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Solve the parametric linear inequality εx−3≤5x+7\varepsilon x - 3 \le 5x + 7, where ε\varepsilon is a real parameter approaching 55 from the left (i.e., ε→5−\varepsilon \to 5^-). Express the solution set in terms of ε\varepsilon in interval notation, and determine the limiting behavior of the lower bound as ε→5−\varepsilon \to 5^-.

Answer and steps

Final answer

Solution set: [10ε−5,∞)\left[ \frac{10}{\varepsilon - 5}, \infty \right); Limiting behavior of lower bound: lim⁡ε→5−10ε−5=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = -\infty](streamdown:incomplete-link)

Steps

01

Collect variable terms and constants on opposite sides

To solve the linear inequality for xx, follow these steps:

Start with the given inequality: εx−3≤5x+7\varepsilon x - 3 \le 5x + 7

Subtract 5x5x from both sides and add 33 to both sides: εx−5x≤7+3\varepsilon x - 5x \le 7 + 3

Factor out xx on the left-hand side: (ε−5)x≤10(\varepsilon - 5)x \le 10

02

Determine the sign of the coefficient (ε−5)(\varepsilon - 5)

We are given that ε\varepsilon approaches 55 from the left (ε→5−\varepsilon \to 5^-), which means ε<5\varepsilon < 5.

Subtracting 55 from both sides gives: ε−5<0\varepsilon - 5 < 0

Key Pitfall: When dividing or multiplying both sides of an inequality by a negative quantity, the inequality sign must reverse direction.

03

Solve for xx and write in interval notation

Divide both sides by (ε−5)(\varepsilon - 5), reversing the inequality sign from ≤\le to ≥\ge: x≥10ε−5x \ge \frac{10}{\varepsilon - 5}

In interval notation, the solution set is: [10ε−5,∞)\left[ \frac{10}{\varepsilon - 5}, \infty \right)

---](streamdown:incomplete-link)

04

Determine the limiting behavior of the lower bound

The lower bound of the solution interval is: L(ε)=10ε−5L(\varepsilon) = \frac{10}{\varepsilon - 5}

As ε→5−\varepsilon \to 5^-:

  • The numerator remains constant at 10>010 > 0.
  • The denominator (ε−5)(\varepsilon - 5) approaches 00 through strictly negative values (denoted as 0−0^-).

Taking the one-sided limit: lim⁡ε→5−10ε−5=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = -\infty

Thus, as ε\varepsilon approaches 55 from the left, the lower bound moves infinitely far to the left on the real number line, expanding the solution set to all real numbers (−∞,∞)(-\infty, \infty) in the limit.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets that ε - 5 is negative when ε approaches 5 from the left and fails to reverse the inequality sign upon division.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    \varepsilon x - 5x ≤\le 7 + 3   ⟹  \implies (\varepsilon - 5)x ≤\le 10

  2. Step 2

    x ≤\le 10ε−5\frac{10}{\varepsilon - 5}

    Correct

    x ≥\ge 10ε−5\frac{10}{\varepsilon - 5}

    Because ε→5−\varepsilon \to 5^-, we know ε<5\varepsilon < 5, which makes the divisor ε−5\varepsilon - 5 strictly negative. Dividing an inequality by a negative quantity requires reversing the inequality sign from ≤\le to ≥\ge.

  3. Step 3

    (−∞,10ε−5]\left(-\infty, \frac{10}{\varepsilon - 5}\right]

    Correct

    [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right)](streamdown:incomplete-link)

    This interval followed your unreversed inequality x≤10ε−5x \le \frac{10}{\varepsilon - 5}. Since the corrected inequality is x≥10ε−5x \ge \frac{10}{\varepsilon - 5}, the value 10ε−5\frac{10}{\varepsilon - 5} is actually the lower bound, giving [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right).](streamdown:incomplete-link)

  4. Step 4

    As ε→5−, ε−5→0−  ⟹  lim⁡[10ε−5]=−∞\text{As } \varepsilon \to 5^-, \ \varepsilon - 5 \to 0^- \implies \lim \left[\frac{10}{\varepsilon - 5}\right] = -\infty

WHAT YOU DID WELL
Compliments from Your Teacher

You collected the variable terms and factored out xx accurately.

You correctly recognized that ε−5→0−\varepsilon - 5 \to 0^- as ε→5−\varepsilon \to 5^-.

You evaluated the one-sided limit of the rational boundary expression with complete accuracy.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Parametric Inequality Sign Reversal

Dividing both sides of (ε−5)x≤10(\varepsilon - 5)x \le 10 by the negative quantity (ε−5)(\varepsilon - 5) without changing the direction of the inequality sign.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating a variable parameter coefficient algebraically without checking whether its limiting condition ( ε<5 \,\varepsilon < 5\,) forces it to be strictly negative before performing division.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the sign before dividing

  • If k<0k < 0, dividing kx≤ck x \le c by kk yields x≥ckx \ge \frac{c}{k}.

