What do you think of this approach to solving mixture algebra problems?

There is a category of algebra problem that involves combining two rates to get a third rate. It might involve mixing chemicals with different concentrations or selling two products with different prices or traveling at two different speeds.

There is a single equation involving weighted average that can be used to solve these problems. Let me show the steps to get to this equation. It may look a little scary at first, but what we end up with is very simple.

Suppose we want to solve a problem involving traveling at two different speeds v1 and v2 for corresponding times t1 and t2 for total distance of d.

We have v1 t1 + v2 t2 = d.
Divide both sides by t1 + t2.
t1/(t1+t2) v1 + t2/(t1+t2) v2=d/(t1+t2)

Setting t1/(t1+t2) = w1 and t2/(t1+t2) to w2, we get the universally applicable equation of:
w1 v1 + w2 v2 = v3.
w1 and w2 are two weighting factors representing the portion of the total time traveling at the two speeds. w1 and w2 are fractions between 0 and 1 which add to 1. v3 is just the average overall speed.

We can generalize the equation to:
w1 r1 + w2 r2 = r3 where the r terms stand for rates. Using the fact that w1 + w2 = 1, we can write this using one variable.
w1 r1 + (1-w1)r2 = r3

Here is a video from Khan Academy that shows the standard way of solving a rate problem.

Here is my approach.
We have solutions of 25% concentration and 10% concentration being combined to give a 15% solution.
We get
.25 w + .1(1 – w) = .15
.25w – .1 w + .1 = .15
.15w = .05
w = ⅓
The 25% solution is ⅓ of the total. Since there are 50 ounces of it, the total is 50 / (⅓) = 150 ounces and there are 150 – 50 = 100 ounces of the 10% solution.

Another way of finishing the problem is to say that since the 25% solution is ⅓ of the total, the 10% solution is ⅔ of the total. There is therefore (⅔) / (⅓) * 50 ounces = 100 ounces.