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?

 A: Be careful. Too many steps at once maybe.
$$\frac53\left(3-\frac x5\right)=x.$$
LCD in parenthesis
$$\frac53\left(\frac{15-x}5\right)=x.$$
LCD on RHS
$$\frac{5(15-x)}{15}=\frac{15x}{15}.$$
A: Multiplying both by $$\frac{3}{5}$$ we get
$$3-\frac{x}{5}=\frac{3}{5}x$$ so $$3=\frac{4}{5}x$$ thus
$$x=\frac{15}{4}$$
A: You don't want to multiply the left-hand numerator by 15.
Although you multiply top and bottom of the right hand side by 15, you don't  fo it to the left because the denominator is already 15.  So the numerator on the left is just $5(15-x)$
A: your first nominator is wrong, $$(3-\frac{x}{5})=\frac{15*3-3x}{15}$$ 
but then you multiply by $$\frac{5}{3}=\frac{25}{15}$$ id would give you 15^2 in the nominator. (best way is always to multiply with the common nominator , here 15) 
trula
A: 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
$$\frac53\left(3-\frac{x}5\right)=\frac{5\cdot5}{15}\cdot\frac{15\cdot3-3x}{15}$$
