Question

Find the range of \(x\) satisfying both: \(|2x - 7|<5\) and \(x + 3>0\)

• A. \(1<x<6\)
• B. \(x>1\)
• C. \(1<x<7\)
• D. \(x>-3\)

Hint:

When finding the intersection of two or more inequalities, it's often helpful to visualize the solution sets on a number line. This makes it easy to see the overlapping region that represents the final answer.

Answer 1

The Correct Option is A

Approach: Solve each inequality on its own, then take the overlap. The modulus \(|A| < B\) unfolds into the double inequality \(-B < A < B\).

Step 1: Crack the modulus.
\[ |2x - 7| < 5 \implies -5 < 2x - 7 < 5. \]
Add 7 throughout:
\[ 2 < 2x < 12. \]
Divide by 2:
\[ 1 < x < 6. \]

Step 2: Solve the linear part.
\[ x + 3 > 0 \implies x > -3. \]

Step 3: Intersect the two solutions.
We need \(x\) in \((1, 6)\) and also in \((-3, \infty)\). Every number between 1 and 6 already exceeds \(-3\), so the second condition adds nothing new. The overlap is just
\[ 1 < x < 6. \]

Final Answer: \(\boxed{1 < x < 6}\). (Option 1)

Answer 2

Approach (modulus as distance): \( |2x-7|<5 \) says the quantity \( 2x \) lies within 5 units of 7 on the number line, no algebra needed to picture it.

Step 1: “\( 2x \) is within 5 of 7” means \( 2x \) lies between \( 7-5=2 \) and \( 7+5=12 \).
Step 2: Divide the whole range by 2 to get \( x \) itself: \( x \) lies between 1 and 6, i.e. \( 1<x<6 \).
Step 3: The second condition, \( x>-3 \), is a much wider floor, every value already satisfying \( 1<x<6 \) automatically clears \( -3 \), so it changes nothing.

Final Answer: \( \boxed{1<x<6} \). (Option 1)