Hello, zasi!

We already know that the final mixure is 60 ml of 3% boric acid.A chemistry class has a bottle of 5% boric acid solution and a bottle of 2% boric acid solution.

It need 60 milliliters of a 3% boric acid solution for an experiment.

How much of each solution is needed?

. . That is, it contains: .0.03 × 60 .= .1.8 ml of boric acid. .[1]

Let x = number of ml of the 5% solution.

Let y = number of ml of the 2% solution.

Then: .x + y .= .60 .[2]

The x ml of 5% boric acid contains: 0.05x ml of boric acid.

The y ml of 2% boric acid contains: 0.02y ml of boric acid.

. . Hence, the mixture will contain: .0.05x + 0.02y ml of boric acid.

But [1] says the mixture contains 1.8 ml of boric acid.

There is our second equation! . 0.05x + 0.02y .= .1.8

Multiply by 100: .5x + 2y .= .180 .[3]

Solve the system formed by [2] and [3].