Finite Math Examples

$x + y = 10$ , $3 + 2 y = 24$

Step 1

Move all terms not containing a variable to the right side of the equation.

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Step 1.1

Subtract $3$ from both sides of the equation.

$x + y = 10$

$2 y = 24 - 3$

Step 1.2

Subtract $3$ from $24$ .

$x + y = 10$

$2 y = 21$

$x + y = 10$

$2 y = 21$

Step 2

Write the system as a matrix.

$[\begin{array}{c} 1 & 1 & 10 \\ 0 & 2 & 21 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

$[\begin{array}{c} 1 & 1 & 10 \\ \frac{0}{2} & \frac{2}{2} & \frac{21}{2} \end{array}]$

Step 3.1.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 1 & 10 \\ 0 & 1 & \frac{21}{2} \end{array}]$

Step 3.2

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

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Step 3.2.1

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

$[\begin{array}{c} 1 - 0 & 1 - 1 & 10 - \frac{21}{2} \\ 0 & 1 & \frac{21}{2} \end{array}]$

Step 3.2.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{1}{2} \\ 0 & 1 & \frac{21}{2} \end{array}]$

Step 4

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

$x = - \frac{1}{2}$

$y = \frac{21}{2}$

Step 5

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

$(- \frac{1}{2}, \frac{21}{2})$