Finite Math Examples

$x + y = 1$ , $0.9 x + 0.2 y = x$ , $0.1 x + 0.8 y = y$

Step 1

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

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Step 1.1

Subtract $x$ from both sides of the equation.

$x + y = 1$

$0.9 x + 0.2 y - x = 0$

$0.1 x + 0.8 y = y$

Step 1.2

Subtract $x$ from $0.9 x$ .

$x + y = 1$

$- 0.1 x + 0.2 y = 0$

$0.1 x + 0.8 y = y$

Step 1.3

Subtract $y$ from both sides of the equation.

$x + y = 1$

$- 0.1 x + 0.2 y = 0$

$0.1 x + 0.8 y - y = 0$

Step 1.4

Subtract $y$ from $0.8 y$ .

$x + y = 1$

$- 0.1 x + 0.2 y = 0$

$0.1 x - 0.2 y = 0$

$x + y = 1$

$- 0.1 x + 0.2 y = 0$

$0.1 x - 0.2 y = 0$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 1 \\ - 0.1 & 0.2 & 0 \\ 0.1 & - 0.2 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} + 0.1 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} + 0.1 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ - 0.1 + 0.1 \cdot 1 & 0.2 + 0.1 \cdot 1 & 0 + 0.1 \cdot 1 \\ 0.1 & - 0.2 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.3 & 0.1 \\ 0.1 & - 0.2 & 0 \end{array}]$

Step 3.2

Perform the row operation $R_{3} = R_{3} - 0.1 R_{1}$ to make the entry at $3, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{3} = R_{3} - 0.1 R_{1}$ to make the entry at $3, 1$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.3 & 0.1 \\ 0.1 - 0.1 \cdot 1 & - 0.2 - 0.1 \cdot 1 & 0 - 0.1 \cdot 1 \end{array}]$

Step 3.2.2

Simplify $R_{3}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 0.3 & 0.1 \\ 0 & - 0.3 & - 0.1 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{0.3}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{0.3}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 1 & 1 \\ \frac{0}{0.3} & \frac{0.3}{0.3} & \frac{0.1}{0.3} \\ 0 & - 0.3 & - 0.1 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0.\overline{3} \\ 0 & - 0.3 & - 0.1 \end{array}]$

Step 3.4

Perform the row operation $R_{3} = R_{3} + 0.3 R_{2}$ to make the entry at $3, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{3} = R_{3} + 0.3 R_{2}$ to make the entry at $3, 2$ a $0$ .

$[\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 & 0.\overline{3} \\ 0 + 0.3 \cdot 0 & - 0.3 + 0.3 \cdot 1 & - 0.1 + 0.3 \cdot 0.\overline{3} \end{array}]$

Step 3.4.2

Simplify $R_{3}$ .

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

Step 3.5

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

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Step 3.5.1

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

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

Step 3.5.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 0.\overline{6} \\ 0 & 1 & 0.\overline{3} \\ 0 & 0 & 0 \end{array}]$

Step 4

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

$x = 0.\overline{6}$

$y = 0.\overline{3}$

$0 = 0$

Step 5

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

$(0.\overline{6}, 0.\overline{3})$