Answer :
Sure! To find an equivalent expression for [tex]\(\frac{81 t^2 + 27 t}{9 t^2 + 9 t}\)[/tex] for [tex]\(t \neq 0\)[/tex], let's simplify the fraction step-by-step.
1. Factor the numerator and the denominator:
The numerator is [tex]\(81 t^2 + 27 t\)[/tex]. We can factor out the greatest common factor, which is [tex]\(27 t\)[/tex]:
[tex]\[ 81 t^2 + 27 t = 27 t (3 t + 1) \][/tex]
The denominator is [tex]\(9 t^2 + 9 t\)[/tex]. We can factor out the greatest common factor, which is [tex]\(9 t\)[/tex]:
[tex]\[ 9 t^2 + 9 t = 9 t (t + 1) \][/tex]
2. Rewrite the fraction using the factored forms:
[tex]\[ \frac{81 t^2 + 27 t}{9 t^2 + 9 t} = \frac{27 t (3 t + 1)}{9 t (t + 1)} \][/tex]
3. Simplify the fraction by canceling out the common terms:
Both the numerator and the denominator contain the term [tex]\(t\)[/tex] (where [tex]\(t \neq 0\)[/tex]) and a constant factor that can be canceled out:
[tex]\[ \frac{27 t (3 t + 1)}{9 t (t + 1)} = \frac{27 (3 t + 1)}{9 (t + 1)} = 3 \cdot \frac{3 t + 1}{t + 1} \][/tex]
Thus, the expression simplifies to:
[tex]\[ \boxed{3 \cdot \frac{3 t + 1}{t + 1}} \][/tex]
So, for [tex]\(t \neq 0\)[/tex], the given expression [tex]\(\frac{81 t^2 + 27 t}{9 t^2 + 9 t}\)[/tex] is equivalent to [tex]\(\boxed{3 \cdot \frac{3 t + 1}{t + 1}}\)[/tex].
1. Factor the numerator and the denominator:
The numerator is [tex]\(81 t^2 + 27 t\)[/tex]. We can factor out the greatest common factor, which is [tex]\(27 t\)[/tex]:
[tex]\[ 81 t^2 + 27 t = 27 t (3 t + 1) \][/tex]
The denominator is [tex]\(9 t^2 + 9 t\)[/tex]. We can factor out the greatest common factor, which is [tex]\(9 t\)[/tex]:
[tex]\[ 9 t^2 + 9 t = 9 t (t + 1) \][/tex]
2. Rewrite the fraction using the factored forms:
[tex]\[ \frac{81 t^2 + 27 t}{9 t^2 + 9 t} = \frac{27 t (3 t + 1)}{9 t (t + 1)} \][/tex]
3. Simplify the fraction by canceling out the common terms:
Both the numerator and the denominator contain the term [tex]\(t\)[/tex] (where [tex]\(t \neq 0\)[/tex]) and a constant factor that can be canceled out:
[tex]\[ \frac{27 t (3 t + 1)}{9 t (t + 1)} = \frac{27 (3 t + 1)}{9 (t + 1)} = 3 \cdot \frac{3 t + 1}{t + 1} \][/tex]
Thus, the expression simplifies to:
[tex]\[ \boxed{3 \cdot \frac{3 t + 1}{t + 1}} \][/tex]
So, for [tex]\(t \neq 0\)[/tex], the given expression [tex]\(\frac{81 t^2 + 27 t}{9 t^2 + 9 t}\)[/tex] is equivalent to [tex]\(\boxed{3 \cdot \frac{3 t + 1}{t + 1}}\)[/tex].
