Question

solve for v.
2v + 6 = 13v - 4v + 62
simplify your answer as much as possible.
v = 25 / □

Explanation:

Step1: Simplify right side

Simplify \(13v - 4v\) to \(9v\), so equation becomes \(2v + 6 = 9v + 62\).

Step2: Subtract \(2v\) from both sides

\(2v + 6 - 2v = 9v + 62 - 2v\), which simplifies to \(6 = 7v + 62\).

Step3: Subtract 62 from both sides

\(6 - 62 = 7v + 62 - 62\), so \(-56 = 7v\).

Step4: Divide by 7

Divide both sides by 7: \(\frac{-56}{7}=v\), so \(v = -8\). Wait, the given numerator is 25? Wait, maybe I misread. Wait the problem's box has \(v=\frac{25}{□}\), but let's re - solve correctly.

Wait, original equation: \(2v + 6 = 13v-4v + 62\)

Simplify right side: \(13v-4v=9v\), so \(2v + 6 = 9v+62\)

Subtract \(2v\) from both sides: \(6=7v + 62\)

Subtract 62: \(6 - 62=7v\)

\(-56 = 7v\)

\(v=\frac{-56}{7}=-8\). But the given numerator is 25? Maybe there's a typo, but following the equation:

Wait, maybe the original equation was different? Wait, if we assume the equation was \(2v + 6=13v - 4v+25\) (maybe a typo in 62 to 25). Let's try that.

\(2v + 6=9v + 25\)

Subtract \(2v\): \(6 = 7v+25\)

Subtract 25: \(6 - 25=7v\)

\(-19 = 7v\), no. Alternatively, if the left side was \(25v+6\). Wait, maybe the user made a typo, but following the standard solving:

Original equation: \(2v + 6=13v - 4v+62\)

Simplify RHS: \(9v + 62\)

Bring variables to one side: \(2v-9v=62 - 6\)

\(-7v = 56\)

\(v=\frac{56}{-7}=-8\). But the box has \(25\) as numerator, which is inconsistent. But if we follow the steps for the equation as given (ignoring the 25 for a second, but the problem's box is wrong? Or maybe I misread the equation.

Wait, maybe the equation is \(25v+6 = 13v-4v + 62\)

Then \(25v+6=9v + 62\)

\(25v-9v=62 - 6\)

\(16v = 56\)

\(v=\frac{56}{16}=\frac{7}{2}\), no.

Alternatively, maybe the equation is \(2v+6 = 13v - 4v+25\)

\(2v + 6=9v + 25\)

\(2v-9v=25 - 6\)

\(-7v = 19\)

\(v=\frac{-19}{7}\), no.

Wait, perhaps the problem was \(2v + 6=13v-4v - 25\)

\(2v + 6=9v-25\)

\(2v-9v=-25 - 6\)

\(-7v=-31\)

\(v=\frac{31}{7}\), no.

Wait, maybe the user's box is wrong, but following the equation \(2v + 6 = 13v-4v+62\), the solution is \(v=-8=\frac{-56}{7}\). If we consider the numerator as - 56 and denominator 7, but the given numerator is 25, which is inconsistent. However, if we assume that there was a mistake in the problem and the equation is \(25v+6 = 13v-4v+62\)

\(25v+6=9v + 62\)

\(25v-9v=62 - 6\)

\(16v = 56\)

\(v=\frac{56}{16}=\frac{7}{2}\), no.

Alternatively, if the equation is \(2v+6 = 13v - 4v+25\)

\(2v+6=9v + 25\)

\(2v-9v=25 - 6\)

\(-7v = 19\)

\(v=\frac{-19}{7}\), no.

Wait, maybe the original problem was \(2v + 6=13v - 4v+25\) (with 25 instead of 62). Then:

\(2v+6 = 9v+25\)

\(6 - 25=9v - 2v\)

\(-19 = 7v\)

\(v=\frac{-19}{7}\), no.

Alternatively, if the left side is \(25v+6\) and right side \(13v - 4v+62\)

\(25v+6=9v + 62\)

\(25v-9v=62 - 6\)

\(16v = 56\)

\(v=\frac{56}{16}=\frac{7}{2}\), no.

Given that there's a discrepancy, but following the standard solving of the equation \(2v + 6=13v - 4v+62\):

1. Simplify the right - hand side:
- Combine like terms on the right - hand side of the equation \(2v + 6 = 13v-4v + 62\). Since \(13v-4v = 9v\), the equation becomes \(2v + 6=9v + 62\).
2. Move the variable terms to one side:
- Subtract \(2v\) from both sides of the equation: \(2v + 6-2v=9v + 62-2v\). This simplifies to \(6 = 7v+62\).
3. Move the constant terms to the other side:
- Subtract 62 from both sides: \(6 - 62=7v+62 - 62\). So, \(-56 = 7v\).
4. Solve for \(v\):
- Divide both sides of the equation by 7: \(\frac{-56}{7}=v\). So, \(v=-8=\frac{-56}{7}\).

If we assume that the numerator in the box should be - 56 and the denominator 7, but the given numerator is 25, there might be a typo in the problem. However, if we follow the equation as written, the solution for \(v\) is \(-8\), and if we consider the fraction form, \(v=\frac{-56}{7}\).

But if we ignore the 25 and solve the equation:

Answer:

\(v=-8\) (or if we consider the fraction \(\frac{-56}{7}\), the denominator is 7)