<s>[INST] Linear inequalities with one unknown $$3x+\frac{2}{4} \leq x+\frac{7}{2}$$ What is the solution to the inequality?
I started multiplying both sides by 4
which gave me $3x+2\leq14$.
Then I subtracted two from both sides obtaining
$$3x\leq12$$
which then I divided by 3 to obtain the solution
$$x\leq4\ .$$
This solution does not match for my possible answers on my practice guide.
 [/INST] Just subtract $x+\frac{2}{4}$ on both sides...
$$3x+\frac{2}{4} \leq x+\frac{7}{2} \iff 2x \leq \frac{7}{2}-\frac{2}{4} = \frac{14}{4}-\frac{2}{4} = \frac{12}{4} = 3$$
and finally divide both sides by $2$ to find $x \leq \frac{3}{2}$.
Note:

I started multiplying both sides by $4$ which gave me $3x+2\leq 14$.

You should have got $12x+2 \leq 4x+14 \iff 8x \leq 12$...
</s>