$0.x3+0.3x=\frac{7}{9} \implies x=?$ An old book problem

Question

I was trying to solve some of problems in an old very old book , which not signed $$0.3\times x ,\\0.\overline{3}\times x \\ 0.\overline{3x} $$ to solve below equation . I tried like below .
My question : How many way we can read this question ?
Am I right ? $$0.x3+0.3x=\frac{7}{9}$$

$$\quad{0.\overline{x3}+0.\overline{x3}=\frac{7}{9}\\(\frac{\overline{x3}}{100}+\frac{\overline{x3}}{10000}+...)+ (\frac{\overline{3x}}{100}+\frac{\overline{3x}}{10000}+...)\\ (\dfrac{\dfrac{\overline{3x}}{100}}{1-\frac{1}{100}})+(\dfrac{\dfrac{\overline{x3}}{100}}{1-\frac{1}{100}})=\\\frac{\overline{x3}}{99}+\frac{\overline{3x}}{99}=\\ \frac{\overline{x3}+\overline{3x}}{99}=\\ \frac{10x+3+30+x}{99}=\\\frac{x+3}{9} \to \\\frac{x+3}{9}=\frac{7}{9} \\x=4}$$

Answer 1

$$\quad{0.\overline{x3}+0.\overline{3x}=\frac{7}{9}\\0.\overline{x}\times 3+0.\overline{3x}=\frac{7}{9}\\ 0.\overline{x3}+0.\overline{3}\times x=\frac{7}{9} \\0.\overline{x}\times 3+0.\overline{3}\times x=\frac{7}{9}\\or \\0.3\times x+0.x\times 3=\frac{7}{9}\\0.(3x)+0.(x3)=\frac{7}{9}}$$ is there an other possible reading ?