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The product of two numbers is 32. The first number, [tex]\( x \)[/tex], is one-half of the second number, [tex]\( y \)[/tex].

A. [tex]\(\begin{cases} x y = 32 \\ x = \frac{1}{2} y \end{cases}\)[/tex]

B. [tex]\(\begin{cases} x y = 32 \\ x = y - \frac{1}{2} \end{cases}\)[/tex]

C. [tex]\(\begin{cases} x + y = 32 \\ x = y - \frac{1}{2} \end{cases}\)[/tex]

D. [tex]\(\begin{cases} x + y = 32 \\ x = \frac{1}{2} y \end{cases}\)[/tex]



Answer :

GinnyAnswer
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]

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