Question

l 6-115. solve each equation.
a. ( 3(2 + x) = 4 - (x - 2) )
b. ( \frac{z}{2} + \frac{z}{3} - 1 = \frac{z}{6} + 3 )

Explanation:

Step1: Expand both sides

For part a:
$3(2+x) = 4 - (x-2)$
$6 + 3x = 4 - x + 2$

Step2: Simplify right-hand side

$6 + 3x = 6 - x$

Step3: Isolate x terms

$3x + x = 6 - 6$
$4x = 0$

Step4: Solve for x

$x = \frac{0}{4} = 0$

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Step1: Eliminate denominators (part b)

Multiply all terms by 6 (LCM of 2,3,6):
$6\times\frac{x}{2} + 6\times\frac{x}{3} - 6\times1 = 6\times\frac{x}{6} + 6\times3$
$3x + 2x - 6 = x + 18$

Step2: Simplify left-hand side

$5x - 6 = x + 18$

Step3: Isolate x terms

$5x - x = 18 + 6$
$4x = 24$

Step4: Solve for x

$x = \frac{24}{4} = 6$

Answer:

a. $x = 0$
b. $x = 6$