Finite Math Examples

$x + 2 y = 10$ , $y = 4 x - 4$

Step 1

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

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Step 1.1

Subtract $4 x$ from both sides of the equation.

$x + 2 y = 10$

$y - 4 x = - 4$

Step 1.2

Reorder $y$ and $- 4 x$ .

$x + 2 y = 10$

$- 4 x + y = - 4$

$x + 2 y = 10$

$- 4 x + y = - 4$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 2 & 10 \\ - 4 & 1 & - 4 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.1.1

Perform the row operation $R_{2} = R_{2} + 4 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 2 & 10 \\ - 4 + 4 \cdot 1 & 1 + 4 \cdot 2 & - 4 + 4 \cdot 10 \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 10 \\ 0 & 9 & 36 \end{array}]$

Step 3.2

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

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Step 3.2.1

Multiply each element of $R_{2}$ by $\frac{1}{9}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 2 & 10 \\ \frac{0}{9} & \frac{9}{9} & \frac{36}{9} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 2 & 10 \\ 0 & 1 & 4 \end{array}]$

Step 3.3

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

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Step 3.3.1

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

$[\begin{array}{c} 1 - 2 \cdot 0 & 2 - 2 \cdot 1 & 10 - 2 \cdot 4 \\ 0 & 1 & 4 \end{array}]$

Step 3.3.2

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.

$(2, 4)$