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β = 100, Transistor Q1 is PNP and Q2 is NPN

\$\beta = 100\$, Transistor Q1 is PNP and Q2 is NPN. Trying to find \$V_1, V_2, V_3, V_4\$, and \$V_5\$. I assumed active mode for both.

\$\beta = 100\$
\$V_{BE} = 0.7 \mathrm{V}\$

Then $$ \alpha = \frac{\beta}{\beta + 1} = \frac{100}{101} = 0.99 $$ and $$ I_{B1} = (1 – \alpha)I_{E1} \iff I_{B1} = 0.01I_{E1} $$ Now, $$ \frac{V_1}{100000} = \frac{0.01(3 - V_2)}{9100} = 91(3 – V_2) $$ Then $$ \frac{(V_2 - 0.7)}{100000} = 273 – 91V_2 $$ $$ V_2 – 0.7 = 27300000 – 9100000V_2 \iff 9100001V_2 = 27300000.7 $$ thus $$ V_2 = 2.999999747 \mathrm{V} $$ and $$ V_1 = V_2 – 0.7 = 3 – 0.7 = 2.299999747 \mathrm{V} $$ Then $$ \begin{split} I_{E1} &= \frac{3 - V_2}{9100} = \frac{3 - 2.999999747}{9100} = 2.3 \cdot 10^{-3} \mathrm{A} \\ I_{B1} &= \frac{V_1}{100000} = \frac{2.3}{100000} = 2.3 \cdot 10^{-5} \mathrm{A}\\ \\ I_{C1} &= I_{E1} – I_{B1}\\ &= 2.3\cdot 10^{-3} – 2.3 \cdot 10^{-5}\\ &= 2.277 \cdot 10^{-3} \mathrm{A}\\ \end{split} $$ Now $$ I_{C1} = \frac{V_3 + 3}{9100} + I_{B2} $$ thus $$ \begin{split} 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_4 + 3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 - 0.7 + 3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{V_3 + 3}{9100} + \frac{V_3 + 2.3}{4300}\\ 2.277 \cdot 10^{-3} &= \frac{4300V_3 + 12900}{39130000} + \frac{9100V_3 + 20930}{39130000}\\ 2.277 \cdot 10^{-3} &= \frac{134V_3 +338.3}{391300}\\ 890.9901 &= 134V_3 + 338.3\\ 552.6901 &= 134V_3\\ V_3 &= 4.124552985 \mathrm{V}\\ \end{split} $$ and \$V_4 = V_3 – 0.7 = 4.124552985 – 0.7 = 3.424552985 \mathrm{V}\$. Then $$ \begin{split} I_{E2} &= \frac{V_4 + 3}{4300} = \frac{3.424552985+ 3}{4300} = 1.49408209 \cdot 10^{-3} \mathrm{A}\\ I_{C2} &= \alpha I_{E2} = 0.99(1.49408209 \cdot 10^{-3}) = 1.479141269 \cdot 10^{-3} \mathrm{A}\\ I_{B2} &= I_{E2} – I_{C2} = 1.49408209 \cdot 10^{-3} – 1.479141269 \cdot 10^{-3} = 1.494082134 \cdot 10^{-5} \mathrm{A} \end{split} $$ and $$ \begin{split} I_{C2} &= \frac{3 - V_5}{5100}\\ 1.479141269 \cdot 10^{-3} &= \frac{3 - V_5}{5100} \\ 7.543620472 &= 3 – V_5\\ V_5 &= -4.543620472 \mathrm{V} \end{split} $$

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This is more of a comment than an answer, but I don't have enough reputation to comment just yet.

A partial answer to "where did I go wrong" is a simple arithmetic error early on in the equations:

V_1/100000 = 0.01(3 - V_2)/9100
V_1/100000 = 91(3 – V2)

In this step it appears you multiplied 0.01 by 9100 rather than divide. Once this changes, everything downstream needs to change too.

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