Question

2.
a. $\frac{1}{4}(12 - 4x)=3 - x$
b. $x - 3 = 3 - x$
c. $x - 3 = 3 + x$

Explanation:

Step1: Expand left-hand side

$\frac{1}{4}(12-4x) = 3 - x$
$3 - x = 3 - x$

Step2: Simplify the equation

Subtract $3$ from both sides: $-x = -x$
Add $x$ to both sides: $0 = 0$
This is always true, so all real numbers are solutions.

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Step1: Isolate $x$ terms on left

$x - 3 = 3 - x$
Add $x$ to both sides: $2x - 3 = 3$

Step2: Isolate the constant term

Add $3$ to both sides: $2x = 6$

Step3: Solve for $x$

Divide by 2: $x = \frac{6}{2} = 3$

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Step1: Isolate $x$ terms on left

$x - 3 = 3 + x$
Subtract $x$ from both sides: $-3 = 3$

Step2: Evaluate the result

$-3 = 3$ is false, so there is no solution.

Answer:

a. All real numbers are solutions
b. $x=3$
c. No solution