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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.

1 answer

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.