

Reorder $- 3 x$ and $- k$ .

$- k - 3 x = 0$

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

Step 1.5

Subtract $k \cdot 2$ from both sides of the equation.

$- k - 3 x = 0$

$x \cdot 1 - x \cdot 2 - k \cdot 2 = 0$

Step 1.6

Simplify each term.

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Step 1.6.1

Multiply $x$ by $1$ .

$- k - 3 x = 0$

$x - x \cdot 2 - k \cdot 2 = 0$

Step 1.6.2

Multiply $2$ by $- 1$ .

$- k - 3 x = 0$

$x - 2 x - k \cdot 2 = 0$

Step 1.6.3

Multiply $2$ by $- 1$ .

$- k - 3 x = 0$

$x - 2 x - 2 k = 0$

$- k - 3 x = 0$

$x - 2 x - 2 k = 0$

Step 1.7

Subtract $2 x$ from $x$ .

$- k - 3 x = 0$

$- x - 2 k = 0$

Step 1.8

Reorder $- x$ and $- 2 k$ .

$- k - 3 x = 0$

$- 2 k - x = 0$

$- k - 3 x = 0$

$- 2 k - x = 0$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 1 & - 3 & 0 \\ - 2 & - 1 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} - - 1 & - - 3 & - 0 \\ - 2 & - 1 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 3 & 0 \\ - 2 & - 1 & 0 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & 3 & 0 \\ - 2 + 2 \cdot 1 & - 1 + 2 \cdot 3 & 0 + 2 \cdot 0 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 3 & 0 \\ 0 & 5 & 0 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 & 3 & 0 \\ \frac{0}{5} & \frac{5}{5} & \frac{0}{5} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$k = 0$

$x = 0$

Step 5

The solution is the set of ordered pairs that make the system true.

$(0, 0)$