Not quite; you made a
sign error. If you check your answer, you find that
[imath]2\cdot\frac{29}{9}-3\left(2\cdot\frac{29}{9}-3\right)=\frac{58}{9}-3\left(\frac{58}{9}-\frac{27}{9}\right)=\frac{58}{9}-3\left(\frac{31}{9}\right)=\frac{58}{9}-\frac{93}{9}=\frac{-65}{9}[/imath]
[imath]2\cdot\frac{29}{9}-5\left(3\cdot\frac{29}{9}-4\right)=\frac{58}{9}-5\left(\frac{87}{9}-\frac{36}{9}\right)=\frac{58}{9}-5\left(\frac{51}{9}\right)=\frac{58}{9}-\frac{255}{9}=\frac{-197}{9}[/imath]
So it doesn't work. It's good to add a solution to an abandoned thread, but we have to check our work.
Here is my work:
[imath]2y-3(2y-3) = 2y-5(3y-4)\\2y-(6y-9) = 2y-(15y-20)\\2y-6y+9 = 2y-15y+20\\-4y+9 = -13y+20\\-4y+13y+9 = 20\\9y+9 = 20\\9y = 11\\y = \frac{11}{9}[/imath]
When I check this, I get
[imath]2\cdot\frac{11}{9}-3\left(2\cdot\frac{11}{9}-3\right)=\frac{22}{9}-3\left(\frac{22}{9}-\frac{27}{9}\right)=\frac{22}{9}-3\left(\frac{-5}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]
[imath]2\cdot\frac{11}{9}-5\left(3\cdot\frac{11}{9}-4\right)=\frac{22}{9}-5\left(\frac{33}{9}-\frac{36}{9}\right)=\frac{22}{9}-5\left(\frac{-3}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]
So this solution works.