Algebra Examples

$3 x + 3 y = 3$ , $x < y$

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

$x + Z = y$

$3 x + 3 y - 3 = 0$

Move $y$ to the left side of the equation because it contains a variable.

$x + Z - y = 0$

$3 x + 3 y - 3 = 0$

Move $3$ to the right side of the equation because it does not contain a variable.

$x + Z - y = 0$

$3 x + 3 y = 3$

Write the system of equations in matrix form.

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

Row reduce.

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Perform the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

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

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 3 \cdot R_{1} + R_{2}$ .

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

$R_{2} = - 3 \cdot R_{1} + R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{2} = \frac{1}{6} R_{2}$ on $R_{2}$ (row $2$ ) in order to convert some elements in the row to $1$ .

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Replace $R_{2}$ (row $2$ ) with the row operation $R_{2} = \frac{1}{6} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ \frac{1}{6} R_{2} & \frac{1}{6} R_{2} & \frac{1}{6} R_{2} & \frac{1}{6} R_{2} \end{array}]$

$R_{2} = \frac{1}{6} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = \frac{1}{6} R_{2}$ .

$[\begin{array}{c} 1 & - 1 & 0 & 0 \\ (\frac{1}{6}) \cdot (0) & (\frac{1}{6}) \cdot (6) & (\frac{1}{6}) \cdot (0) & (\frac{1}{6}) \cdot (3) \end{array}]$

$R_{2} = \frac{1}{6} R_{2}$

Simplify $R_{2}$ (row $2$ ).

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

Perform the row operation $R_{1} = R_{2} + R_{1}$ on $R_{1}$ (row $1$ ) in order to convert some elements in the row to $0$ .

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Replace $R_{1}$ (row $1$ ) with the row operation $R_{1} = R_{2} + R_{1}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} R_{2} + R_{1} & R_{2} + R_{1} & R_{2} + R_{1} & R_{2} + R_{1} \\ 0 & 1 & 0 & \frac{1}{2} \end{array}]$

$R_{1} = R_{2} + R_{1}$

Replace $R_{1}$ (row $1$ ) with the actual values of the elements for the row operation $R_{1} = R_{2} + R_{1}$ .

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

$R_{1} = R_{2} + R_{1}$

Simplify $R_{1}$ (row $1$ ).

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

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

$x = \frac{1}{2}$

$y = \frac{1}{2}$

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