Algebra Examples

$x + y = - 2$ , $53 x - 8 y = 0$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 1 & - 2 \\ 53 & - 8 & 0 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

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Perform the row operation $R_{2} = - 53 \cdot R_{1} + R_{2}$ 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 $R_{2} = - 53 \cdot R_{1} + R_{2}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 1 & - 2 \\ - 53 \cdot R_{1} + R_{2} & - 53 \cdot R_{1} + R_{2} & - 53 \cdot R_{1} + R_{2} \end{array}]$

$R_{2} = - 53 \cdot R_{1} + R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - 53 \cdot R_{1} + R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 2 \\ (- 53) \cdot (1) + 53 & (- 53) \cdot (1) - 8 & (- 53) \cdot (- 2) + 0 \end{array}]$

$R_{2} = - 53 \cdot R_{1} + R_{2}$

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

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & - 61 & 106 \end{array}]$

Perform the row operation $R_{2} = - \frac{1}{61} R_{2}$ 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 $R_{2} = - \frac{1}{61} R_{2}$ in order to convert some elements in the row to the desired value $1$ .

$[\begin{array}{c} 1 & 1 & - 2 \\ - \frac{1}{61} R_{2} & - \frac{1}{61} R_{2} & - \frac{1}{61} R_{2} \end{array}]$

$R_{2} = - \frac{1}{61} R_{2}$

Replace $R_{2}$ (row $2$ ) with the actual values of the elements for the row operation $R_{2} = - \frac{1}{61} R_{2}$ .

$[\begin{array}{c} 1 & 1 & - 2 \\ (- \frac{1}{61}) \cdot (0) & (- \frac{1}{61}) \cdot (- 61) & (- \frac{1}{61}) \cdot (106) \end{array}]$

$R_{2} = - \frac{1}{61} R_{2}$

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

$[\begin{array}{c} 1 & 1 & - 2 \\ 0 & 1 & - \frac{106}{61} \end{array}]$

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

$[\begin{array}{c} - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} & - 1 \cdot R_{2} + R_{1} \\ 0 & 1 & - \frac{106}{61} \end{array}]$

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

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

$[\begin{array}{c} (- 1) \cdot (0) + 1 & (- 1) \cdot (1) + 1 & (- 1) \cdot (- \frac{106}{61}) - 2 \\ 0 & 1 & - \frac{106}{61} \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & - \frac{16}{61} \\ 0 & 1 & - \frac{106}{61} \end{array}]$

Step 3

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

$x = - \frac{16}{61}$

$y = - \frac{106}{61}$

Step 4

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

$(- \frac{16}{61}, - \frac{106}{61})$

Step 5

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} - \frac{16}{61} \\ - \frac{106}{61} \end{array}]$

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