Solve {l}{{(x+1)}-{(x-1)}=3-x+3}{y=4}{z=y}{a=z}{b=a}{text{Solvefor}ctext{where}}{c=b}

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

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

$$c=4$$

Solution Steps

Consider the first equation. To find the opposite of x-1, find the opposite of each term.

$$x+1-x+1=3-x+3$$

Combine x and -x to get 0.

$$1+1=3-x+3$$

Add 1 and 1 to get 2.

$$2=3-x+3$$

Add 3 and 3 to get 6.

$$2=6-x$$

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

$$6-x=2$$

Subtract 6 from both sides.

$$-x=2-6$$

Subtract 6 from 2 to get -4.

$$-x=-4$$

Divide both sides by -1.

$$x=\frac{-4}{-1}$$

Fraction \frac{-4}{-1} can be simplified to 4 by removing the negative sign from both the numerator and the denominator.

$$x=4$$

The system is now solved.

$$x=4 y=4 z=4 a=4 b=4 c=4$$