Finite Math Examples

$8 x + 5 y + u = 7$ , $x + 6 y + v = 9$ , $- 32 x - 2 y + m = 0$

Step 1

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

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Step 1.1

Move $5 y$ .

$8 x + u + 5 y = 7$

$x + 6 y + v = 9$

$- 32 x - 2 y + m = 0$

Step 1.2

Reorder $8 x$ and $u$ .

$u + 8 x + 5 y = 7$

$x + 6 y + v = 9$

$- 32 x - 2 y + m = 0$

Step 1.3

Move $6 y$ .

$u + 8 x + 5 y = 7$

$x + v + 6 y = 9$

$- 32 x - 2 y + m = 0$

Step 1.4

Reorder $x$ and $v$ .

$u + 8 x + 5 y = 7$

$v + x + 6 y = 9$

$- 32 x - 2 y + m = 0$

Step 1.5

Move $- 2 y$ .

$u + 8 x + 5 y = 7$

$v + x + 6 y = 9$

$- 32 x + m - 2 y = 0$

Step 1.6

Reorder $- 32 x$ and $m$ .

$u + 8 x + 5 y = 7$

$v + x + 6 y = 9$

$m - 32 x - 2 y = 0$

$u + 8 x + 5 y = 7$

$v + x + 6 y = 9$

$m - 32 x - 2 y = 0$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 0 & 1 & 0 & 8 & 5 & 7 \\ 0 & 0 & 1 & 1 & 6 & 9 \\ 1 & 0 & 0 & - 32 & - 2 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Swap $R_{3}$ with $R_{1}$ to put a nonzero entry at $1, 1$ .

$[\begin{array}{c} 1 & 0 & 0 & - 32 & - 2 & 0 \\ 0 & 0 & 1 & 1 & 6 & 9 \\ 0 & 1 & 0 & 8 & 5 & 7 \end{array}]$

Step 3.2

Swap $R_{3}$ with $R_{2}$ to put a nonzero entry at $2, 2$ .

$[\begin{array}{c} 1 & 0 & 0 & - 32 & - 2 & 0 \\ 0 & 1 & 0 & 8 & 5 & 7 \\ 0 & 0 & 1 & 1 & 6 & 9 \end{array}]$

Step 4

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

$m - 32 x - 2 y = 0$

$u + 8 x + 5 y = 7$

$v + x + 6 y = 9$

Step 5

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

$(32 x + 2 y, 7 - 8 x - 5 y, 9 - x - 6 y, x, y)$