Question

Identify the system of equations used to solve the following problem. How much of a 60% saline solution would you need to add to pure water to create 10 L of a 50% saline solution?

Answer 1

The first step is to choose variables.
You need to use some amount of 60% saline solution and some amount of water.
Let the amount of saline solution be x, and let the amount of water be y.

You are creating 10 L of solution.
Using x and y as the amounts you will be mixing, and 10 as the end result, can you come up with an equation for the amounts of solutions and water?

Answer 2

You will use x amount of 60% solution and y amount of water.
Altogether, you will make 10 L of 50% solution.

The first equation is:
x + y = 10

Do you understand it so far?

Answer 3

yes, im just terrible with %'s

Answer 4

Ok. The first equation just dealt with the amounts of the 60% solution and water needed to produce 10 L of 50% solution.
Now we need to deal with the salt content.

First, look at the final result.
10 L of 50% solution contains 50% * 10L = 0.50 * 10L = 5 L of salt.

Answer 5

The final solution will will be 10 L of 50% solution which contains 5 L of salt.

The first solution is 60% solution.
We will use x liters of it.
How much salt is there in x liters of 60% solution?
What is 60% of x written as an algebraic expression using x?

Answer 6

60% of x is 60% * x = 0.6 * x = 0.6x

The amount of salt in x liters of 60% solution is 0.6x.
Water contributes no slat, so we get 0% * y = 0
The total salt in the end solution is 5 L
Now we have all we need to write the second equation

0.6x + 0y = 5

or simply

0.6x = 5

That means the system of equations is:

x + y = 10
0.6x = 5