## Precalculus (6th Edition) Blitzer

We have $10$ milliliters of solution with $34\%$ of concentration, and $90$ milliliters of solution having $4\%$ concentration.
Let us assume x milliliters of solution 1 must be mixed with y milliliters of solution 2 in order to obtain 100 milliliters solution with 7% of silver nitrate concentration. So, the equation becomes $x+y=100$ Now, consider the concentration of silver nitrate in solution 1 as 34% and in solution 2 as 4%; And show the total amount of silver nitrate in the mixture as: $\frac{34}{100}x+\frac{4}{100}y$ We get that the mixture has 7% of silver nitrate, hence the equation becomes: \begin{align} & \frac{34}{100}x+\frac{4}{100}y=\frac{7}{100}\times 100 \\ & 34x+4y=700 \end{align} From the equation, we have: $x+y=100$, $y=100-x$ Put $y=100-x$ in the equation $34x+4y=700$ as given below: \begin{align} & 34x+4\left( 100-x \right)=700 \\ & 34x+400-4x=700 \\ & 30x=300 \\ & x=10 \end{align} Therefore, use $y=100-x$, to obtain the value of y. \begin{align} & y=100-10 \\ & =90 \end{align} Hence, we have $10$ milliliters of solution with $34\%$ of concentration, and $90$ milliliters of solution having $4\%$ concentration.