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. .
Let x = number of ml of the 5% solution.
Let y = number of ml of the 2% solution.
Then: .x + y .= .60 .
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  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 .
Solve the system formed by  and .