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Which value of [tex]n[/tex] makes this equation true?

[tex] \frac{3n+3}{5} = \frac{5n-1}{9} [/tex]

A. [tex]n = -16[/tex]
B. [tex]n = -2[/tex]
C. [tex]n = 2[/tex]
D. [tex]n = 16[/tex]

Sagot :

To solve the equation [tex]\(\frac{3n + 3}{5} = \frac{5n - 1}{9}\)[/tex], we need to find the value of [tex]\(n\)[/tex] that satisfies this equation. Here's a detailed, step-by-step solution:

1. First, clear the fractions by finding a common denominator.
- The common denominator for 5 and 9 is 45.
- Multiply both sides of the equation by 45 to eliminate the denominators.

[tex]\[ \begin{aligned} 45 \cdot \frac{3n + 3}{5} & = 45 \cdot \frac{5n - 1}{9} \\ 45 \cdot \frac{3n + 3}{5} & = 45 \cdot \frac{5n - 1}{9} \end{aligned} \][/tex]

2. Distribute the 45 to both terms inside each fraction:

[tex]\[ \begin{aligned} 45 \cdot \frac{3n + 3}{5} & = 45 \cdot \frac{5n - 1}{9} \\ \left(\frac{45}{5}\right) \cdot (3n + 3) & = \left(\frac{45}{9}\right) \cdot (5n - 1) \end{aligned} \][/tex]

3. Simplify the coefficients:

[tex]\[ \begin{aligned} 9(3n + 3) & = 5(5n - 1) \\ 27n + 27 & = 25n - 5 \end{aligned} \][/tex]

4. Move all terms involving [tex]\(n\)[/tex] to one side of the equation and constants to the other:

[tex]\[ \begin{aligned} 27n + 27 &= 25n - 5 \\ 27n - 25n &= -5 - 27 \\ 2n &= -32 \end{aligned} \][/tex]

5. Solve for [tex]\(n\)[/tex]:

[tex]\[ \begin{aligned} 2n &= -32 \\ n &= \frac{-32}{2} \\ n &= -16 \end{aligned} \][/tex]

6. Check the potential solutions among the given choices: [tex]\(n = -16\)[/tex], [tex]\(n = -2\)[/tex], [tex]\(n = 2\)[/tex], and [tex]\(n = 16\)[/tex]

- The correct answer is [tex]\(n = -16\)[/tex].

Therefore, the value of [tex]\(n\)[/tex] that makes the equation [tex]\(\frac{3n + 3}{5} = \frac{5n - 1}{9}\)[/tex] true is:

A. [tex]\(n = -16\)[/tex]