  • Always write down whether the divisor is positive or negative next to the step before carrying out the division.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Observe that ε→5−\varepsilon \to 5^- implies ε<5\varepsilon < 5, so ε−5<0\varepsilon - 5 < 0.

  2. Divide (ε−5)x≤10(\varepsilon - 5)x \le 10 by the negative expression (ε−5)(\varepsilon - 5) and flip the inequality sign: x≥10ε−5x \ge \frac{10}{\varepsilon - 5}.

  3. Express this solution as the half-bounded interval [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right).](streamdown:incomplete-link)

  4. Take the limit of the lower bound: lim⁡ε→5−10ε−5=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = -\infty.

Final answer

Solution set: [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right); Limiting behavior of lower bound: lim⁡ε→5−10ε−5=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = -\infty](streamdown:incomplete-link)

NEXT STEPS
Final Takeaways from Your Teacher

  • Before dividing or multiplying an inequality by any algebraic factor containing a parameter, explicitly write down its sign.

  • Practice identifying whether one-sided limits like t→a−t \to a^- or t→a+t \to a^+ make a difference expression (t−a)(t - a) positive or negative.

OverviewACEpath Teacher's Diagnosis

02 The student correctly finds the lower bound but misinterprets 10/0^- as positive infinity instead of negative infinity.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (ε−5)x≤10(\varepsilon - 5)x \le 10

  2. Step 2

    Since \text{Since } \varepsilon < 5,   \varepsilon - 5 < 0   ⟹  \implies x ≥\ge 10ε−5\frac{10}{\varepsilon - 5}

  3. Step 3

    [10ε−5,∞)\left[ \frac{10}{\varepsilon - 5}, \infty \right)   ⟹  \implies Lower bound is \text{Lower bound is } 10ε−5\frac{10}{\varepsilon - 5}](streamdown:incomplete-link)

  4. Step 4

    As \text{As } \varepsilon →\to 5^-,   10ε−5\frac{10}{\varepsilon - 5} →\to 100\frac{10}{0} = ∞\infty

    Correct

    $\lim_{\varepsilon →\to 5^-}$ \frac{10}{\varepsilon - 5} = -\infty_

    Because ε→5−\varepsilon \to 5^-, we have ε<5\varepsilon < 5, so the denominator ε−5\varepsilon - 5 approaches 00 through negative values (0−0^-). A positive numerator divided by small negative values approaches −∞-\infty, not +∞+\infty.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly collected terms and factored the linear inequality.

You recognized that ε−5<0\varepsilon - 5 < 0 and properly flipped the inequality sign.

You expressed the solution set in interval notation and extracted the lower bound.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
One-Sided Limit Sign Error

You wrote 100=∞\frac{10}{0} = \infty without accounting for the fact that the denominator approaches zero from the negative side.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating one-sided limits of rational expressions as dividing by an unsigned zero rather than tracking the sign of the approaching values (0−0^- vs. 0+0^+).

REASONING PATH
Your ACEpath teacher’s suggested approach
Track Denominator Sign in One-Sided Limits

  • For a non-zero numerator c>0c > 0, if denominator →0−\to 0^-, the quotient approaches −∞-\infty; if denominator →0+\to 0^+, the quotient approaches +∞+\infty.

  • Check the sign of (ε−5)(\varepsilon - 5) when ε<5\varepsilon < 5: since ε−5<0\varepsilon - 5 < 0, evaluate 100−=−∞\frac{10}{0^-} = -\infty.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Observe that as ε→5−\varepsilon \to 5^-, ε<5\varepsilon < 5, which means ε−5<0\varepsilon - 5 < 0.

  2. The denominator approaches 00 through strictly negative numbers (0−0^-).

  3. Calculate lim⁡ε→5−10ε−5=100−=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = \frac{10}{0^-} = -\infty.

Final answer

Solution set: [10ε−5,∞)\left[ \frac{10}{\varepsilon - 5}, \infty \right); Limiting behavior of lower bound: −∞-\infty](streamdown:incomplete-link)

NEXT STEPS
Final Takeaways from Your Teacher

  • Whenever taking a limit where the denominator goes to 00, always explicitly write whether it approaches 0+0^+ or 0−0^-.

  • Apply the standard sign rule: positivenegative=negative\frac{\text{positive}}{\text{negative}} = \text{negative} to conclude infinite limits.

OverviewACEpath Teacher's Diagnosis

03 The student mistakenly treats the non-strict inequality as strict, resulting in an open interval for the solution set.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    (ε−5)x≤10  ⟹  ε−5<0 because ε<5(\varepsilon - 5)x \le 10 \implies \varepsilon - 5 < 0 \text{ because } \varepsilon < 5

  2. Step 2

    x>10ε−5x > \frac{10}{\varepsilon - 5}

    Correct

    x≥10ε−5x \ge \frac{10}{\varepsilon - 5}

    When dividing both sides of a non-strict inequality (≤\le) by a negative number, the direction reverses to ≥\ge, not >>. The boundary point remains included.

