Solve {l}{8{(x+1)}=4{(x+3)}}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { 8 {(x + 1)} = 4 {(x + 3)} }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=1$$

Solution Steps

Consider the first equation. Use the distributive property to multiply 8 by x+1.

$$8x+8=4\left(x+3\right)$$

Use the distributive property to multiply 4 by x+3.

$$8x+8=4x+12$$

Subtract 4x from both sides.

$$8x+8-4x=12$$

Combine 8x and -4x to get 4x.

$$4x+8=12$$

Subtract 8 from both sides.

$$4x=12-8$$

Subtract 8 from 12 to get 4.

$$4x=4$$

Divide both sides by 4.

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

Divide 4 by 4 to get 1.

$$x=1$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=1$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=1$$

The system is now solved.

$$x=1 y=1 z=1$$