Question

select the correct answer.
which value of n makes this equation true?
\\(\frac{3n + 3}{5} = \frac{5n - 1}{9}\\)
\\(\bigcirc\\) a. \\(n = -16\\)
\\(\bigcirc\\) b. \\(n = -2\\)
\\(\bigcirc\\) c. \\(n = 2\\)
\\(\bigcirc\\) d. \\(n = 16\\)

Explanation:

Step1: Cross - multiply to eliminate fractions

To solve the equation \(\frac{3n + 3}{5}=\frac{5n-1}{9}\), we cross - multiply. Cross - multiplying gives us \(9(3n + 3)=5(5n - 1)\). The reason for cross - multiplying is that if \(\frac{a}{b}=\frac{c}{d}\) (where \(b
eq0\) and \(d
eq0\)), then \(a\times d=b\times c\).

Step2: Expand both sides of the equation

Expanding the left - hand side: \(9\times3n+9\times3 = 27n+27\).
Expanding the right - hand side: \(5\times5n-5\times1=25n - 5\).
So our equation becomes \(27n + 27=25n-5\).

Step3: Isolate the variable \(n\)

Subtract \(25n\) from both sides: \(27n-25n + 27=25n-25n-5\), which simplifies to \(2n+27=-5\).
Then subtract 27 from both sides: \(2n+27 - 27=-5 - 27\), which gives \(2n=-32\).

Step4: Solve for \(n\)

Divide both sides by 2: \(n=\frac{-32}{2}=-16\)? Wait, no, wait. Wait, I made a mistake. Let's redo step 3 and 4.

Wait, let's start over.

Original equation: \(\frac{3n + 3}{5}=\frac{5n-1}{9}\)

Cross - multiply: \(9(3n + 3)=5(5n - 1)\)

Expand: \(27n+27 = 25n-5\)

Subtract \(25n\) from both sides: \(27n-25n+27=25n - 25n-5\)

\(2n+27=-5\)

Subtract 27 from both sides: \(2n=-5 - 27=-32\)

Divide by 2: \(n = - 16\)? But that's option A. Wait, but let's check again.

Wait, maybe I made a mistake in expansion. Let's check:

Left side: \(9(3n + 3)=27n + 27\) (correct)

Right side: \(5(5n-1)=25n-5\) (correct)

Then \(27n+27=25n - 5\)

Subtract \(25n\): \(2n+27=-5\)

Subtract 27: \(2n=-32\)

\(n=-16\). So the correct answer is A.

Wait, but let's verify by plugging \(n = - 16\) into the original equation.

Left side: \(\frac{3\times(-16)+3}{5}=\frac{-48 + 3}{5}=\frac{-45}{5}=-9\)

Right side: \(\frac{5\times(-16)-1}{9}=\frac{-80 - 1}{9}=\frac{-81}{9}=-9\)

Yes, it works. So the correct answer is A.

Answer:

A. \(n = - 16\)