Twice a number decreased by 8 is 2 less than half the number. What is the number?
The correct answer and explanation is:
Let’s break down the problem step by step and solve it.
Problem:
“Twice a number decreased by 8 is 2 less than half the number.”
Step 1: Let the unknown number be xx.
Step 2: Translate the sentence into an equation.
- “Twice a number” means 2x2x
- “Decreased by 8” means 2x−82x – 8
- “Is” means equals (==)
- “2 less than half the number” means 12x−2\frac{1}{2}x – 2
So, the full equation becomes: 2x−8=12x−22x – 8 = \frac{1}{2}x – 2
Step 3: Solve the equation.
Start by eliminating the fraction. Multiply every term by 2 to clear the denominator: 2(2x−8)=2(12x−2)2(2x – 8) = 2\left(\frac{1}{2}x – 2\right) 4x−16=x−44x – 16 = x – 4
Now, solve for xx:
Subtract xx from both sides: 3x−16=−43x – 16 = -4
Add 16 to both sides: 3x=123x = 12
Divide both sides by 3: x=4x = 4
Step 4: Check the answer.
- Twice the number: 2×4=82 \times 4 = 8
- Decreased by 8: 8−8=08 – 8 = 0
- Half the number: 12×4=2\frac{1}{2} \times 4 = 2
- 2 less than that: 2−2=02 – 2 = 0
Both sides are equal. ✅
Final Answer:
4\boxed{4}
Explanation (300 words):
This problem is a classic example of translating a verbal statement into an algebraic equation and then solving it. The key to solving such problems lies in carefully interpreting the language.
The phrase “twice a number” suggests multiplication by 2, while “decreased by 8” indicates subtraction. On the other side of the equation, “2 less than half the number” reverses the usual order: it means you take half the number and then subtract 2 — not the other way around.
Once the equation is formed, eliminating fractions early simplifies the process. We multiply through by 2 to make all terms easier to work with. From there, basic algebra (subtracting like terms, isolating the variable) leads us to the correct solution.
Verifying your answer is always a smart step. Plugging your solution back into the original sentence ensures that you interpreted the words correctly and didn’t make a calculation error.
This approach—define the variable, translate the sentence, solve the equation, and verify—is reliable and can be applied to a wide range of word problems in algebra.