Finite Math Examples

$[\begin{array}{c} - 3 & - 3 \\ 5 & 1 \end{array}] [\begin{array}{c} x \cdot 1 \\ x \cdot 2 \end{array}] = [\begin{array}{c} - 21 \\ 19 \end{array}]$

Step 1

Multiply $x$ by $1$ .

$[\begin{array}{c} - 3 & - 3 \\ 5 & 1 \end{array}] [\begin{array}{c} x \\ x \cdot 2 \end{array}] = [\begin{array}{c} - 21 \\ 19 \end{array}]$

Step 2

Move $2$ to the left of $x$ .

$[\begin{array}{c} - 3 & - 3 \\ 5 & 1 \end{array}] [\begin{array}{c} x \\ 2 \cdot x \end{array}] = [\begin{array}{c} - 21 \\ 19 \end{array}]$

Step 3

Multiply $[\begin{array}{c} - 3 & - 3 \\ 5 & 1 \end{array}] [\begin{array}{c} x \\ 2 x \end{array}]$ .

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Step 3.1

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 $2 \times 1$ .

Step 3.2

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

$[\begin{array}{c} - 3 x - 3 (2 x) \\ 5 x + 1 (2 x) \end{array}] = [\begin{array}{c} - 21 \\ 19 \end{array}]$

Step 3.3

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

$[\begin{array}{c} - 9 x \\ 7 x \end{array}] = [\begin{array}{c} - 21 \\ 19 \end{array}]$

Step 4

Write as a linear system of equations.

$- 9 x = - 21$

$7 x = 19$

Step 5

Solve the system of equations.

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Step 5.1

Divide each term in $- 9 x = - 21$ by $- 9$ and simplify.

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Step 5.1.1

Divide each term in $- 9 x = - 21$ by $- 9$ .

$\frac{- 9 x}{- 9} = \frac{- 21}{- 9}$

$7 x = 19$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $- 9$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{- 9 x}{- 9} = \frac{- 21}{- 9}$

$7 x = 19$

Step 5.1.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 21}{- 9}$

$7 x = 19$

$x = \frac{- 21}{- 9}$

$7 x = 19$

$x = \frac{- 21}{- 9}$

$7 x = 19$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Cancel the common factor of $- 21$ and $- 9$ .

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Step 5.1.3.1.1

Factor $- 3$ out of $- 21$ .

$x = \frac{- 3 \cdot 7}{- 9}$

$7 x = 19$

Step 5.1.3.1.2

Cancel the common factors.

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Step 5.1.3.1.2.1

Factor $- 3$ out of $- 9$ .

$x = \frac{- 3 \cdot 7}{- 3 \cdot 3}$

$7 x = 19$

Step 5.1.3.1.2.2

Cancel the common factor.

$x = \frac{- 3 \cdot 7}{- 3 \cdot 3}$

$7 x = 19$

Step 5.1.3.1.2.3

Rewrite the expression.

$x = \frac{7}{3}$

$7 x = 19$

$x = \frac{7}{3}$

$7 x = 19$

$x = \frac{7}{3}$

$7 x = 19$

$x = \frac{7}{3}$

$7 x = 19$

$x = \frac{7}{3}$

$7 x = 19$

Step 5.2

Replace all occurrences of $x$ with $\frac{7}{3}$ in each equation.

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Step 5.2.1

Replace all occurrences of $x$ in $7 x = 19$ with $\frac{7}{3}$ .

$7 (\frac{7}{3}) = 19$

$x = \frac{7}{3}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Multiply $7 (\frac{7}{3})$ .

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Step 5.2.2.1.1

Combine $7$ and $\frac{7}{3}$ .

$\frac{7 \cdot 7}{3} = 19$

$x = \frac{7}{3}$

Step 5.2.2.1.2

Multiply $7$ by $7$ .

$\frac{49}{3} = 19$

$x = \frac{7}{3}$

$\frac{49}{3} = 19$

$x = \frac{7}{3}$

$\frac{49}{3} = 19$

$x = \frac{7}{3}$

$\frac{49}{3} = 19$

$x = \frac{7}{3}$

Step 5.3

Since $\frac{49}{3} = 19$ is not true, there is no solution.

$No solution$