Find the Variables [[-1,4x-2,-5],[0,3z-9,y-1]]=[[-1,2,-5],[0,3,2]]

$[\begin{array}{c} - 1 & 4 x - 2 & - 5 \\ 0 & 3 z - 9 & y - 1 \end{array}] = [\begin{array}{c} - 1 & 2 & - 5 \\ 0 & 3 & 2 \end{array}]$

Step 1

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

$- 1 = - 1$

$4 x - 2 = 2$

$- 5 = - 5$

$0 = 0$

$3 z - 9 = 3$

$y - 1 = 2$

Step 2

Solve for $x$ in $4 x - 2 = 2$ .

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

Add $2$ to both sides of the equation.

$4 x = 2 + 2$

$3 z - 9 = 3$

$y - 1 = 2$

Add $2$ and $2$ .

$4 x = 4$

$3 z - 9 = 3$

$y - 1 = 2$

$4 x = 4$

$3 z - 9 = 3$

$y - 1 = 2$

Divide each term in $4 x = 4$ by $4$ and simplify.

Divide each term in $4 x = 4$ by $4$ .

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

$3 z - 9 = 3$

$y - 1 = 2$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

$3 z - 9 = 3$

$y - 1 = 2$

Divide $x$ by $1$ .

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

$3 z - 9 = 3$

$y - 1 = 2$

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

$3 z - 9 = 3$

$y - 1 = 2$

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

$3 z - 9 = 3$

$y - 1 = 2$

Simplify the right side.

Divide $4$ by $4$ .

$x = 1$

$3 z - 9 = 3$

$y - 1 = 2$

$x = 1$

$3 z - 9 = 3$

$y - 1 = 2$

$x = 1$

$3 z - 9 = 3$

$y - 1 = 2$

$x = 1$

$3 z - 9 = 3$

$y - 1 = 2$

Step 3

Replace all occurrences of $x$ with $1$ in each equation.

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

Add $9$ to both sides of the equation.

$3 z = 3 + 9$

$x = 1$

$y - 1 = 2$

Add $3$ and $9$ .

$3 z = 12$

$x = 1$

$y - 1 = 2$

$3 z = 12$

$x = 1$

$y - 1 = 2$

Divide each term in $3 z = 12$ by $3$ and simplify.

Divide each term in $3 z = 12$ by $3$ .

$\frac{3 z}{3} = \frac{12}{3}$

$x = 1$

$y - 1 = 2$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 z}{3} = \frac{12}{3}$

$x = 1$

$y - 1 = 2$

Divide $z$ by $1$ .

$z = \frac{12}{3}$

$x = 1$

$y - 1 = 2$

$z = \frac{12}{3}$

$x = 1$

$y - 1 = 2$

$z = \frac{12}{3}$

$x = 1$

$y - 1 = 2$

Simplify the right side.

Divide $12$ by $3$ .

$z = 4$

$x = 1$

$y - 1 = 2$

$z = 4$

$x = 1$

$y - 1 = 2$

$z = 4$

$x = 1$

$y - 1 = 2$

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

Add $1$ to both sides of the equation.

$y = 2 + 1$

$z = 4$

$x = 1$

Add $2$ and $1$ .

$y = 3$

$z = 4$

$x = 1$

$y = 3$

$z = 4$

$x = 1$

$y = 3$

$z = 4$

$x = 1$

Step 4

List all of the solutions.

$y = 3, z = 4, x = 1$

Find the Variables [[-1,4x-2,-5],[0,3z-9,y-1]]=[[-1,2,-5],[0,3,2]]

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