Find the Variables [[6,3],[8,4]]*[[x],[y]]=[[9],[12]]

$[\begin{array}{c} 6 & 3 \\ 8 & 4 \end{array}] \cdot [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 9 \\ 12 \end{array}]$

Step 1

Simplify the left side of the equation.

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

$[\begin{array}{c} 6 \cdot x + 3 \cdot y \\ 8 \cdot x + 4 \cdot y \end{array}] = [\begin{array}{c} 9 \\ 12 \end{array}]$

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

$[\begin{array}{c} 6 x + 3 y \\ 8 x + 4 y \end{array}] = [\begin{array}{c} 9 \\ 12 \end{array}]$

Step 2

The matrix equation can be written as a set of equations.

$6 x + 3 y = 9$

$8 x + 4 y = 12$

Step 3

Solve for $x$ in $6 x + 3 y = 9$ .

Subtract $3 y$ from both sides of the equation.

$6 x = 9 - 3 y$

$8 x + 4 y = 12$

Divide each term in $6 x = 9 - 3 y$ by $6$ and simplify.

Divide each term in $6 x = 9 - 3 y$ by $6$ .

$\frac{6 x}{6} = \frac{9}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{9}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Divide $x$ by $1$ .

$x = \frac{9}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

$x = \frac{9}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

$x = \frac{9}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Simplify the right side.

Simplify each term.

Cancel the common factor of $9$ and $6$ .

Factor $3$ out of $9$ .

$x = \frac{3 (3)}{6} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Cancel the common factors.

Factor $3$ out of $6$ .

$x = \frac{3 \cdot 3}{3 \cdot 2} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Cancel the common factor.

$x = \frac{3 \cdot 3}{3 \cdot 2} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Rewrite the expression.

$x = \frac{3}{2} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

$x = \frac{3}{2} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

$x = \frac{3}{2} + \frac{- 3 y}{6}$

$8 x + 4 y = 12$

Cancel the common factor of $- 3$ and $6$ .

Factor $3$ out of $- 3 y$ .

$x = \frac{3}{2} + \frac{3 (- y)}{6}$

$8 x + 4 y = 12$

Cancel the common factors.

Factor $3$ out of $6$ .

$x = \frac{3}{2} + \frac{3 (- y)}{3 (2)}$

$8 x + 4 y = 12$

Cancel the common factor.

$x = \frac{3}{2} + \frac{3 (- y)}{3 \cdot 2}$

$8 x + 4 y = 12$

Rewrite the expression.

$x = \frac{3}{2} + \frac{- y}{2}$

$8 x + 4 y = 12$

$x = \frac{3}{2} + \frac{- y}{2}$

$8 x + 4 y = 12$

$x = \frac{3}{2} + \frac{- y}{2}$

$8 x + 4 y = 12$

Move the negative in front of the fraction.