Question

A chemist is weakening an 80% solution by adding a 40% solution. How much of each should he use to get 100 milliliters of a 50% solution? the chemist’s problem is modeled by the system: {0.80x+0.40y=0.50⋅100 (x+y=100 where x is the number of milliliters of 80% solution and y is the number of milliliters of 40% solution.

Answer 1

X- 25 y- 75because we have to make a equation

Answer 2

Answer: The chemist needs to mix 25 milliliters of the 80% solution with 75 milliliters of the 40% solution to get 100 milliliters of a 50% solution.

Explanation:

Let x be the number of milliliters of 80% solution needed, and y be the number of milliliters of 40% solution needed.

From the first equation, we can write:

0.80x + 0.40y = 0.50(100)

Simplifying this equation, we get:

0.80x + 0.40y = 50

Dividing both sides by 0.20, we get:

4x + 2y = 250

From the second equation, we know that:

x + y = 100

Solving this equation for y, we get:

y = 100 - x

Substituting this expression for y into the first equation, we get:

4x + 2(100 - x) = 250

Simplifying this equation, we get:

2x + 200 = 250

Subtracting 200 from both sides, we get:

2x = 50

Dividing both sides by 2, we get:

x = 25

Substituting this value for x into the equation y = 100 - x, we get:

y = 75

Therefore, the chemist needs to mix 25 milliliters of the 80% solution with 75 milliliters of the 40% solution to get 100 milliliters of a 50% solution.