  3. Step 3

    (10ε−5,∞)\left(\frac{10}{\varepsilon - 5}, \infty\right)

    Correct

    [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right)](streamdown:incomplete-link)

    Because of the dropped equality in the previous step, your interval notation used a parenthesis rather than a square bracket for the included lower bound.

  4. Step 4

    As ε→5−,10ε−5→−∞\text{As } \varepsilon \to 5^-, \frac{10}{\varepsilon - 5} \to -\infty

WHAT YOU DID WELL
Compliments from Your Teacher

You collected like terms and factored out xx correctly.

You correctly deduced that ε−5<0\varepsilon - 5 < 0 because ε\varepsilon approaches 55 from below.

You correctly recognized that dividing by a negative quantity reverses the inequality direction.

You correctly analyzed the one-sided limit of the lower bound as ε→5−\varepsilon \to 5^-.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Strictness toggle error

You dropped the equality condition when reversing the inequality sign, writing >> instead of ≥\ge.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing sign reversal with switching between strict and non-strict inequality types. Reversing an inequality flips the orientation (≤\le becomes ≥\ge, and << becomes >>), but preserves whether the boundary value is included.

REASONING PATH
Your ACEpath teacher’s suggested approach
Preserve Equality Inclusivity

  • Dividing or multiplying by a negative number flips direction: ≤⟷≥\le \longleftrightarrow \ge and <⟷>< \longleftrightarrow >. It never changes a non-strict inequality into a strict one.

  • Always check whether the original inequality includes equality (≤\le or ≥\ge); if it does, keep the line under the inequality symbol after reversing.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Start from (ε−5)x≤10(\varepsilon - 5)x \le 10.

  2. Divide both sides by (ε−5)<0(\varepsilon - 5) < 0, flipping ≤\le to ≥\ge: x≥10ε−5x \ge \frac{10}{\varepsilon - 5}.

  3. Write the solution set in interval notation with a closed bracket: [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right).](streamdown:incomplete-link)

  4. Evaluate the limit of the lower bound: lim⁡ε→5−10ε−5=−∞\lim_{\varepsilon \to 5^-} \frac{10}{\varepsilon - 5} = -\infty.

Final answer

Solution set: [10ε−5,∞)\left[\frac{10}{\varepsilon - 5}, \infty\right); Limiting behavior of lower bound: −∞-\infty](streamdown:incomplete-link)

NEXT STEPS
Final Takeaways from Your Teacher

  • Double-check that every ≤\le remains non-strict (≥\ge) whenever you multiply or divide by a negative quantity.

  • Verify interval endpoints: non-strict inequalities always require closed brackets [[ or ]] at finite boundaries.

Lock it in

Question

When solving an inequality of the form Ax≤BA x \le B, under what condition must the direction of the inequality sign flip?

Answer

Flip the sign whenever dividing or multiplying by AA if A<0A < 0. If A>0A > 0, the sign stays the same.

Similar question

01

Solve the inequality (a−2)x>6(a - 2)x > 6 for xx, given that a<2a < 2. Express the solution in interval notation.

Answer

Since a<2a < 2, we have a−2<0a - 2 < 0. Dividing both sides by the negative quantity (a−2)(a - 2) reverses the inequality sign: x<6a−2x < \frac{6}{a - 2}. In interval notation: (−∞,6a−2)\left(-\infty, \frac{6}{a - 2}\right).

02

Solve the inequality (3−m)x+4≥10(3 - m)x + 4 \ge 10, where m→3+m \to 3^+. Express the solution set in interval notation and evaluate the limit of the bound as m→3+m \to 3^+.

Answer

Subtract 44 to get (3−m)x≥6(3 - m)x \ge 6. Since m→3+m \to 3^+, m>3m > 3, so 3−m<03 - m < 0. Dividing by (3−m)(3 - m) flips the inequality: x≤63−mx \le \frac{6}{3 - m}. The solution set is (−∞,63−m]\left(-\infty, \frac{6}{3 - m}\right]. As m→3+m \to 3^+, 3−m→0−3 - m \to 0^-, so the upper bound satisfies lim⁡m→3+63−m=−∞\lim_{m \to 3^+} \frac{6}{3 - m} = -\infty.

03

Solve the compound linear inequality system {(k−1)x≤4,2x+1>5}\lbrace (k - 1)x \le 4, \quad 2x + 1 > 5 \rbrace for xx in terms of kk, where k→1−k \to 1^-. State the interval notation for the resulting solution set.

Answer

From the second inequality, 2x>4  ⟹  x>22x > 4 \implies x > 2. For the first inequality, since k→1−k \to 1^-, k<1k < 1, which means k−1<0k - 1 < 0. Dividing by (k−1)(k - 1) flips the sign: x≥4k−1x \ge \frac{4}{k - 1}. As k→1−k \to 1^-, 4k−1→−∞\frac{4}{k - 1} \to -\infty, meaning 4k−1<2\frac{4}{k - 1} < 2 for all kk sufficiently close to 11. The intersection of x≥4k−1x \ge \frac{4}{k - 1} and x>2x > 2 is simply x>2x > 2. In interval notation: (2,∞)(2, \infty).

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