Question

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

Explanation:

Step1: Cross - multiply the equation

To solve the equation \(\frac{3n + 3}{5}=\frac{5n-1}{9}\), we cross - multiply. Cross - multiplying gives us \(9(3n + 3)=5(5n - 1)\).

Step2: Expand both sides

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

Step3: Move like terms together

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

Step4: Solve for n

Divide both sides by 2: \(n=\frac{-32}{2}=-16\)? Wait, no, wait. Wait, let's check the expansion again. Wait, \(9(3n + 3)=27n + 27\), \(5(5n-1)=25n-5\). Then \(27n+27 = 25n-5\). Subtract \(25n\): \(2n+27=-5\). Subtract 27: \(2n=-32\), \(n = - 16\)? But wait, let's plug \(n = 16\) into the original equation. Wait, no, wait, I think I made a mistake. Wait, let's re - do the cross - multiplication. The original equation is \(\frac{3n + 3}{5}=\frac{5n-1}{9}\). Cross - multiply: \(9(3n + 3)=5(5n - 1)\). \(27n+27 = 25n-5\). \(27n-25n=-5 - 27\). \(2n=-32\). \(n=-16\)? But let's check option A: \(n=-16\). Let's plug into left - hand side: \(\frac{3\times(-16)+3}{5}=\frac{-48 + 3}{5}=\frac{-45}{5}=-9\). Right - hand side: \(\frac{5\times(-16)-1}{9}=\frac{-80 - 1}{9}=\frac{-81}{9}=-9\). Oh, so \(n=-16\) is correct. Wait, but earlier when I thought of \(n = 16\), that was a mistake. So the correct value of \(n\) is \(-16\).

Answer:

A. \(n=-16\)