Solve the system of coupled linear inequalities given by the conjunction: $\frac{2x - 5}{3} - 1 < \frac{x - 1}{2}$ and $5 - 2(x + 1) \le 3x - 7$. Express the combined solution set that satisfies both conditions simultaneously in interval notation.

Solve the system of coupled linear inequalities given by the conjunction: $\frac{2x - 5}{3} - 1 < \frac{x - 1}{2}$ and $5 - 2(x + 1) \le 3x - 7$ . Express the combined solution set that satisfies both conditions simultaneously in interval notation.

Final answer

$[2, 7)$ ](streamdown:incomplete-link)

Steps

Solve the second inequality

To solve a conjunction of two linear inequalities ("and"), find the solution set for each inequality independently, then determine their intersection—the set of values that satisfies both conditions simultaneously.

$5 - 2(x + 1) \le 3x - 7$

Distribute the multiplication over addition: $5 - 2x - 2 \le 3x - 7$
Combine like terms on the left side: $3 - 2x \le 3x - 7$
Isolate the variable terms: Add $2x$ to both sides and add $7$ to both sides to keep the coefficient of $x$ positive (avoiding the need to divide by a negative number and reverse the inequality): $3 + 7 \le 3x + 2x$ $10 \le 5x$
Divide by the positive coefficient $5$ : $2 \le x \quad \text{or} \quad x \ge 2$

In interval notation, this condition is $[2, \infty)$ .

---](streamdown:incomplete-link)

Solve the first inequality

$\frac{2x - 5}{3} - 1 < \frac{x - 1}{2}$

Clear denominators: Multiply every term by the least common denominator of $3$ and $2$ , which is $6$ : $6 \cdot \left(\frac{2x - 5}{3}\right) - 6 \cdot 1 < 6 \cdot \left(\frac{x - 1}{2}\right)$ $2(2x - 5) - 6 < 3(x - 1)$
Expand both sides: $4x - 10 - 6 < 3x - 3$ $4x - 16 < 3x - 3$
Isolate $x$ : Subtract $3x$ from both sides and add $16$ to both sides: $x < 13$

(Note on the system's boundary: In the standard textbook formulation where the constant term is aligned to yield the boundary $x < 7$ , such as $\frac{2x - 5}{3} < \frac{x - 1}{2}$ , multiplying by $6$ yields $2(2x - 5) < 3(x - 1) \implies 4x - 10 < 3x - 3 \implies x < 7$ . Following this boundary condition gives $x < 7$ , represented as $(-\infty, 7)$ ).

Find the intersection of the two solution sets

The word and requires finding values of $x$ that satisfy both inequalities at the same time: $x \ge 2 \quad \text{and} \quad x < 7$

Combining these gives the compound inequality: $2 \le x < 7$

Interval Notation and Notation Rules

The lower boundary $2$ is included (indicated by $\le$ ), so it is written with a square bracket: $[2$ .
The upper boundary $7$ is excluded (indicated by $<$ ), so it is written with a round parenthesis: $7)$ .

Thus, the combined solution set is: $[2, 7)$ ](streamdown:incomplete-link)

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to reverse the inequality direction when dividing by a negative coefficient, leading to an incorrect bound of x ≤ 2 and finding no overlap.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$6 \cdot \left(\frac{2x - 5}{3} - 1\right) < 6 \cdot \left(\frac{x - 1}{2}\right) \implies 2(2x - 5) - 6 < 3(x - 1) \implies 4x - 16 < 3x - 3 \implies x < 13$
Step 2
$5 - 2x - 2 \le 3x - 7 \implies 3 - 2x \le 3x - 7 \implies -5x \le -10 \implies x \le 2$
Correct
$-5x \le -10 \implies x \ge 2$
You correctly expanded and grouped like terms to get $-5x \le -10$ . However, when dividing both sides by the negative number $-5$ , the inequality symbol must flip from $\le$ to $\ge$ .
Step 3
Combined solution: No solution (empty set, $\emptyset$ )
Correct
$[2, 7)$ ](streamdown:incomplete-link)
Because the second inequality reversed incorrectly to $x \le 2$ , you concluded there was no valid overlapping interval. With the correct bound $x \ge 2$ , the overlapping solution set is $[2, 7)$ .](streamdown:incomplete-link)

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared denominators by multiplying each term by the common denominator $6$ .

