To solve this problem, we will use the given system of equations:
[tex]\[
\left\{
\begin{array}{l}
x \cdot y = 32 \\
x = \frac{1}{2} y
\end{array}
\right.
\][/tex]
1. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
From the second equation, [tex]\( x = \frac{1}{2} y \)[/tex], we can express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[
y = 2x
\][/tex]
2. Substitute [tex]\( y \)[/tex] in the first equation:
We substitute [tex]\( y = 2x \)[/tex] into the first equation, [tex]\( x \cdot y = 32 \)[/tex]:
[tex]\[
x \cdot (2x) = 32
\][/tex]
3. Simplify and solve for [tex]\( x \)[/tex]:
Simplify the equation:
[tex]\[
2x^2 = 32
\][/tex]
Divide both sides by 2:
[tex]\[
x^2 = 16
\][/tex]
Take the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[
x = \sqrt{16}
\][/tex]
Since we are looking for a positive solution given the context:
[tex]\[
x = 4
\][/tex]
4. Solve for [tex]\( y \)[/tex]:
Use [tex]\( y = 2x \)[/tex] to find [tex]\( y \)[/tex]:
[tex]\[
y = 2 \cdot 4 = 8
\][/tex]
So, the two numbers are:
[tex]\[
x = 4 \quad \text{and} \quad y = 8
\][/tex]
