substitution/elimination method help

Here's mine:

$$24I_A - 15I_B + 0I_C = 5 \space ------------> (1)$$ $$15I_A + 12I_B -20I_C = 5 \space ------------> (2) $$ $$0I_A + 20I_B - 27I_C = -5 \space -- --------> (3) $$

Now using the Elimination Method:

Looking at equations (1) and (2):

$$24I_A - 15I_B + 0I_C = 5 \space ---------> (1)$$ $$15I_A + 12I_B -20I_C = 5\space ----------> (2) $$

Multiply the equation (2) by $\frac{24}{15}$
$$\frac{24}{15} (15I_A - 12I_B -20I_C = 5)$$ $$24I_A +\frac{96}{5}I_B -32I_C = 8$$

We can now subtract equation (1) and (2) to eliminate $I_A$

$$24I_A - 15I_B + 0I_C = 5 \space ---------> (1)$$ $$-(24I_A +\frac{96}{5}I_B -32I_C = 8) \space ---------> (2)$$
which is equal to $$-\frac{171}{5} I_B + 32I_C = -3 \space--------->(4)$$

Now looking at equations (4) and (3):
$$-\frac{171}{5} I_B + 32I_C = -3 \space--------->(4)$$ $$0I_A + 20I_B - 27I_C = -5 \space -- --------> (3) $$

We get the $I_C$ in equation (3):
$$0I_A + 20I_B - 27I_C = -5 \space ---------> (3) $$ $$I_C = \frac{5+20I_B}{27}$$

Then substitute it to equation (4):

$$-\frac{171}{5} I_B + 32I_C = -3 \space--------->(4)$$ $$-\frac{171}{5} I_B + 32\left(\frac{5+20I_B}{27}\right) = -3 \space--------->(4)$$ $$-\frac{171}{5} I_B + \frac{160+640I_B}{27} = -3 \space--------->(4)$$ $$-\frac{171}{5} I_B + \frac{160}{27} + \frac{640}{27} I_B = - 3$$

Now getting the $I_B.$ Turns out, $I_B = \frac{1205}{1417}$

Now getting the $I_C$:
$$I_C = \frac{5+20I_B}{27}$$ $$I_C = \frac{5+20 \left(\frac{1205}{1417}\right)}{27}$$ $$I_C = \frac{1155}{1417}$$

Now, we can get $I_A:$ Looking at equation (1):

$$24I_A - 15I_B + 0I_C = 5$$ $$24I_A - 15\left(\frac{1205}{1417}\right) + 0I_C = 5$$ $$I_A = \frac{3145}{4215}$$

Therefore, $I_A = \frac{3145}{4215},$ $I_B = \frac{1205}{1417},$ $I_C = \frac{1155}{1417}$