Finite Math Examples

$[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & \frac{- 1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 1

Move the negative in front of the fraction.

$[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 2

Multiply $[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} 1 & 2 \\ 3 & 1 \end{array}]$ .

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Step 2.1

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. In this case, the first matrix is $2 \times 2$ and the second matrix is $2 \times 2$ .

Step 2.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} \frac{1}{7} \cdot 1 + \frac{2}{7} \cdot 3 & \frac{1}{7} \cdot 2 + \frac{2}{7} \cdot 1 \\ \frac{3}{7} \cdot 1 - \frac{1}{7} \cdot 3 & \frac{3}{7} \cdot 2 - \frac{1}{7} \cdot 1 \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 2.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} 1 & \frac{4}{7} \\ 0 & \frac{5}{7} \end{array}] [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 3

Multiply $[\begin{array}{c} 1 & \frac{4}{7} \\ 0 & \frac{5}{7} \end{array}] [\begin{array}{c} x \\ y \end{array}]$ .

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Step 3.1

Step 3.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} 1 x + \frac{4}{7} y \\ 0 x + \frac{5}{7} y \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 3.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} x + \frac{4 y}{7} \\ \frac{5 y}{7} \end{array}] = [\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$

Step 4

Multiply $[\begin{array}{c} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & - \frac{1}{7} \end{array}] [\begin{array}{c} - 1 \\ 4 \end{array}]$ .

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Step 4.1

Step 4.2

Multiply each row in the first matrix by each column in the second matrix.

$[\begin{array}{c} x + \frac{4 y}{7} \\ \frac{5 y}{7} \end{array}] = [\begin{array}{c} \frac{1}{7} \cdot - 1 + \frac{2}{7} \cdot 4 \\ \frac{3}{7} \cdot - 1 - \frac{1}{7} \cdot 4 \end{array}]$

Step 4.3

Simplify each element of the matrix by multiplying out all the expressions.

$[\begin{array}{c} x + \frac{4 y}{7} \\ \frac{5 y}{7} \end{array}] = [\begin{array}{c} 1 \\ - 1 \end{array}]$

Step 5

Write as a linear system of equations.

$x + \frac{4 y}{7} = 1$

$\frac{5 y}{7} = - 1$

Step 6

Solve the system of equations.

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Step 6.1

Solve for $y$ in $\frac{5 y}{7} = - 1$ .

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Step 6.1.1

Multiply both sides of the equation by $\frac{7}{5}$ .

$\frac{7}{5} \cdot \frac{5 y}{7} = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2

Simplify both sides of the equation.

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Step 6.1.2.1

Simplify the left side.

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Step 6.1.2.1.1

Simplify $\frac{7}{5} \cdot \frac{5 y}{7}$ .

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Step 6.1.2.1.1.1

Cancel the common factor of $7$ .

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Step 6.1.2.1.1.1.1

Cancel the common factor.

$\frac{7}{5} \cdot \frac{5 y}{7} = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.1.1.1.2

Rewrite the expression.

$\frac{1}{5} \cdot (5 y) = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

$\frac{1}{5} \cdot (5 y) = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.1.1.2

Cancel the common factor of $5$ .

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Step 6.1.2.1.1.2.1

Factor $5$ out of $5 y$ .

$\frac{1}{5} \cdot (5 (y)) = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.1.1.2.2

Cancel the common factor.

$\frac{1}{5} \cdot (5 y) = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.1.1.2.3

Rewrite the expression.

$y = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

$y = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

$y = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

$y = \frac{7}{5} \cdot - 1$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.2

Simplify the right side.

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Step 6.1.2.2.1

Simplify $\frac{7}{5} \cdot - 1$ .

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Step 6.1.2.2.1.1

Multiply $\frac{7}{5} \cdot - 1$ .

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Step 6.1.2.2.1.1.1

Combine $\frac{7}{5}$ and $- 1$ .

$y = \frac{7 \cdot - 1}{5}$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.2.1.1.2

Multiply $7$ by $- 1$ .

$y = \frac{- 7}{5}$

$x + \frac{4 y}{7} = 1$

$y = \frac{- 7}{5}$

$x + \frac{4 y}{7} = 1$

Step 6.1.2.2.1.2

Move the negative in front of the fraction.

$y = - \frac{7}{5}$

$x + \frac{4 y}{7} = 1$

$y = - \frac{7}{5}$

$x + \frac{4 y}{7} = 1$

$y = - \frac{7}{5}$

$x + \frac{4 y}{7} = 1$

$y = - \frac{7}{5}$

$x + \frac{4 y}{7} = 1$

$y = - \frac{7}{5}$

$x + \frac{4 y}{7} = 1$

Step 6.2

Replace all occurrences of $y$ with $- \frac{7}{5}$ in each equation.

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Step 6.2.1

Replace all occurrences of $y$ in $x + \frac{4 y}{7} = 1$ with $- \frac{7}{5}$ .

$x + \frac{4 (- \frac{7}{5})}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Simplify the numerator.

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Step 6.2.2.1.1.1

Multiply $- 1$ by $4$ .

$x + \frac{- 4 (\frac{7}{5})}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.1.2

Combine $- 4$ and $\frac{7}{5}$ .

$x + \frac{\frac{- 4 \cdot 7}{5}}{7} = 1$

$y = - \frac{7}{5}$

$x + \frac{\frac{- 4 \cdot 7}{5}}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.2

Multiply $- 4$ by $7$ .

$x + \frac{\frac{- 28}{5}}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.3

Move the negative in front of the fraction.

$x + \frac{- \frac{28}{5}}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.4

Multiply the numerator by the reciprocal of the denominator.

$x - \frac{28}{5} \cdot \frac{1}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.5

Cancel the common factor of $7$ .

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Step 6.2.2.1.5.1

Move the leading negative in $- \frac{28}{5}$ into the numerator.

$x + \frac{- 28}{5} \cdot \frac{1}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.5.2

Factor $7$ out of $- 28$ .

$x + \frac{7 (- 4)}{5} \cdot \frac{1}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.5.3

Cancel the common factor.

$x + \frac{7 \cdot - 4}{5} \cdot \frac{1}{7} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.5.4

Rewrite the expression.

$x + \frac{- 4}{5} = 1$

$y = - \frac{7}{5}$

$x + \frac{- 4}{5} = 1$

$y = - \frac{7}{5}$

Step 6.2.2.1.6

Move the negative in front of the fraction.

$x - \frac{4}{5} = 1$

$y = - \frac{7}{5}$

$x - \frac{4}{5} = 1$

$y = - \frac{7}{5}$

$x - \frac{4}{5} = 1$

$y = - \frac{7}{5}$

$x - \frac{4}{5} = 1$

$y = - \frac{7}{5}$

Step 6.3

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.1

Add $\frac{4}{5}$ to both sides of the equation.

$x = 1 + \frac{4}{5}$

$y = - \frac{7}{5}$

Step 6.3.2

Write $1$ as a fraction with a common denominator.

$x = \frac{5}{5} + \frac{4}{5}$

$y = - \frac{7}{5}$

Step 6.3.3

Combine the numerators over the common denominator.

$x = \frac{5 + 4}{5}$

$y = - \frac{7}{5}$

Step 6.3.4

Add $5$ and $4$ .

$x = \frac{9}{5}$

$y = - \frac{7}{5}$

$x = \frac{9}{5}$

$y = - \frac{7}{5}$

Step 6.4

Solve the system of equations.

$x = \frac{9}{5} y = - \frac{7}{5}$

Step 6.5

List all of the solutions.

$x = \frac{9}{5}, y = - \frac{7}{5}$