Solve {l}{4*{(x+3)}=6*{(x-2)}}{y=3}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { 4 \cdot {(x + 3)} = 6 \cdot {(x - 2)} }\\ { y = 3 }\\ { z = y }\\ { a = z }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for x, y, z, a, b, c

$$c=3$$

Solution Steps

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

$$4x+12=6\left(x-2\right)$$

Use the distributive property to multiply 6 by x-2.

$$4x+12=6x-12$$

Subtract 6x from both sides.

$$4x+12-6x=-12$$

Combine 4x and -6x to get -2x.

$$-2x+12=-12$$

Subtract 12 from both sides.

$$-2x=-12-12$$

Subtract 12 from -12 to get -24.

$$-2x=-24$$

Divide both sides by -2.

$$x=\frac{-24}{-2}$$

Divide -24 by -2 to get 12.

$$x=12$$

The system is now solved.

$$x=12 y=3 z=3 a=3 b=3 c=3$$