Solve for $$ x $$ $$ \frac{x-3}{4}+\frac{x}{8} \geq \frac{2 x-3}{6} $$

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UpStudy AI · Accepted Answer

We start with the inequality
$$
\frac{x-3}{4}+\frac{x}{8} \geq \frac{2x-3}{6}.
$$

**Step 1. Clear the denominators**

The denominators are $$4$$, $$8$$, and $$6$$. The least common multiple (LCM) of these numbers is $$24$$. Multiply every term by $$24$$:
$$
24\left(\frac{x-3}{4}\right) + 24\left(\frac{x}{8}\right) \geq 24\left(\frac{2x-3}{6}\right).
$$

**Step 2. Simplify each term**

Calculate each term:
- $$24\left(\frac{x-3}{4}\right) = 6(x-3) = 6x - 18$$,
- $$24\left(\frac{x}{8}\right) = 3x$$,
- $$24\left(\frac{2x-3}{6}\right) = 4(2x-3) = 8x - 12$$.

The inequality becomes:
$$
6x - 18 + 3x \geq 8x - 12.
$$

**Step 3. Combine like terms**

Combine the terms on the left:
$$
9x - 18 \geq 8x - 12.
$$

**Step 4. Solve for $$ x $$**

Subtract $$8x$$ from both sides:
$$
9x - 8x - 18 \geq -12 \quad \Rightarrow \quad x - 18 \geq -12.
$$

Add $$18$$ to both sides:
$$
x \geq 6.
$$

The solution to the inequality is:
$$
x \geq 6.
$$
$$ x \geq 6 $$

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