Finite Math Examples

$5 x \cdot 1 - 3 x \cdot 2 = 20$ , $7 x \cdot 1 - 5 x \cdot 2 = 20$

Step 1

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

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $5$ by $1$ .

$5 x - 3 x \cdot 2 = 20$

$7 x \cdot 1 - 5 x \cdot 2 = 20$

Step 1.1.2

Multiply $2$ by $- 3$ .

$5 x - 6 x = 20$

$7 x \cdot 1 - 5 x \cdot 2 = 20$

$5 x - 6 x = 20$

$7 x \cdot 1 - 5 x \cdot 2 = 20$

Step 1.2

Subtract $6 x$ from $5 x$ .

$- x = 20$

$7 x \cdot 1 - 5 x \cdot 2 = 20$

Step 1.3

Simplify each term.

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Step 1.3.1

Multiply $7$ by $1$ .

$- x = 20$

$7 x - 5 x \cdot 2 = 20$

Step 1.3.2

Multiply $2$ by $- 5$ .

$- x = 20$

$7 x - 10 x = 20$

$- x = 20$

$7 x - 10 x = 20$

Step 1.4

Subtract $10 x$ from $7 x$ .

$- x = 20$

$- 3 x = 20$

$- x = 20$

$- 3 x = 20$

Step 2

Write the system as a matrix.

$[\begin{array}{c} - 1 & 20 \\ - 3 & 20 \end{array}]$

Step 3

Find the reduced row echelon form.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & - 20 \\ - 3 & 20 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 3.2.1

Perform the row operation $R_{2} = R_{2} + 3 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & - 20 \\ - 3 + 3 \cdot 1 & 20 + 3 \cdot - 20 \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 20 \\ 0 & - 40 \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $- \frac{1}{40}$ to make the entry at $2, 2$ a $1$ .

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Step 3.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{40}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & - 20 \\ - \frac{1}{40} \cdot 0 & - \frac{1}{40} \cdot - 40 \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & - 20 \\ 0 & 1 \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} + 20 R_{2}$ to make the entry at $1, 2$ a $0$ .

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Step 3.4.1

Perform the row operation $R_{1} = R_{1} + 20 R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 + 20 \cdot 0 & - 20 + 20 \cdot 1 \\ 0 & 1 \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

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

Step 4

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

$x = 0$

$0 = 1$

Step 5

The system is inconsistent so there is no solution.

$No solution$