I'm having issues solving this problem, with the CCVS. Here is what I have so far:
$$-(24 - V_1) / 250 - (60I_b) / 150 = V_1 / 50$$
Which simplifies down to: \$-28.8 = 120I_b + 4.8V_1.\$
But without a second equation, I don't see how I can solve this. Thoughts?
It looks as though you are trying to use KCL and that the middle node at the top of your circuit is being called \$V_1\$ in your first equation in your question. (Potentially confusing, as you also have a labeled voltage source on the left with the same name.)
If so, then it is really:
$$\frac{24\:\text{V}-V_1}{250\:\Omega}+\frac{60\cdot I_b-V_1}{150\:\Omega}=\frac{V_1}{50\:\Omega}$$
This is not too far away from what you wrote. But I think you missed something in that second term on the left side and, perhaps, also got a sign wrong.
The above KCL I just wrote should be correct, if I'm not mistaken.
Also, you know that \$I_b=\frac{24\:\text{V}-V_1}{250\:\Omega}\$. So all you need to do is substitute that into the above equation, for \$I_b\$, and solve for the only remaining variable, \$V_1\$.
A properly arranged circuit for LTspice may be drawn out like this:
The above will solve out in LTspice and provide exactly the same value that you should get using the substitution I suggested.
Given this is essentially just a DC circuit for which you are trying to find the operating point, containing only linear elements and each source can be reduced to a simple set of resistor dividers, you can use the superposition theory to solve these circuits.
Essentially you analyse each voltage or current source independently, while setting all other voltage sources to 0 and current sources to open circuit. In your case you have only two voltage sources, so you need to solve two circuits:
Start by finding the current through each of the resistors in the above two circuits. The first circuit will give you exact values. The second circuit will give you values in terms of \$I_b\$ due to your variable source.
Once calculated, you can work out the actual current through each of the resistors in the original circuit by adding together the currents from the two superimposed circuits. Note that when adding these currents, make sure to get the sign correct. I've marked the polarities on the two circuits. You can pick which circuit to use as the reference, just be consistent across all resistors. Given that \$I_b\$ and \$I_{R1}\$ in the original circuit are the same, I would use the first of the two sub-circuits as being all positive currents.
This should give you three equations, one for each of \$I_{R1}\$, \$I_{R2}\$, and \$I_{R3}\$:
$$ I_{R1} = I_{R1\mathrm{(top)}} - I_{R1\mathrm{(bot)}}\\ I_{R2} = I_{R2\mathrm{(top)}} + I_{R2\mathrm{(bot)}}\\ I_{R3} = I_{R3\mathrm{(top)}} - I_{R3\mathrm{(bot)}} $$
You know through KCL, that all currents through a node add up to zero, so looking at the top middle node, you can say that:
$$ I_{R1} = I_{R2} + I_{R3} $$
By the same logic, if you take the node between \$V1\$ and \$R1\$, you can also say that:
$$ I_{R1} = I_b $$
Once you plug in all the values, this should ultimately give you a single equations in terms of \$I_b\$, simply solve to find the value for this current. Once you know \$I_b\$ you can work out the voltage source operating point and the voltages across each resistor.
