Finite Math Examples

$2 y - 3 x = 2$ , $4 y = 6 x + 4$

Step 1

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

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Step 1.1

Reorder $2 y$ and $- 3 x$ .

$- 3 x + 2 y = 2$

$4 y = 6 x + 4$

Step 1.2

Subtract $6 x$ from both sides of the equation.

$- 3 x + 2 y = 2$

$4 y - 6 x = 4$

Step 1.3

Reorder $4 y$ and $- 6 x$ .

$- 3 x + 2 y = 2$

$- 6 x + 4 y = 4$

$- 3 x + 2 y = 2$

$- 6 x + 4 y = 4$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 3 & 2 & 2 \\ - 6 & 4 & 4 \end{array}]$

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 $1$ .

$[\begin{array}{c} - \frac{1}{3} \cdot - 3 & - \frac{1}{3} \cdot 2 & - \frac{1}{3} \cdot 2 \\ - 6 & 4 & 4 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{2}{3} \\ - 6 & 4 & 4 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 6 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - \frac{2}{3} & - \frac{2}{3} \\ - 6 + 6 \cdot 1 & 4 + 6 (- \frac{2}{3}) & 4 + 6 (- \frac{2}{3}) \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

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

Step 4

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

$x - \frac{2}{3} y = - \frac{2}{3}$

$0 = 0$

Step 5

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

$(- \frac{2}{3} + \frac{2 y}{3}, y)$