This is a lot of equations, so it can get very confusing if you don’t do it systematically. I would start by getting rid of $i_7$ from all the equations, then $i_6$, and so on. If we can write the numerator of $R$ in terms of $i_0$ then it’ll cancel with the $i_0$ in the denominator and we’d be done, so try to get everything in terms of $i_0$.

The basic approach here is to use one of the equations to get rid of one of the variables from all the equations. Then that equation you just used can be discarded; you don’t need it any more. Then you pick another variable to eliminate.

It doesn’t really matter which variable you eliminate first (although it can’t be $R$). My instinct would be not to eliminate $i_0$ because I don’t want the denominator of $R$ to get messy, but it’s fine if you chose to do so. It’ll be less work to pick a variable that we already have a formula for, e.g. equation 3 is a formula for $i_7$. I’ll start with that. There are three other equations that involve $i_7$: equations 4, 7, and 8. Let’s substitute $i_7 = i_4 + i_5$ into each of these: \begin{align*} i_6 + 2i_7 &= i_0 \tag{$4$}\\ i_6 + 2(i_4 + i_5) &= i_0\\ i_6 + 2i_4 + 2i_5 &= i_0\tag{$9$} \end{align*} After simplifying, I’d give this new equation a label since we’ll need it later. Now we won’t need equation 4 again. You can cross it out if you’re working on paper.

Moving on to equation 7: \begin{align*} i_6 &= i_5 + i_7 \tag{$7$}\\ &= i_5 + (i_4 + i_5)\\ i_6&= i_4 + 2i_5\tag{$10$} \end{align*} and equation 8: \begin{align*} R &= \frac{i_1 + i_4 + i_7}{i_0} \tag{$8$}\\ &= \frac{i_1 + i_4 + (i_4 + i_5)}{i_0}\\ R&= \frac{i_1 + 2i_4 + i_5}{i_0} \tag{$11$} \end{align*} Now we no longer need equations 3, 4, 7, or 8. All that information has gone into equations 9, 10, and 11. So far, this is what we have: \begin{align*} i_0 &= 2i_1 + i_2 \tag{$1$}\\ i_1 &= i_3 + i_4 \tag{$2$}\\ i_2 &= i_1 + i_3 \tag{$5$}\\ i_4 &= i_3 + i_5 \tag{$6$}\\ i_6 + 2i_4 + 2i_5 &= i_0\tag{$9$}\\ i_6 &= i_4 + 2i_5\tag{$10$}\\ R &= \frac{i_1 + 2i_4 + i_5}{i_0} \tag{$11$} \end{align*} Let’s pick another variable to eliminate. I’d go for $i_6$ just to make it systematic, working backwards from $i_7$. So let’s substitute equation 10 into 9: \begin{align*} i_6 + 2i_4 + 2i_5 &= i_0\tag{$9$}\\ (i_4 + 2i_5) + 2i_4 + 2i_5 &= i_0\\ 3i_4 + 4i_5 &= i_0 \tag{$12$} \end{align*} We don’t need equations 10 or 9 any more; you can cross them out. I’d like to eliminate $i_5$ now. From equation 6 we can get $i_5 = i_4 - i_3$, then sub this into 11 and 12: \begin{align*} R &= \frac{i_1 + 2i_4 + i_5}{i_0} \tag{$11$}\\ &= \frac{i_1 + 2i_4 + (i_4 - i_3)}{i_0}\\ R &= \frac{i_1 + 3i_4 - i_3}{i_0} \tag{$13$} \end{align*} \begin{align*} 3i_4 + 4i_5 &= i_0 \tag{$12$}\\ 3i_4 + 4(i_4 - i_3) &= i_0\\ 7i_4 - 4i_3 &= i_0 \tag{$14$} \end{align*} Now equations 6, 9, 10, 11, and 12 have been used so we can cross them out. So far we have: \begin{align*} i_0 &= 2i_1 + i_2 \tag{$1$}\\ i_1 &= i_3 + i_4 \tag{$2$}\\ i_2 &= i_1 + i_3 \tag{$5$}\\ R &= \frac{i_1 + 3i_4 - i_3}{i_0} \tag{$13$}\\ 7i_4 - 4i_3 &= i_0 \tag{$14$} \end{align*} From 2 we get $i_4 = i_1 - i_3$ which we can sub into 13 and 14 to get: \begin{align*} R &= \frac{4i_1 - 4i_3}{i_0} \tag{$15$}\\ 7i_1 - 11i_3 &= i_0 \tag{$16$} \end{align*} Now we only have four equations: \begin{align*} i_0 &= 2i_1 + i_2 \tag{$1$}\\ i_2 &= i_1 + i_3 \tag{$5$}\\ R &= \frac{4i_1 - 4i_3}{i_0} \tag{$15$}\\ 7i_1 - 11i_3 &= i_0 \tag{$16$} \end{align*} From 5, $i_3 = i_2 - i_1$ which we can sub into 15 and 16 to get: \begin{align*} R &= \frac{8i_1 - 4i_2}{i_0}\tag{$17$}\\ 18i_1 - 11i_2 &= i_0\tag{$18$} \end{align*} From equation 1, $i_2 = i_0 - 2i_1$ which we can sub into 17 and 18: \begin{align*} R &= \frac{16i_1 - 4i_0}{i_0} \tag{$19$}\\ 18i_1 - 11(i_0 - 2i_1) &= i_0\\ 40i_1 - 11i_0 &= i_0\\ 40i_1 &= 12i_0\\ i_1 &= \frac{3}{10} i_0 \tag{$20$} \end{align*} We’re down to just two equations. Sub 20 into 19: \begin{align*} R &= \frac{16\times\frac{3}{10} i_0 - 4i_0}{i_0}\\ &= \frac{\frac{24}{5} i_0 - 4i_0}{i_0}\\ &= \frac{24}{5} - 4\\ &= \frac{4}{5} \end{align*}