## Precalculus (6th Edition) Blitzer

The jeweler needs to use $\text{8 ounces}$ of $16 \%$ gold content and $\text{24 ounces}$ of $28 \%$ gold content.
Let $x$ represent the ounces of $16 \%$ gold content. Let $y$ represent the ounces of $28 \%$ gold content. The amount of pure gold in each solution is found by multiplying the amount of ounces by the concentration rate. This information can be organized in a table. There are two unknown quantities; therefore, a system of two independent equations relating $x$ and $y$ is set up. Amount of $16 \%$ gold + Amount of $28 \%$ gold = Amount of $25 \%$ gold And: Amount of pure $16 \%$ gold + Amount of pure $28 \%$ gold = Amount of pure $25 \%$ gold Consider the equation, \begin{align} & x+y=32 \\ & x=32-y \end{align} …… (1) And $0.16x+0.28y=8$ …… (2) Substitute $32-y$for $x$ in the equation $\left( 2 \right)$ to get, \begin{align} & 0.16\left( 32-y \right)+0.28y=8 \\ & 5.12-0.16y+0.28y=8 \\ & 0.12y=2.88 \end{align} Divide the above equation by $0.12$ to get, \begin{align} & \frac{0.12y}{0.12}=\frac{8}{0.12} \\ & y=24 \end{align} Substitute $24$ for $y$ in the equation $\left( 1 \right)$ to get, \begin{align} & x=32-24 \\ & x=8 \\ \end{align} Check: $\left( 8,24 \right)$ Put $x=8$ and $y=24$ in the equation (1), \begin{align} x+y\overset{?}{\mathop{=}}\,32 & \\ \left( 8 \right)+\left( 24 \right)\overset{?}{\mathop{=}}\,32 & \\ 32=32 & \\ \end{align} And put $x=8$ and $y=24$ in the equation (2), \begin{align} 0.16\left( 8 \right)+0.28\left( 24 \right)\overset{?}{\mathop{=}}\,8 & \\ 1.28+6.72\overset{?}{\mathop{=}}\,8 & \\ 8=8 & \\ \end{align} The ordered pair $\left( 8,24 \right)$ satisfies both equations. Hence, the jeweler needs to use $\text{8 ounces}$ of $16 \%$ gold content and $\text{24 ounces}$ of $28 \%$ gold content.