Finite Math Examples

$y - 3 < 5$ , $y = 9$

Step 1

Introduce slack variables $u$ and $v$ to replace the inequalities with equations.

$y - 3 + Z = 5$

$y - 9 = 0$

Step 2

Add $3$ to both sides of the equation.

$y + Z = 5 + 3, y - 9 = 0$

Step 3

Add $5$ and $3$ .

$y + Z = 8, y - 9 = 0$

Step 4

Add $9$ to both sides of the equation.

$y + Z = 8, y = 9$

Step 5

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 0 & 8 \\ 1 & 0 & 9 \end{array}]$

Step 6

Find the reduced row echelon form.

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Step 6.1

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

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Step 6.1.1

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

$[\begin{array}{c} 1 & 0 & 8 \\ 1 - 1 & 0 - 0 & 9 - 8 \end{array}]$

Step 6.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 8 \\ 0 & 0 & 1 \end{array}]$

Step 6.2

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

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Step 6.2.1

Perform the row operation $R_{1} = R_{1} - 8 R_{2}$ to make the entry at $1, 3$ a $0$ .

$[\begin{array}{c} 1 - 8 \cdot 0 & 0 - 8 \cdot 0 & 8 - 8 \cdot 1 \\ 0 & 0 & 1 \end{array}]$

Step 6.2.2

Simplify $R_{1}$ .

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

Step 7

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

$y = 0$

$0 = 1$

Step 8

Since $0 \neq 1$ , there are no solutions.

No solution