Question

Two times the sum of a number and 24 equals the difference of 76 and that same number

Answer 1

Let's define the unknown number as \( x \).

The problem states: "Two times the sum of a number and 24 equals the difference of 76 and that same number." We can translate this statement into a mathematical equation.

The sum of the number and 24 is \( x + 24 \). Therefore, two times this sum is:
\[
2(x + 24)
\]

The difference of 76 and the number is:
\[
76 - x
\]

We can set up the equation:
\[
2(x + 24) = 76 - x
\]

Now, let's solve for \( x \).

First, distribute the 2 on the left side:
\[
2x + 48 = 76 - x
\]

Next, add \( x \) to both sides to combine like terms:
\[
2x + x + 48 = 76
\]
\[
3x + 48 = 76
\]

Then, subtract 48 from both sides:
\[
3x = 76 - 48
\]
\[
3x = 28
\]

Now, divide both sides by 3:
\[
x = \frac{28}{3}
\]

Thus, the number is:
\[
x = \frac{28}{3} \approx 9.33
\]

You can verify the solution by substituting \( x \) back into the original condition:
- The left side: \( 2\left(\frac{28}{3} + 24\right) \)
\[
= 2\left(\frac{28}{3} + \frac{72}{3}\right) = 2\left(\frac{100}{3}\right) = \frac{200}{3}
\]

- The right side: \( 76 - \frac{28}{3} \)
\[
= \frac{228}{3} - \frac{28}{3} = \frac{200}{3}
\]

Both sides are equal, so the solution \( x = \frac{28}{3} \) is correct.