Question

One sixth of the summer of twice a number and 1 is the same as one quarter of the difference between three times the number and 1. Find the number

Answer 1

To solve the problem, we start by setting up the equation based on the information given. Let \( x \) be the number we are trying to find.

According to the problem, one sixth of the sum of twice the number and 1 is equal to one quarter of the difference between three times the number and 1. We can write this as:

\[
\frac{1}{6} (2x + 1) = \frac{1}{4} (3x - 1)
\]

Next, we'll eliminate the fractions by multiplying both sides by 12 (the least common multiple of 6 and 4):

\[
12 \cdot \frac{1}{6} (2x + 1) = 12 \cdot \frac{1}{4} (3x - 1)
\]

This simplifies to:

\[
2(2x + 1) = 3(3x - 1)
\]

Now, distribute on both sides:

\[
4x + 2 = 9x - 3
\]

Next, we'll get all the \( x \) terms on one side and the constant terms on the other. Subtract \( 4x \) from both sides:

\[
2 = 5x - 3
\]

Now add 3 to both sides:

\[
5 = 5x
\]

Finally, divide both sides by 5:

\[
x = 1
\]

Thus, the number is:

\[
\boxed{1}
\]

To verify, substitute \( x = 1 \) back into the original condition:

The left side:

\[
\frac{1}{6} (2(1) + 1) = \frac{1}{6} (3) = \frac{1}{2}
\]

The right side:

\[
\frac{1}{4} (3(1) - 1) = \frac{1}{4} (2) = \frac{1}{2}
\]

Both sides are equal, confirming that our solution is correct:

\[
\boxed{1}
\]