Finite Math Examples

$[\begin{array}{c} - 3 & - 3 \\ - 5 & - 2 \end{array}] [\begin{array}{c} x \cdot 1 \\ x \cdot 2 \end{array}] + [\begin{array}{c} - 4 \\ 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 1

Simplify each term.

Tap for more steps...

Step 1.1

Multiply $x$ by $1$ .

$[\begin{array}{c} - 3 & - 3 \\ - 5 & - 2 \end{array}] [\begin{array}{c} x \\ x \cdot 2 \end{array}] + [\begin{array}{c} - 4 \\ 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 1.2

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

$[\begin{array}{c} - 3 & - 3 \\ - 5 & - 2 \end{array}] [\begin{array}{c} x \\ 2 \cdot x \end{array}] + [\begin{array}{c} - 4 \\ 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 1.3

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

Tap for more steps...

Step 1.3.1

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 1.3.2

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

$[\begin{array}{c} - 3 x - 3 (2 x) \\ - 5 x - 2 (2 x) \end{array}] + [\begin{array}{c} - 4 \\ 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 1.3.3

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

$[\begin{array}{c} - 9 x \\ - 9 x \end{array}] + [\begin{array}{c} - 4 \\ 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 2

Add the corresponding elements.

$[\begin{array}{c} - 9 x - 4 \\ - 9 x + 20 \end{array}] = [\begin{array}{c} - 16 \\ 15 \end{array}]$

Step 3

Write as a linear system of equations.

$- 9 x - 4 = - 16$

$- 9 x + 20 = 15$

Step 4

Solve the system of equations.

Tap for more steps...

Step 4.1

Solve for $x$ in $- 9 x - 4 = - 16$ .

Tap for more steps...

Step 4.1.1

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

Tap for more steps...

Step 4.1.1.1

Add $4$ to both sides of the equation.

$- 9 x = - 16 + 4$

$- 9 x + 20 = 15$

Step 4.1.1.2

Add $- 16$ and $4$ .

$- 9 x = - 12$

$- 9 x + 20 = 15$

$- 9 x = - 12$

$- 9 x + 20 = 15$

Step 4.1.2

Divide each term in $- 9 x = - 12$ by $- 9$ and simplify.

Tap for more steps...

Step 4.1.2.1

Divide each term in $- 9 x = - 12$ by $- 9$ .

$\frac{- 9 x}{- 9} = \frac{- 12}{- 9}$

$- 9 x + 20 = 15$

Step 4.1.2.2

Simplify the left side.

Tap for more steps...

Step 4.1.2.2.1

Cancel the common factor of $- 9$ .

Tap for more steps...

Step 4.1.2.2.1.1

Cancel the common factor.

$\frac{- 9 x}{- 9} = \frac{- 12}{- 9}$

$- 9 x + 20 = 15$

Step 4.1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 12}{- 9}$

$- 9 x + 20 = 15$

$x = \frac{- 12}{- 9}$

$- 9 x + 20 = 15$

$x = \frac{- 12}{- 9}$

$- 9 x + 20 = 15$

Step 4.1.2.3

Simplify the right side.

Tap for more steps...

Step 4.1.2.3.1

Cancel the common factor of $- 12$ and $- 9$ .

Tap for more steps...

Step 4.1.2.3.1.1

Factor $- 3$ out of $- 12$ .

$x = \frac{- 3 \cdot 4}{- 9}$

$- 9 x + 20 = 15$

Step 4.1.2.3.1.2

Cancel the common factors.

Tap for more steps...

Step 4.1.2.3.1.2.1

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

$x = \frac{- 3 \cdot 4}{- 3 \cdot 3}$

$- 9 x + 20 = 15$

Step 4.1.2.3.1.2.2

Cancel the common factor.

$x = \frac{- 3 \cdot 4}{- 3 \cdot 3}$

$- 9 x + 20 = 15$

Step 4.1.2.3.1.2.3

Rewrite the expression.

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

$x = \frac{4}{3}$

$- 9 x + 20 = 15$

Step 4.2

Replace all occurrences of $x$ with $\frac{4}{3}$ in each equation.

Tap for more steps...

Step 4.2.1

Replace all occurrences of $x$ in $- 9 x + 20 = 15$ with $\frac{4}{3}$ .

$- 9 (\frac{4}{3}) + 20 = 15$

$x = \frac{4}{3}$

Step 4.2.2

Simplify the left side.

Tap for more steps...

Step 4.2.2.1

Simplify $- 9 (\frac{4}{3}) + 20$ .

Tap for more steps...

Step 4.2.2.1.1

Simplify each term.

Tap for more steps...

Step 4.2.2.1.1.1

Cancel the common factor of $3$ .

Tap for more steps...

Step 4.2.2.1.1.1.1

Factor $3$ out of $- 9$ .

$3 (- 3) (\frac{4}{3}) + 20 = 15$

$x = \frac{4}{3}$

Step 4.2.2.1.1.1.2

Cancel the common factor.

$3 \cdot (- 3 (\frac{4}{3})) + 20 = 15$

$x = \frac{4}{3}$

Step 4.2.2.1.1.1.3

Rewrite the expression.

$- 3 \cdot 4 + 20 = 15$

$x = \frac{4}{3}$

$- 3 \cdot 4 + 20 = 15$

$x = \frac{4}{3}$

Step 4.2.2.1.1.2

Multiply $- 3$ by $4$ .

$- 12 + 20 = 15$

$x = \frac{4}{3}$

$- 12 + 20 = 15$

$x = \frac{4}{3}$

Step 4.2.2.1.2

Add $- 12$ and $20$ .

$8 = 15$

$x = \frac{4}{3}$

$8 = 15$

$x = \frac{4}{3}$

$8 = 15$

$x = \frac{4}{3}$

$8 = 15$

$x = \frac{4}{3}$

Step 4.3

Since $8 = 15$ is not true, there is no solution.

$No solution$