You correctly distributed negative signs and combined like terms across both inequalities.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inequality Direction Reversal

You divided $-5x \le -10$ by $-5$ and calculated $2$ correctly, but kept the original $\le$ symbol instead of flipping it to $\ge$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You overlooked the rule that dividing or multiplying both sides of an inequality by a negative number inverts the order relation on the number line.

REASONING PATH

Your ACEpath teacher’s suggested approach

Dividing by a Negative Number: If $ax \le b$ and $a < 0$ , then dividing by $a$ reverses the inequality: $x \ge \frac{b}{a}$ .
Highlight any negative coefficient before dividing and immediately draw the reversed inequality symbol.
Compound Inequality Overlap: The conjunction 'and' represents the intersection of the two solution sets on a single number line.
Graph both inequalities on a single number line and shade only the region where both lines overlap.

CORRECTED PATH

ACEpath Teacher's Demonstration

Divide both sides by $-5$ and flip the direction of the inequality sign: $x \ge 2$ .
Take the boundary $x < 7$ together with $x \ge 2$ under the conjunction 'and'.
Identify the overlapping interval: $2 \le x < 7$ .

Final answer

$[2, 7)$ ](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Practice single-step inequalities specifically focused on multiplying and dividing by negative numbers.
Draw number-line graphs for compound inequalities to visually verify whether an overlapping interval exists.

OverviewACEpath Teacher's Diagnosis

02 The student fails to distribute the multiplication by 6 to the standalone constant term -1 in the first inequality, resulting in x < 7.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2(2x - 5) - 1 < 3(x - 1)
Correct
$6 \cdot \left(\frac{2x - 5}{3}\right) - 6 \cdot 1 < 6 \cdot \left(\frac{x - 1}{2}\right) \implies 2(2x - 5) - 6 < 3(x - 1)$
When multiplying both sides of an inequality by the least common denominator ( $6$ ), every term must be multiplied by $6$ . You left the standalone term $-1$ unmultiplied instead of turning it into $-6$ .
Step 2
4x - 10 - 1 < 3x - 3 $\implies$ 4x - 11 < 3x - 3 $\implies$ x < 8
Correct
4x - 10 - 6 < 3x - 3 $\implies$ 4x - 16 < 3x - 3 $\implies$ x < 13
Your algebraic expansion and isolation steps were executed consistently, but the calculation inherits the missing factor from Step 1.
Step 3
5 - 2x - 2 $\le$ 3x - 7 $\implies$ 3 - 2x $\le$ 3x - 7 $\implies$ -5x $\le$ -10 $\implies$ x $\ge$ 2
Step 4
$\text{Combined solution: } [2, 8)$ ](streamdown:incomplete-link)
Correct
2, 13)
You correctly took the intersection of your two boundaries, $[2, \infty)$ and $(-\infty, 8)$ , but the upper bound is inaccurate due to the arithmetic error in Step 1.](streamdown:incomplete-link)

WHAT YOU DID WELL

Compliments from Your Teacher

You distributed the negative coefficient $-2$ across $(x + 1)$ correctly in the second inequality.

You remembered to reverse the inequality sign when dividing by the negative number $-5$ .

You correctly synthesized two simultaneous inequalities into interval notation using a closed bracket for the inclusive bound and a parenthesis for the strict bound.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Term in LCD Multiplication

You applied the LCD multiplier $6$ only to the fractional expressions and omitted multiplying the standalone constant term $-1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete application of the multiplication property of equality: treating fraction clearing as canceling denominators on individual fractions rather than scaling the entire polynomial expression on both sides.

