Answer
The jeweler needs to use $\text{8 ounces}$ of $16 \%$ gold content and $\text{24 ounces}$ of $28 \%$ gold content.
Work Step by Step
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.
