<s>[INST] Solving $\frac53\left(3-\frac{x}{5}\right) = x$ Although I had already solved the following equation, I can't figure out why I fail when passing thru a $15$ denominator I get a bad result.
$$\frac53\cdot\left(3-\frac{x}{5}\right) = x$$
Transformation to get $15$ denominator:
$$\begin{align}
\frac{5\cdot 5(15\cdot 3-3x)}{15} &= \frac{15\cdot x}{15} \tag{1}\\[4pt]
25\cdot(45-3x) &= 15x \tag{2}\\
1125 - 75x &= 15x \tag{3}\\
1125 &= 90x \tag{4}\\[4pt]
x &= \frac{1125}{90} \tag{5}
\end{align}$$
Using alternative approaches, $x = 15/4 = 3.75$, which I believe is the right answer.

What am I doing wrong on the above try?

 [/INST] Your first line of working is wrong because
$$\frac53\left(3-\frac{x}5\right)\ne\frac{5\cdot5(15\cdot3-3x)}{15}$$
you should actually get
$$\frac53\left(3-\frac{x}5\right)=\frac{5\cdot5}{15}\cdot\frac{15\cdot3-3x}{15}$$
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