Let \( x \) be the mL of the 30% solution and \( y \) be the mL of the 10% solution

February 22, 2026

$Let \( x \) be the mL of the 30% solution and \( y \) be the mL of the 10% solution$

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Substituting known values: \( 155 = \frac{10}{2} (2 \times 5 + 9d) \).
Simplifying: \( 155 = 5 (10 + 9d) \) → \( 155 = 50 + 45d \) → \( 105 = 45d \) → \( d = \frac{105}{45} = \frac{7}{3} \).
#### \(\frac{7}{3}\)1. A laboratory experiment requires a solution with 150 mL of a 20% saline solution. The technician has a 30% saline solution and a 10% saline solution available. How many mL of each solution should be mixed to achieve the desired concentration?
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