Does the conservation of linear momentum for a ball hitting a wall and changing direction? The conservation of linear momentum essentially says:
$$\sum m \vec{v_1}=\sum m \vec{v_2}$$
But if I take the absolute value of both sides, and drop summation, is it also equivalent to say that:
$$mv = mv$$
Here is where my confusion would take place:
If a ball is travelling 45 degrees to the horizontal at 3 m/s and collides with a wall (that doesn't move at all), then travels in the +x axis, what speed is it moving in?
I know $mv_1=mv_2$ for the ball, because the velocity of the wall is 0 in both cases.
So is it correct to say that momentum is only conserved in the x axis, making the velocity of the ball in the x axis $$m * 3 * \sin45 = m * v_2 \implies v = 3 * \sin45$$
Or is it rather $$m * 3 = m * v \implies v = 3$$
 A: The steps in the OP's suggested solution are deeply flawed.
First:  you cannot take the absolute values of the two sides of a vector equation. Momentum has both size and direction, and both must be taken into account when doing an addition.  Would you examine your bank statement for the month, while treating the deposits and withdrawals differently.  Is a credit balance the same as a debit balance?
Secondly:  momentum is only conserved for a system in the absence of any external force.  if you treat the ball as your system, the force exerted by the wall on the ball constitutes an external force for this system, and thus conservation of momentum  for the ball simply does not apply.
What if you consider the ball plus wall as the system?  Then the forces are internal, and conservation of momentum applies.  However, you must also assume that the wall recoils, however slightly, from the collision with the ball.  If the ground exerts any horizontal force on the wall, then you have an external force again, and no conservation of momentum.
The simplest method is to assume a collision between a ball, mass $m_b$ and wall, mass $m_w$.  Allow them to collide elastically, and conserve both kinetic energy and linear momentum. Keep track of the positive and negative signs on the velocities.  Finally, see what happens to the final velocity of the ball, as the mass of the wall tends to infinity.
A: 
But if I take the absolute value of both sides, and drop summation

It might be dangerous just to "drop" summation. You have to include all particles moving before and all moving after collision. In your case of only a ball and a wall it would reduce to:
$$\sum m v_{1x}=\sum m v_{2x} \Rightarrow\\
m_{ball} v_{1x,ball}+m_{wall} v_{1x,wall}=m_{ball} v_{2x,ball}+m_{wall} v_{2x,wall} \Rightarrow\\
m_{ball} v_{1x,ball}=m_{ball} v_{2x,ball}$$
Which of course still reduces to your expression since the wall has no speed in any case.

So is it correct to say that momentum is only conserved in the x axis, making the velocity of the ball in the x axis

Momentum is conserved in all directions. As you can see I have added a small $x$ to indicate my chosen component. The law of conservation of momentum is very easily applied if we simply look at each component by itself. This above equation is for the x-direction. For the y-direction the equation would be:
$$\sum m v_{1y}=\sum m v_{2y} \Rightarrow\\
m_{ball} v_{1y,ball}+m_{wall} v_{1y,wall}=m_{ball} v_{2y,ball}+m_{wall} v_{2y,wall} \Rightarrow\\
m_{ball} v_{1y,ball}=m_{ball} v_{2y,ball}$$
Perfect elastic collision like this with a rigid im-movable wall will not change anything in the speed of the ball in any direction (it only changes direction - but these are the magnetudes).
