Solve Using a Matrix by Elimination w+z=320 , w+x=510 , x-y=140 , y-z=50

$w + z = 320$ , $w + x = 510$ , $x - y = 140$ , $y - z = 50$

Step 1

Write the system of equations in matrix form.

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 1 & 1 & 0 & 0 & 510 \\ 0 & 1 & - 1 & 0 & 140 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

Step 2

Find the reduced row echelon form of the matrix.

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} & - 1 \cdot R_{1} + R_{2} \\ 0 & 1 & - 1 & 0 & 140 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ (- 1) \cdot (1) + 1 & (- 1) \cdot (0) + 1 & (- 1) \cdot (0) + 0 & (- 1) \cdot (1) + 0 & (- 1) \cdot (320) + 510 \\ 0 & 1 & - 1 & 0 & 140 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 1 & - 1 & 0 & 140 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} & - 1 \cdot R_{2} + R_{3} \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ (- 1) \cdot (0) + 0 & (- 1) \cdot (1) + 1 & (- 1) \cdot (0) - 1 & (- 1) \cdot (- 1) + 0 & (- 1) \cdot (190) + 140 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 0 & - 1 & 1 & - 50 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ - 1 \cdot R_{3} & - 1 \cdot R_{3} & - 1 \cdot R_{3} & - 1 \cdot R_{3} & - 1 \cdot R_{3} \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

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

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ (- 1) \cdot (0) & (- 1) \cdot (0) & (- 1) \cdot (- 1) & (- 1) \cdot (1) & (- 1) \cdot (- 50) \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

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

Simplify $R_{3}$ (row $3$ ).

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 0 & 1 & - 1 & 50 \\ 0 & 0 & 1 & - 1 & 50 \end{array}]$

Perform the row operation $R_{4} = - 1 \cdot R_{3} + R_{4}$ on $R_{4}$ (row $4$ ) in order to convert some elements in the row to $0$ .

Replace $R_{4}$ (row $4$ ) with the row operation $R_{4} = - 1 \cdot R_{3} + R_{4}$ in order to convert some elements in the row to the desired value $0$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 0 & 1 & - 1 & 50 \\ - 1 \cdot R_{3} + R_{4} & - 1 \cdot R_{3} + R_{4} & - 1 \cdot R_{3} + R_{4} & - 1 \cdot R_{3} + R_{4} & - 1 \cdot R_{3} + R_{4} \end{array}]$

$R_{4} = - 1 \cdot R_{3} + R_{4}$

Replace $R_{4}$ (row $4$ ) with the actual values of the elements for the row operation $R_{4} = - 1 \cdot R_{3} + R_{4}$ .

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 0 & 1 & - 1 & 50 \\ (- 1) \cdot (0) + 0 & (- 1) \cdot (0) + 0 & (- 1) \cdot (1) + 1 & (- 1) \cdot (- 1) - 1 & (- 1) \cdot (50) + 50 \end{array}]$

$R_{4} = - 1 \cdot R_{3} + R_{4}$

Simplify $R_{4}$ (row $4$ ).

$[\begin{array}{c} 1 & 0 & 0 & 1 & 320 \\ 0 & 1 & 0 & - 1 & 190 \\ 0 & 0 & 1 & - 1 & 50 \\ 0 & 0 & 0 & 0 & 0 \end{array}]$

Step 3

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

$w + z = 320$

$x - z = 190$

$y - z = 50$

Step 4

Subtract $z$ from both sides of the equation.

$w = 320 - z$

$x - z = 190$

$y - z = 50$

Step 5

Add $z$ to both sides of the equation.

$x = 190 + z$

$w = 320 - z$

$y - z = 50$

Step 6

Add $z$ to both sides of the equation.

$y = 50 + z$

$w = 320 - z$

$x = 190 + z$

Step 7

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

$(320 - z, 190 + z, 50 + z, z)$

Solve Using a Matrix by Elimination w+z=320 , w+x=510 , x-y=140 , y-z=50

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