Nodal analysis for resistive circuit
When performing nodal analysis you want to solve for the node voltages, not the branch currents. After you know the node voltages you can use Ohm's Law to calculate the branch current in each resistor.

I'll redraw your circuit as shown in Figure 1. Note that there are two nodes whose voltages are unknown: \$V_A\$ and \$V_B\$. Therefore, you'll need to derive two equations to solve for these two unknown voltages. Use Kirchhoff's Current Law to derive these two equations&mdash;i.e., the sum of the currents at each node is zero.



Figure 1.

Recall from Ohm's Law that

$$
I_R = \frac{V_R}{R}
$$

Use Kirchhoff's Current Law and nodal analysis to derive an equation at node \$V_A\$:

$$
\frac{V_A-V_B}{R2} + \frac{V_A-V_B}{R3} + \frac{V_A-0}{R4} + \frac{V_A-0}{R5}=0
$$

Use Kirchhoff's Current Law and nodal analysis to derive an equation at node \$V_B\$:

$$
\frac{V_B-5}{R1} + \frac{V_B-V_A}{R2} + \frac{V_B-V_A}{R3} = 0
$$

(NB: Pay close attention to how I defined the voltages across the resistors in each equation. When performing nodal analysis at node \$V_A\$ I always subtract the neighboring node voltage from \$V_A\$.  Likewise, when performing nodal analysis at \$V_B\$ I always subtract the neighboring node voltage from \$V_B\$. The advantage of this method is that you do not need to know in advance the actual direction of current flow in each resistor.)

Now use whatever method you like to solve for \$V_A\$ and \$V_B\$. Once you know those two node voltages you can use Ohm's Law to solve for the current in each resistor.

HINT

This analysis can be simplified if you realize that resistors R2 and R3 are connected in parallel and therefore they can be replaced by a single resistor Rx whose value is the parallel combination of R2 and R3.  Likewise, resistors R4 and R5 are connected in parallel and they can be replaced by a single resistor Ry whose value is the parallel combination of R4 and R5.  Your two equations are now

$$
\frac{V_A-V_B}{Rx} + \frac{V_A-0}{Ry} = 0 \\
\frac{V_B-5}{R1} + \frac{V_B-V_A}{Rx} = 0
$$