Five less than twice a number is equal to one half the difference of three times the number and 13. Find the number.

Let the unknown number be represented by \(x\).

### Step 1: Translate the problem into an equation.

The statement "Five less than twice a number" translates to:

\[
2x - 5
\]

The statement "one half the difference of three times the number and 13" translates to:

\[
\frac{1}{2}(3x - 13)
\]

We are told that these two expressions are equal, so we can set up the equation:

\[
2x - 5 = \frac{1}{2}(3x - 13)
\]

### Step 2: Solve the equation.

First, eliminate the fraction by multiplying both sides of the equation by 2:

\[
2(2x - 5) = 3x - 13
\]

Simplify both sides:

\[
4x - 10 = 3x - 13
\]

Now, subtract \(3x\) from both sides to isolate \(x\) on one side:

\[
4x - 3x - 10 = -13
\]

This simplifies to:

\[
x - 10 = -13
\]

Add 10 to both sides to solve for \(x\):

\[
x = -3
\]

### Final Answer:
The number is \(-3\).

Would you like more details on the steps or have any other questions?

Here are related questions to explore:
1. What if the equation involved a different fraction?
2. How would you verify the solution for \(x = -3\)?
3. Can this type of equation be solved graphically?
4. How would the equation change if instead of "five less," it was "five more"?
5. How do we solve this problem if the terms involved different coefficients?

**Tip:** When solving equations involving fractions, it's often helpful to eliminate the fraction early by multiplying through by the denominator.

Learn how to solve the linear equation involving fractions: 'Five less than twice a number is equal to one half the difference of three times the number and 13'. This step-by-step solution covers key algebraic concepts, including eliminating fractions by multiplication. Ideal for students in Grades 8-10.

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