## Precalculus (6th Edition) Blitzer

The wine company should mix $100\text{ gallons}$ of $5\%$ California wine with $\text{100 gallons}$ of $9\%$ French wine.
Let $x$ represent the gallons of $5\text{ percent}$ California wine and $y$ represent the gallons of $9\text{ percent}$ French wine. The amount of pure wine in each solution is found by multiplying the amount of gallons by the concentration rate. There are two unknown quantities; therefore, a system of two independent equations with $x$ and $y$ is set up. Amount of wine 5 percent mixture + Amount of wine 9 percent mixture=Amount of wine 7 percent mixture And: Amount of pure wine in 5 percent mixture + Amount of pure wine in 9 percent mixture = Amount of pure wine in 7 percent mixture Consider the equation, \begin{align} & x+y=200 \\ & x=200-y \end{align} …… (1) And $0.05x+0.09y=14$ …… (2) Substitute $200-y$for $x$ in equation $\left( 2 \right)$ to get, \begin{align} & 0.05\left( 200-y \right)+0.09y=14 \\ & 10-0.05y+0.09y=14 \\ & 0.04y=4 \end{align} Divide the above equation by $0.04$ to get, \begin{align} & \frac{0.04y}{0.04}=\frac{4}{0.04} \\ & y=100 \end{align} Substitute $100$ for $y$ in the equation $\left( 1 \right)$ to get, \begin{align} & x=200-100 \\ & x=100 \\ \end{align} Check: $\left( 100,100 \right)$ Put $x=100$ and $y=100$ in the equation (1), \begin{align} x+y=200 & \\ 100+100\overset{?}{\mathop{=}}\,200 & \\ 200=200 & \\ \end{align} And put $x=100$and $y=100$ in the equation (2), \begin{align} 0.05\left( 100 \right)+0.09\left( 100 \right)\overset{?}{\mathop{=}}\,14 & \\ 5+9\overset{?}{\mathop{=}}\,14 & \\ 14=14 & \\ \end{align} The ordered pair $\left( 100,100 \right)$ satisfies both equations. Hence, the wine company should mix $100\text{ gallons}$ of $5\%$ California wine with $\text{100 gallons}$ of $9\%$ French wine.