Finite Math Examples

$- 3 x + 3 y - 18 = 0$ , $- 2 x + 4 y - 20 = 0$

Step 1

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

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Step 1.1

Add $18$ to both sides of the equation.

$- 3 x + 3 y = 18$

$- 2 x + 4 y - 20 = 0$

Step 1.2

Add $20$ to both sides of the equation.

$- 3 x + 3 y = 18$

$- 2 x + 4 y = 20$

$- 3 x + 3 y = 18$

$- 2 x + 4 y = 20$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & 3 & 18 \\ - 2 & 4 & 20 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

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Step 3.1.1

Multiply each element of $R_{1}$ by $- \frac{1}{3}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 3 & - \frac{1}{3} \cdot 18 \\ - 2 & 4 & 20 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 1 & - 6 \\ - 2 & 4 & 20 \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 & - 1 & - 6 \\ - 2 + 2 \cdot 1 & 4 + 2 \cdot - 1 & 20 + 2 \cdot - 6 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{2}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 1 & - 6 \\ \frac{0}{2} & \frac{2}{2} & \frac{8}{2} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 1 & - 6 \\ 0 & 1 & 4 \end{array}]$

Step 3.4

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

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Step 3.4.1

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

$[\begin{array}{c} 1 + 0 & - 1 + 1 \cdot 1 & - 6 + 1 \cdot 4 \\ 0 & 1 & 4 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - 2 \\ 0 & 1 & 4 \end{array}]$

Step 4

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

$x = - 2$

$y = 4$

Step 5

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

$(- 2, 4)$