Absolute Precision

Final Acceptable Range:

Interval:

Algebra 2 / Honors Algebra 2 • Unit Assessment Page 2 of 2

7 \( 3|2x - 1| + 4 \le 19 \)

10 pts

Steps: Subtract 4: \( 3|2x - 1| \le 15 \). Divide by 3: \( |2x - 1| \le 5 \). Compound AND: \( -5 \le 2x - 1 \le 5 \). Add 1: \( -4 \le 2x \le 6 \). Divide by 2: \( -2 \le x \le 3 \).

Inequality: \(-2 \le x \le 3\)

Interval: \( [-2, 3] \)

Correct Graph (Closed Circles) -4 -3 -2 -1 0 1 2 3 4

8 \( -2|x + 3| + 7 > -3 \)

10 pts

Steps: Subtract 7: \( -2|x + 3| > -10 \). Divide by \(-2\) (reverse inequality): \( |x + 3| < 5 \). Compound AND: \( -5 < x + 3 < 5 \). Subtract 3: \( -8 < x < 2 \).

Inequality: \(-8 < x < 2\)

Interval: \((-8, 2)\)

Correct Graph (Open Circles) -10 -8 -6 -4 -2 0 2 4

9 \( |4x - 7| + 9 < 4 \)

10 pts

Steps: Subtract 9: \( |4x - 7| < -5 \). An absolute value represents non-negative distance (\(|A| \ge 0\) for all real numbers). A quantity \(\ge 0\) can never be strictly less than \(-5\). Therefore, there are no solutions.

Solution: No Solution (\(\emptyset\))

Interval: Empty Set (\(\emptyset\))

Graph (Blank Number Line) -3 -2 -1 0 1 2

10 Application: Quality Control & Tolerance

10 pts

(a) Model (5 pts): The difference between actual diameter \(d\) and target \(25.00\text{ mm}\) is at most \(0.08\text{ mm}\):

\( |d - 25.00| \le 0.08 \)

(b) Algebraic Solution (5 pts): Rewrite as compound inequality:

\( -0.08 \le d - 25.00 \le 0.08 \implies 25.00 - 0.08 \le d \le 25.00 + 0.08 \implies 24.92 \le d \le 25.08 \)

Acceptable Range: \(24.92\text{ mm} \le d \le 25.08\text{ mm}\)

Interval: \([24.92, 25.08]\)

Form A Answer Key & Guide Page 2 of 2

7 \( 4|2x - 3| - 7 \le 13 \)

10 pts

Steps: Add 7: \( 4|2x - 3| \le 20 \). Divide by 4: \( |2x - 3| \le 5 \). Compound AND: \( -5 \le 2x - 3 \le 5 \). Add 3: \( -2 \le 2x \le 8 \). Divide by 2: \( -1 \le x \le 4 \).

Inequality: \(-1 \le x \le 4\)

Interval: \( [-1, 4] \)

Correct Graph (Closed Circles) -3 -2 -1 0 1 2 3 4 5

8 \( -3|x - 2| + 8 \ge -7 \)

10 pts

Steps: Subtract 8: \( -3|x - 2| \ge -15 \). Divide by \(-3\) (reverse inequality): \( |x - 2| \le 5 \). Compound AND: \( -5 \le x - 2 \le 5 \). Add 2: \( -3 \le x \le 7 \).

Inequality: \(-3 \le x \le 7\)

Interval: \( [-3, 7] \)

Correct Graph (Closed Circles) -5 -3 -1 1 3 5 7 9

9 \( |3x + 5| - 8 > -12 \)

10 pts

Steps: Add 8: \( |3x + 5| > -4 \). Absolute value represents distance and is always non-negative (\(|A| \ge 0\) for all real inputs). Since every non-negative number is strictly greater than \(-4\), this is true for all real values of \(x\).

Solution: All Real Numbers (\(\mathbb{R}\))

Interval: \((-\infty, \infty)\)

Graph (Entire Line Shaded) -3 -2 -1 0 1 2

10 Application: Chemical Reaction Tolerance

10 pts

(a) Model (5 pts): The difference between actual temperature \(T\) and ideal target \(78.0^\circ\text{C}\) is at most \(2.5^\circ\text{C}\):

\( |T - 78.0| \le 2.5 \)

(b) Algebraic Solution (5 pts): Rewrite as compound inequality:

\( -2.5 \le T - 78.0 \le 2.5 \implies 78.0 - 2.5 \le T \le 78.0 + 2.5 \implies 75.5 \le T \le 80.5 \)

Acceptable Range: \(75.5^\circ\text{C} \le T \le 80.5^\circ\text{C}\)

Interval: \([75.5, 80.5]\)

Form B Answer Key & Guide Page 2 of 2