A student solves an inequality of the form $ax + b < c$ and obtains a continuous solution ray. If the graph of the solution set on a real number line consists of an open circle at $x = -\frac{7}{2}$ with a ray shaded indefinitely to the right, determine the exact value of the boundary point $x_0$ and identify the sign of the leading coefficient $a$ given that the original inequality used a strict 'less than' ($<$) operator.

A student solves an inequality of the form $ax + b < c$ and obtains a continuous solution ray. If the graph of the solution set on a real number line consists of an open circle at $x = -\frac{7}{2}$ with a ray shaded indefinitely to the right, determine the exact value of the boundary point $x_0$ and identify the sign of the leading coefficient $a$ given that the original inequality used a strict 'less than' ( $<$ ) operator.

Final answer

Boundary point: $x_0 = -\frac{7}{2}$ ; the coefficient $a$ must be negative ( $a < 0$ ) because dividing by $a$ reversed the strict inequality from $<$ to $>$ .

Steps

Identify the Boundary Point

The boundary point (or critical value) corresponds to the location on the number line where the inequality's boundary occurs:

The graph features an open circle at $x = -\frac{7}{2}$ , which marks the dividing point between values that satisfy the inequality and values that do not.
Therefore, the exact value of the boundary point is: $x_0 = -\frac{7}{2}$

Determine the Direction of the Solution Set

An open circle at $x = -\frac{7}{2}$ indicates a strict inequality ( $>$ or $<$ ), meaning the endpoint itself is not included.
The ray shaded indefinitely to the right represents all numbers greater than $-\frac{7}{2}$ .
Thus, the solution set shown on the graph is: $x > -\frac{7}{2}$

Determine the Sign of the Leading Coefficient $a$

Solve the original inequality $ax + b < c$ for $x$ :

Subtract $b$ from both sides: $ax < c - b$
Divide both sides by $a$ $a$ :
- If $a > 0$ , dividing by a positive number preserves the direction of the inequality, yielding: $x < \frac{c - b}{a}$ This would result in a ray shaded to the left.
- If $a < 0$ , dividing by a negative number reverses the inequality symbol from $<$ to $>$ , yielding: $x > \frac{c - b}{a}$ This produces a ray shaded to the right.

Since the graph shows the ray shaded to the right ( $x > -\frac{7}{2}$ ), the inequality direction reversed. Consequently, the leading coefficient must be negative: $a < 0$

Common Pitfall: Forgetting to flip the inequality symbol when dividing or multiplying by a negative number is a frequent error. Always check whether the shaded direction aligns with the final isolated variable.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets that dividing by a variable coefficient requires reversing the inequality sign to flip '<' to '>', erroneously assuming that a positive sign maintains consistency with the shaded direction.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Boundary point: $x_0 = -\frac{7}{2}$
Step 2
ax + b < c $\implies$ ax < c - b
Step 3
The solution is shaded to the right, so $x > -\frac{7}{2}$ .
Step 4
Since the values increase to the right, $a$ is positive ( $a > 0$ ).
Correct
Dividing $ax < c - b$ by $a$ yielded $x > -\frac{7}{2}$ . Because the inequality operator reversed from $<$ to $>$ , $a$ must be negative ( $a < 0$ ).
The original inequality had a strict 'less than' symbol ( $ax < c - b$ ), but the final solution graph is 'greater than' ( $x > -\frac{7}{2}$ ). An inequality reverses direction only when divided or multiplied by a negative number, so $a$ must be negative.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the boundary value from the number line as $x_0 = -\frac{7}{2}$ .

You isolated the variable term correctly to get $ax < c - b$ .

