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 Question from Ken, a student: Stephen owns a gardening supply store that sells two brands of planting soil, each of which contains plan food. Brand A contains 10% plant food, while brand B contains 7% plant food. To accommodate a customer, the store will combine the two brands to make 3 lbs of an 8% plant food mixture. How much of brand A and brand B should be used?

Ken, use algebra to solve this one.

Let a = the amount of Brand A in pounds that Stephen needs and let b = the amount of Brand B in pounds.

Then a + b = 3 pounds, right? The total mixture is 3 lbs.

Stephen wants an 8% mixture, so 8% of the 3 lbs is plant food. Let's write that as (3 x 8%). That's the number of pounds of plant food in the final mixture.

The amount of plant food in the Brand A is 10%, so the amount of pounds of plant food he has in "a" pounds of it is (a x 10%).

Similarly, the amount from Brand B will be (b x 7%).

But if you add these two together, you get the mixture, so:
(a x 10%) + (b x 7%) = (3 x 8%).

Turn this into decimals and simplify:

0.10a + 0.07b = 0.24

So we have two equations in two unknowns. This is what we call a "system of equations" or "simultaneous equations". An easy way to solve it is by using "substitution".

Since a + b = 3, then we know that a = 3 - b. But that means that wherever we see "a", we can replace it with "b-3", because they equal each other.

So 0.10a + 0.07b = 0.24 becomes 0.10(b-3) + 0.07b = 0.24.

Now you can solve this for b and once you know that, you can solve a = 3 - b and you have your answer.

Cheers,
Stephen La Rocque.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.