Question

Liters of solution
Percent​ (as a​ decimal)
Liters of pure dye
x
0.20
  
0.20 x
y
0.59
  
0.59 y
260
0.50
  
130
​(Simplify your answers. Use integers or decimals for any numbers in the​ expression.)
Part 2
To get 260 L of a 50​% ​solution,
  
enter your response here liters of the 20​% solution and
  
enter your response here liters of the 59​% solution are needed.

Answer 1

To solve for \( x \) and \( y \) in the context of the problem, we want to create a system of equations based on the total volume of the solution and the total amount of pure dye.

1. We know that:
- The total volume of the solution:
\[
x + y = 260
\]
- The total amount of pure dye:
\[
0.20x + 0.59y = 0.50 \times 260
\]

2. Calculate the right-hand side of the second equation:
\[
0.50 \times 260 = 130
\]
So our second equation becomes:
\[
0.20x + 0.59y = 130
\]

Now we have the following system of equations:

\[
\begin{align*}
1. & \quad x + y = 260 \\
2. & \quad 0.20x + 0.59y = 130
\end{align*}
\]

3. We can solve the first equation for \( y \):
\[
y = 260 - x
\]

4. Substitute \( y \) back into the second equation:
\[
0.20x + 0.59(260 - x) = 130
\]

5. Distributing \( 0.59 \):
\[
0.20x + 153.4 - 0.59x = 130
\]

6. Combine like terms:
\[
-0.39x + 153.4 = 130
\]

7. Subtract \( 153.4 \) from both sides:
\[
-0.39x = 130 - 153.4
\]
\[
-0.39x = -23.4
\]

8. Divide both sides by \( -0.39 \):
\[
x = \frac{-23.4}{-0.39} \approx 60
\]

9. Substitute \( x = 60 \) back into the equation for \( y \):
\[
y = 260 - 60 = 200
\]

### Final Answer
To create a 260 L solution that is 50% dye:
- You need **60 liters** of the **20% solution**.
- You need **200 liters** of the **59% solution**.