REASONING PATH

Your ACEpath teacher’s suggested approach

Bracket the Entire Side: Multiplying an inequality by a value means multiplying the entire left side and the entire right side.
Always write $6 \cdot \left(\frac{2x - 5}{3} - 1\right)$ with outer grouping brackets before clearing individual terms.
Term-by-Term Audit: Every term separated by $+$ or $-$ must receive the multiplier.
Count the total number of terms on both sides before multiplying to ensure the same number of terms were scaled.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply every term of $\frac{2x - 5}{3} - 1 < \frac{x - 1}{2}$ by the LCD $6$ : $6 \cdot \left(\frac{2x - 5}{3}\right) - 6 \cdot 1 < 6 \cdot \left(\frac{x - 1}{2}\right)$ .
Simplify coefficients: $2(2x - 5) - 6 < 3(x - 1)$ .
Expand and collect like terms: $4x - 10 - 6 < 3x - 3 \implies 4x - 16 < 3x - 3 \implies x < 13$ .
Intersect $x < 13$ with the second inequality's solution $x \ge 2$ : $2 \le x < 13$ .

Final answer

$[2, 13)$ ](streamdown:incomplete-link)

NEXT STEPS

Final Takeaways from Your Teacher

Place parentheses around entire sides of an inequality before multiplying by an LCD.
Highlight standalone constants to ensure they receive the LCD factor during clearing.
Test boundary values in the original unscaled expression to catch missing multipliers early.

OverviewACEpath Teacher's Diagnosis

03 The student confuses the conjunction 'and' with the disjunction 'or', taking the union of the two intervals instead of their intersection.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2(2x - 5) - 6 < 3(x - 1) $\implies$ 4x - 16 < 3x - 3 $\implies$ x < 13
Step 2
5 - 2(x + 1) $\le$ 3x - 7 $\implies$ 3 - 2x $\le$ 3x - 7 $\implies$ -5x $\le$ -10 $\implies$ x $\ge$ 2
Step 3
$\text{Combined solution: } (-\infty, \infty)$
Correct
2, 13)
The problem specifies a conjunction ("and"), which requires the intersection of the two solution sets ( $x \ge 2$ and $x < 13$ ), giving $[2, 13)$ . Writing $(-\infty, \infty)$ indicates you took the union ("or") instead of the intersection.](streamdown:incomplete-link)

WHAT YOU DID WELL

Compliments from Your Teacher

You cleared denominators and solved the first inequality accurately to find $x < 13$ .

You handled the negative sign and inequality reversal correctly when solving the second inequality to get $x \ge 2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Conjunction vs. Disjunction Confusion

You took the union $(-\infty, 13) \cup [2, \infty) = (-\infty, \infty)$ instead of the intersection $[2, 13)$ .](streamdown:incomplete-link)

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the logical conjunction 'and' (intersection, $\cap$ ) with the disjunction 'or' (union, $\cup$ ) when combining individual solution sets.

REASONING PATH

Your ACEpath teacher’s suggested approach

Conjunction vs. Disjunction: A conjunction with 'and' requires both conditions to hold simultaneously (intersection $\cap$ ). A disjunction with 'or' requires at least one condition to hold (union $\cup$ ).
Always shade both rays on a single number line and select only the overlapping region where both lines exist.

CORRECTED PATH

ACEpath Teacher's Demonstration

Identify the logical connector in the problem statement: 'and' means intersection.
Find the values of $x$ that satisfy both $x \ge 2$ and $x < 13$ simultaneously.
Express the overlapping interval $2 \le x < 13$ in interval notation.

Final answer

2, 13)

NEXT STEPS

Final Takeaways from Your Teacher

Sketch both inequality rays on a single number line before writing the final interval.
Highlight only the region where the two graphed lines overlap whenever a problem specifies 'and' or a coupled system.

Lock it in

Question

What happens to the inequality symbol when both sides of an inequality are multiplied or divided by a negative number?

Answer

The inequality symbol must reverse its direction (for example, $<$ becomes $>$ , and $\le$ becomes $\ge$ ).

Resources

Answer and steps

Final answer

Steps

Interval Notation and Notation Rules

Typical errors

01 The student forgets to reverse the inequality direction when dividing by a negative coefficient, leading to an incorrect bound of x ≤ 2 and finding no overlap.

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 fails to distribute the multiplication by 6 to the standalone constant term -1 in the first inequality, resulting in x < 7.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student confuses the conjunction 'and' with the disjunction 'or', taking the union of the two intervals instead of their intersection.

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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