Solve {l}{1-{(3*2-9*1)}=1-{(x+2)}}{y=8}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 1 - {(3 \cdot 2 - 9 \cdot 1)} = 1 - {(x + 2)} }\\ { y = 8 }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=8$$

Solution Steps

Consider the first equation. Multiply 3 and 2 to get 6.

$$1-\left(6-9\times 1\right)=1-\left(x+2\right)$$

Multiply 9 and 1 to get 9.

$$1-\left(6-9\right)=1-\left(x+2\right)$$

Subtract 9 from 6 to get -3.

$$1-\left(-3\right)=1-\left(x+2\right)$$

The opposite of -3 is 3.

$$1+3=1-\left(x+2\right)$$

Add 1 and 3 to get 4.

$$4=1-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$4=1-x-2$$

Subtract 2 from 1 to get -1.

$$4=-1-x$$

Swap sides so that all variable terms are on the left hand side.

$$-1-x=4$$

Add 1 to both sides.

$$-x=4+1$$

Add 4 and 1 to get 5.

$$-x=5$$

Divide both sides by -1.

$$x=-5$$

The system is now solved.

$$x=-5 y=8 z=8$$