Finite Math Examples

$x + y = 7$ , $4 x + 3 y = 24$

Step 1

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 7 \\ 4 & 3 & 24 \end{array}]$

Step 2

Find the reduced row echelon form.

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Step 2.1

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 2.1.1

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

$[\begin{array}{c} 1 & 1 & 7 \\ 4 - 4 \cdot 1 & 3 - 4 \cdot 1 & 24 - 4 \cdot 7 \end{array}]$

Step 2.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 7 \\ 0 & - 1 & - 4 \end{array}]$

Step 2.2

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

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Step 2.2.1

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

$[\begin{array}{c} 1 & 1 & 7 \\ - 0 & - - 1 & - - 4 \end{array}]$

Step 2.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 7 \\ 0 & 1 & 4 \end{array}]$

Step 2.3

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

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Step 2.3.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 7 - 4 \\ 0 & 1 & 4 \end{array}]$

Step 2.3.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 3 \\ 0 & 1 & 4 \end{array}]$

Step 3

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

$x = 3$

$y = 4$

Step 4

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

$(3, 4)$