$$ \begin{array}{ll} ext { a) } x+12=6-3 x & ext { () } \frac{x-x}{4} \frac{x}{3}=2 \ ext { b) } 2 x-5=3 x+3 & ext { d) } 3 x-x=1 \frac{1}{2}\end{array} $$

Question

$$ \begin{array}{ll}	ext { a) } x+12=6-3 x & 	ext { () } \frac{x-x}{4} \frac{x}{3}=2 \ 	ext { b) } 2 x-5=3 x+3 & 	ext { d) } 3 x-x=1 \frac{1}{2}\end{array} $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x+12=6-3x$$
- step1: Move the expression to the left side:
  $$x+3x=6-12$$
- step2: Add and subtract:
  $$4x=6-12$$
- step3: Add and subtract:
  $$4x=-6$$
- step4: Divide both sides:
  $$\frac{4x}{4}=\frac{-6}{4}$$
- step5: Divide the numbers:
  $$x=-\frac{3}{2}$$
Solve the equation $$ \frac{x-x}{4} \frac{x}{3}=2 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{x-x}{4}\times \frac{x}{3}=2$$
- step1: Simplify:
  $$0=2$$
- step2: The statement is false:
  $$x \in \varnothing $$
Solve the equation $$ 3x-x=1 \frac{1}{2} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$3x-x=1\frac{1}{2}$$
- step1: Subtract the terms:
  $$2x=\frac{3}{2}$$
- step2: Multiply by the reciprocal:
  $$2x\times \frac{1}{2}=\frac{3}{2}\times \frac{1}{2}$$
- step3: Multiply:
  $$x=\frac{3}{4}$$
Solve the equation $$ 2x-5=3x+3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$2x-5=3x+3$$
- step1: Move the expression to the left side:
  $$2x-3x=3+5$$
- step2: Add and subtract:
  $$-x=3+5$$
- step3: Add and subtract:
  $$-x=8$$
- step4: Change the signs:
  $$x=-8$$
Let's solve each equation step by step:

### a) $$ x + 12 = 6 - 3x $$

1. **Rearranging the equation**:
   $$
   x + 3x = 6 - 12
   $$
   This simplifies to:
   $$
   4x = -6
   $$

2. **Dividing both sides by 4**:
   $$
   x = -\frac{3}{2} \quad \text{or} \quad x = -1.5
   $$

### b) $$ \frac{x-x}{4} \frac{x}{3} = 2 $$

1. **Simplifying the left side**:
   Since $$ x - x = 0 $$, we have:
   $$
   \frac{0}{4} \cdot \frac{x}{3} = 0
   $$
   Thus, the equation becomes:
   $$
   0 = 2
   $$
   This is a contradiction, meaning there are no solutions:
   $$
   x \in \varnothing
   $$

### c) $$ 2x - 5 = 3x + 3 $$

1. **Rearranging the equation**:
   $$
   2x - 3x = 3 + 5
   $$
   This simplifies to:
   $$
   -x = 8
   $$

2. **Multiplying both sides by -1**:
   $$
   x = -8
   $$

### d) $$ 3x - x = 1 \frac{1}{2} $$

1. **Converting $$ 1 \frac{1}{2} $$ to an improper fraction**:
   $$
   1 \frac{1}{2} = \frac{3}{2}
   $$

2. **Rearranging the equation**:
   $$
   2x = \frac{3}{2}
   $$

3. **Dividing both sides by 2**:
   $$
   x = \frac{3}{4} \quad \text{or} \quad x = 0.75
   $$

### Summary of Solutions:
- a) $$ x = -\frac{3}{2} $$ or $$ x = -1.5 $$
- b) No solution ($$ x \in \varnothing $$)
- c) $$ x = -8 $$
- d) $$ x = \frac{3}{4} $$ or $$ x = 0.75 $$
a) $$ x = -1.5 $$ b) No solution c) $$ x = -8 $$ d) $$ x = 0.75 $$

\( \begin{array}{ll}\text { a) } x+12=6-3 x & \text { () } \frac{x-x}{4} \frac{x}{3}=2 \\ \text { b) } 2 x-5=3 x+3 & \text { d) } 3 x-x=1 \frac{1}{2}\end{array} \)

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