Shari's explanation is good. The equation is balanced. I've replicated her work in the form of a spreadsheet showing the individual element count. I use this general approach on difficult equations and find it helps with the solution while minimizing errors. This response is long, but it works for even complex equations and minimizes errors.
Look at the table below. I start by write the equation on top, normally without the coefficients if they are unknown (use pencil). I then write each element in a column on the left (Ca, O, H, P) and then count the number of times that element appears in a single molecule of each compound. In this equation, this is what my initial chart would look like. I arbitrarily assigned "1" for each coefficient, knowing that I need at least 1 of every molecule. That is the minimum.
Look over the table of numbers and note that the number of times that element appears in one molecule of each of the reactants and products. For example, Oxygen (O) appears twice in Ca(OH)2 and four times in H3PO4. I write each number directly below its compound, and then sum each element on both the reactant and product sides. I can see that the equation is not balanced for oxygen, since it appears 6 times on the left (reactants) and 7 times on the right (products). In fact, none of the elements are balanced.
1 Ca(OH)2 + 1 H3PO4 ==> 1 Ca3(PO4)2 + 1 H2O
Ca 1 + 0 = 1 3 + 0 = 3
O 2 + 4 = 6 8 + 1 = 9
H 2 + 3 = 7 0 + 2 = 2
P 0 + 1 = 1 2 + 0 = 2
The next step is to change one of the coefficients to help balance the numbers. I usually start with the most complex formula and assume it has a coefficient of "1" and work from there. In this case, Ill choose Ca3(PO4)2. There has to be at least one molecule, so I pencil in the "1." I then go through each element and make changes to the reactants to try and balance each element. I immediately know I need to start with at least 3 Ca(OH)2 molecules, since it is the only source of Ca, and it only has 1 Ca per molecule. Pencil in a "3" for Ca(OH)2. The Calciums are now balanced. I then change the element count for this molecule, which results in 6 H's and 6 O's.
One molecule of Ca3(PO)4 also contains 8 Oxygen, zero Hydrogen, and 2 Phosphorus atoms, as indicated. To get two Phosphorus atoms, I need to add two molecules of H3PO4, so I pencil in a "2" for that molecule. That results in a new element count for that molecule, as you can see below. Now we have 14 O's, 12 H's, and 2 P's. This means that the Ca and P atoms are now balanced, but we need to correct the H's and O's. We have a toatl of 12 H and 14 O atoms on the reactant side. They have to exist since we assumed just one molecule of Ca3(PO4)2 and this requires the minimum of 3 Ca(OH)2 and 2 H3PO4 molecules.
On the product side, the only place hydrogen can go is into water, H2O. Since we have a total of 12 hydrogens, and water has 2 H's per molecule, we need 6 molecules of water to balance the H's.
This leave O. We have 14 O atoms from the reactants. The Ca3(PO)4 takes 8 of them, leaving 6. But now that we have assigned a 6 to the H2O, we can now account for the O's (14 on both sides).
We're done! As you can see, the element counts are the same on both sides of the equation. Not all equations works are straightforward as this, as you will discover. You can often get to near the end and find that something isn't working. This often occurs when working with molecules that have even numbers of elements on one side, but odd numbers on the other. Painful, but simple to fix. Go back to the original molecule you set to "1," Ca3(PO4)2, and increase it to "2." By iterating back and forth, you will soon see a way of balancing both sides of the equation. It can be difficult sometimes, but just keeping increasing one of the molecule coefficient and rebalancing, you will eventually see a solution. It is easy sometimes to increase by too much, and you will note that the coefficients have a common factor, such as 2. Just divide by the common factor to get the lowest whole number coefficients.
For example, if I started this process by assigning a "2" to the Ca3(PO)4, my final equation would be
6Ca(OH)2 + 4H3PO4 ==> 2Ca3(PO4)2 + 12H2O. This is divisible by 2. so the correct equation is given by dividing each coefficient by "2."
3 Ca(OH)2 + 2 H3PO4 ==> 1 Ca3(PO4)2 + 6 H2O
Ca 3 + 0 = 3 3 + 0 = 3
O 6 + 8 = 14 8 + 6 = 14
H 6 + 6 = 12 0 + 12 = 12
P 0 + 2 = 2 2 + 0 = 2
reactant total product total
Sorry this is so long. Use an Excel spreadsheet for difficult problems and you'll have fun making the changes while everyone else complains. :)