Given the equations:
[tex]\[
\begin{array}{l}
y = \frac{4}{3} + 2 \\
3y = x^{-3}
\end{array}
\][/tex]

Format the following question or task so that it is easier to read. Fix any grammar or spelling errors. Remove phrases that are not part of the question. Do not remove or change LaTeX formatting. Do not change or remove [tex] [/tex] tags. Do not translate the question or any part of the question. If the question is nonsense, rewrite it so that it makes sense.



Answer :

Let's break down the given expressions step-by-step.

1. Evaluating [tex]\( y \)[/tex]:
[tex]\[ y = \frac{4}{3} + 2 \][/tex]

To simplify this, first evaluate [tex]\(\frac{4}{3}\)[/tex]:
[tex]\[ \frac{4}{3} \approx 1.3333333333333333 \][/tex]

Next, add 2 to this value:
[tex]\[ y = 1.3333333333333333 + 2 = 3.333333333333333 \][/tex]

So, we have:
[tex]\[ y \approx 3.333333333333333 \][/tex]

2. Relating [tex]\( y \)[/tex] to [tex]\( x \)[/tex]:
[tex]\[ 3y = x^{-3} \][/tex]

Substitute the value of [tex]\( y \)[/tex] that we found into the equation:
[tex]\[ 3 \times 3.333333333333333 \approx 10.0 \][/tex]

Hence,
[tex]\[ x^{-3} \approx 10.0 \][/tex]

In summary,
- The value of [tex]\( y \)[/tex] is approximately [tex]\( 3.333333333333333 \)[/tex].
- The expression [tex]\( 3y \)[/tex] simplifies to approximately [tex]\( 10.0 \)[/tex], thus [tex]\( x^{-3} \approx 10.0 \)[/tex].