Finite Math Examples

$5 x \cdot 1 - 2 x \cdot 2 = - 9$ , $15 x \cdot 1 - 6 x \cdot 2 = - 27$

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 $1$ .

$5 x - 2 x \cdot 2 = - 9$

$15 x \cdot 1 - 6 x \cdot 2 = - 27$

Step 1.1.2

Multiply $2$ by $- 2$ .

$5 x - 4 x = - 9$

$15 x \cdot 1 - 6 x \cdot 2 = - 27$

$5 x - 4 x = - 9$

$15 x \cdot 1 - 6 x \cdot 2 = - 27$

Step 1.2

Subtract $4 x$ from $5 x$ .

$x = - 9$

$15 x \cdot 1 - 6 x \cdot 2 = - 27$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $15$ by $1$ .

$x = - 9$

$15 x - 6 x \cdot 2 = - 27$

Step 1.3.2

Multiply $2$ by $- 6$ .

$x = - 9$

$15 x - 12 x = - 27$

$x = - 9$

$15 x - 12 x = - 27$

Step 1.4

Subtract $12 x$ from $15 x$ .

$x = - 9$

$3 x = - 27$

$x = - 9$

$3 x = - 27$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & - 9 \\ 3 & - 27 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & - 9 \\ 3 - 3 \cdot 1 & - 27 - 3 \cdot - 9 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 9 \\ 0 & 0 \end{array}]$

Step 4

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

$x = - 9$

$0 = 0$

Step 5

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

$(error = error)$