A chemist has one solution containing 30% insecticide and another solution containing 50% insecticide. How much of each solution should the chemist mix to get 200 L of a 42% insecticide?

My teacher explained how to set up the problem:
3x+5y=84
x+y=200
what I don't get is why the first equation is equal to 84, could someone could explain please?

This really enters the boundaries of Chemistry, but I will explain it in simpler terms.

Since solution one contains 30% insecticide, this means 70% is not insecticide but of some other matter.

This applies to solution 2, 50% insecticide and 50% of some other matter.

It tells you to find 42% insecticide therefore when you combine the solution, in which your teacher kindly tells you that x+y=200, you would only want 42% insecticide.

(.42)(200)= 84

Just noticed a typo in my conclusion statement,

should have read:
The chemist will need 80 L of the 30%, and 120 L of the 50% solution.

notice I did use 120L in my check

amount of 30% solution needed ---- x L

amount of 50% solution needed ---- 200-x L

.3x + .5(200-x) = .42(200)
.3x + 100 - .5x = 84
-.2x = -16
x = -16/-.2 = 80

The chemist will need 80 L of the 30%, and 20 L of the 50% solution.
check:
.3(80) + .5(120) = 84

or , using the 2 equations in 2 unknowns ...

3x+5y=84
x+y=200 ---> y = 200-x , note that is how I started
sub that into the 1st
3x + 5(200-x) = 84
3x + 1000 - 5x = 84
-2x = -916
x = 458

this answer makes no sense,
because your first equation makes no sense.
it should have been
.3x + .5y = 84

either your teacher gave you the wrong first equation, or
you copied it incorrectly.

Yes, I was wondering the same thing too, the first equation indeed must be .3x + .5y = 84. This way it will make much more sense.

Sure! I'd be happy to explain why the first equation is set equal to 84 in this problem.

To solve this problem, we are looking for the amounts of the 30% and 50% insecticide solutions that need to be mixed to obtain a 42% insecticide solution. Let's break down the equation step by step.

Let x be the amount (in liters) of the 30% insecticide solution.
Let y be the amount (in liters) of the 50% insecticide solution.

The insecticide content in the 30% solution is 30/100 = 0.3 (or 30%) and in the 50% solution is 50/100 = 0.5 (or 50%).

Now, let's calculate the insecticide amount in each solution. The amount of insecticide in the 30% solution is 0.3x, and the amount of insecticide in the 50% solution is 0.5y.

To obtain a 42% insecticide solution, the total amount of insecticide in the mixture should be 42% of the total volume of the mixture.

So, the equation for the total amount of insecticide in the mixture is:

0.3x + 0.5y = 0.42 * 200

Simplifying the equation:

0.3x + 0.5y = 84

Here, 84 represents 42% of 200 (since 42% of 200 is 84).

Therefore, the reason why the first equation is equal to 84 is that we are setting the total amount of insecticide in the mixture (computed from the amounts of the 30% and 50% solutions) equal to 42% of the total volume (200 L) of the mixture.