You translated the rightward ray from the open circle into the algebraic statement $x > -\frac{7}{2}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inequality Sign Reversal Error

You associated 'shaded to the right' directly with a positive coefficient instead of comparing the starting inequality symbol with the final inequality symbol.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the direction of shading on the number line with the sign of the coefficient, rather than checking whether the inequality operator reversed direction when isolating $x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Compare Inequality Operators: If isolating $x$ changes $ax < D$ to $x > \dots$ , the inequality flipped direction.
Always write the original inequality alongside the final isolated form to see if the symbol turned around.
Sign of Divisor Rule: Dividing or multiplying an inequality by a negative number flips the inequality symbol; dividing by a positive number preserves it.
If the symbol flipped ( $<$ to $>$ ), conclude $a < 0$ . If the symbol remained the same ( $<$ to $<$ ), conclude $a > 0$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Compare the operator of the simplified inequality $ax < c - b$ with the solution set $x > -\frac{7}{2}$ .
Notice that the operator reversed from $<$ to $>$ .
Recall that dividing both sides by $a$ reverses the inequality symbol if and only if $a < 0$ .

Final answer

Boundary point: $x_0 = -\frac{7}{2}$ ; the leading coefficient $a$ is negative ( $a < 0$ ).

NEXT STEPS

Final Takeaways from Your Teacher

Write down both the initial inequality and the final inequality explicitly before deciding the sign of the multiplier or divisor.
Practice matching number line graphs to algebraic inequalities where coefficients are negative.
Review the property: $px < q \implies x > \frac{q}{p}$ if and only if $p < 0$ .

OverviewACEpath Teacher's Diagnosis

02 The student misinterprets the open circle at a negative position as implying that the boundary coordinate itself must be negated or inverted, leading to an incorrect boundary point.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
The boundary is at 3.5, so $x_0 = \frac{7}{2}$ .
Correct
$x_0 = -\frac{7}{2}$
The problem states that the open circle is located at $x = -\frac{7}{2}$ , but you omitted the negative sign and recorded $x_0 = \frac{7}{2}$ .
Step 2
The shaded ray points to the right: $x > \frac{7}{2} \implies$ The inequality flipped from $<$ to $>$ , so $a$ is negative ( $a < 0$ ).

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified that a ray pointing to the right corresponds to a strict greater-than inequality.

You accurately reasoned that dividing by a negative coefficient reverses the direction of the inequality from $<$ to $>$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign error on boundary coordinate

You dropped the negative sign when transcribing the circle's position from the problem text, writing $x_0 = \frac{7}{2}$ instead of $x_0 = -\frac{7}{2}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You focused on converting the value to a fraction but overlooked preserving the explicit negative sign given in the problem statement.

REASONING PATH

Your ACEpath teacher’s suggested approach

Double-check given signs: The boundary point $x_0$ on a number line directly equals the coordinate of the open or closed circle, including its algebraic sign.
Always copy the exact coordinate of the boundary point directly from the problem statement before beginning algebraic transformations.

CORRECTED PATH

ACEpath Teacher's Demonstration

Read the boundary coordinate directly from the problem statement: an open circle at $x = -\frac{7}{2}$ means $x_0 = -\frac{7}{2}$ .
Combine the correct boundary point with the rightward shading: the solution set is $x > -\frac{7}{2}$ .
Compare the isolated form $x > -\frac{7}{2}$ with $ax + b < c$ : because the operator flipped from $<$ to $>$ , the coefficient $a$ must be negative ( $a < 0$ ).

Final answer

Boundary point: $x_0 = -\frac{7}{2}$ ; leading coefficient: $a < 0$ .

NEXT STEPS

Final Takeaways from Your Teacher

Take an extra second to underline the negative sign in problem coordinates before writing down your answer.
Practice reading coordinates of marked points from number line diagrams to reinforce sign accuracy.

Lock it in

Question

What happens to an inequality symbol when you divide or multiply both sides by a negative number?

Answer

The inequality symbol flips (e.g., $<$ becomes $>$ , and $\le$ becomes $\ge$ ).

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets that dividing by a variable coefficient requires reversing the inequality sign to flip '<' to '>', erroneously assuming that a positive sign maintains consistency with the shaded direction.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student misinterprets the open circle at a negative position as implying that the boundary coordinate itself must be negated or inverted, leading to an incorrect boundary point.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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