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In the rational linear equation ax+12x+5=32\frac{a x + 1}{2x + 5} = \frac{3}{2}, determine the behavior of the solution xx as the positive parameter aa approaches the critical asymptotic limit a→3+a \to 3^+. Express the solution xx in terms of aa and state the limiting value lim⁡a→3+x\lim_{a \to 3^+} x.

Answer and steps

Final answer

x=132a−6x = \frac{13}{2a - 6}; lim⁡a→3+x=+∞\lim_{a \to 3^+} x = +\infty

Steps

To determine the behavior of the solution xx, first solve the equation for xx in terms of the parameter aa.

Step 1: Clear the denominators

Given the equation: ax+12x+5=32\frac{a x + 1}{2x + 5} = \frac{3}{2}

Cross-multiply, assuming the denominator 2x+5≠02x + 5 \neq 0: 2(ax+1)=3(2x+5)2(a x + 1) = 3(2x + 5)

Step 2: Expand and isolate terms containing xx

Distribute the constants on both sides: 2ax+2=6x+152ax + 2 = 6x + 15

Subtract 6x6x and 22 from both sides to group the xx-terms on the left: 2ax−6x=15−22ax - 6x = 15 - 2

Factor out xx on the left-hand side: x(2a−6)=13x(2a - 6) = 13

Step 3: Solve for xx

Assuming 2a−6≠02a - 6 \neq 0 (that is, a≠3a \neq 3): x=132a−6x = \frac{13}{2a - 6}

Check for extraneous solutions: The excluded value for xx is given by 2x+5=0  ⟹  x=−522x + 5 = 0 \implies x = -\frac{5}{2}. Setting 132a−6=−52\frac{13}{2a - 6} = -\frac{5}{2} yields −10a+30=26  ⟹  a=25=0.4-10a + 30 = 26 \implies a = \frac{2}{5} = 0.4. Since we are examining values near a=3a = 3, x=−52x = -\frac{5}{2} does not interfere with the domain.

Step 4: Evaluate the limit as a→3+a \to 3^+

As aa approaches 33 from the right (a>3a > 3):

  • The numerator remains a fixed positive constant: 1313.
  • The denominator 2a−62a - 6 is strictly positive and approaches 00: 2a−6→0+2a - 6 \to 0^+

Dividing a positive constant by a quantity approaching zero from the positive side yields: lim⁡a→3+x=lim⁡a→3+132a−6=+∞\lim_{a \to 3^+} x = \lim_{a \to 3^+} \frac{13}{2a - 6} = +\infty

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student makes a sign error when moving constant terms across the equality, obtaining 17 instead of 13 in the numerator.

Sample Student Work

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

    2(ax + 1) = 3(2x + 5)   ⟹  \implies 2ax + 2 = 6x + 15

  2. Step 2

    2ax - 6x = 15 + 2

    Correct

    2ax - 6x = 15 - 2

    When subtracting 22 from both sides to isolate terms, the sign must change: +2+2 on the left becomes −2-2 on the right, giving 15−2=1315 - 2 = 13, not 15+2=1715 + 2 = 17.

  3. Step 3

    2(a - 3)x = 17   ⟹  \implies x = 172(a−3)\frac{17}{2(a - 3)}

    Correct

    2(a - 3)x = 13   ⟹  \implies x = 132(a−3)\frac{13}{2(a - 3)}

    Your factoring and division method was algebraically sound, but the value carries forward the incorrect constant 1717 from the previous step.

  4. Step 4

    $\lim_{a →\to 3^+}$ x = +\infty_

WHAT YOU DID WELL
Compliments from Your Teacher

You cleared the rational equation correctly by cross-multiplying denominators.

You expanded terms correctly across parentheses using the distributive property.

You factored out xx systematically as 2(a−3)x2(a - 3)x.

You correctly reasoned that a positive constant divided by 0+0^+ diverges to +∞+\infty.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Transposition sign error

You wrote +2+2 instead of −2-2 on the right side when moving the constant term across the equality sign.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Losing track of the subtraction property of equality during multi-step transposition, inadvertently adding the term rather than inverting its sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Inverting Signs across the Equals Sign

  • Adding a term to one side requires subtracting that same term from the other side: A+c=B  ⟹  A=B−cA + c = B \implies A = B - c.

  • Circle positive constant terms before moving them and explicitly write the subtraction step on both sides.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Subtract 22 from both sides: 2ax−6x=15−2=132ax - 6x = 15 - 2 = 13.

  2. Factor the left-hand side: 2(a−3)x=132(a - 3)x = 13.

  3. Divide by 2(a−3)2(a - 3): x=132a−6x = \frac{13}{2a - 6}.

Final answer

x = 132a−6\frac{13}{2a - 6};  $\lim_{a →\to 3^+}$ x = +\infty_

NEXT STEPS
Final Takeaways from Your Teacher

  • Write out the explicit balance operation (e.g., −2-2 under both sides) whenever moving constants to prevent sign flips.

  • Substitute a simple test value for aa (such as a=4a = 4) into both your isolated expression and the original equation to verify arithmetic accuracy.

OverviewACEpath Teacher's Diagnosis

02 The student ignores the directional one-sided approach from the right and misinterprets division by zero as undefined rather than approaching positive infinity.

Sample Student Work

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

    2(ax+1) = 3(2x+5)   ⟹  \implies 2ax + 2 = 6x + 15

  2. Step 2

    2(a-3)x = 13   ⟹  \implies x = 132(a−3)\frac{13}{2(a-3)}

  3. Step 3

    $\lim_{a →\to 3^+} x = \frac{13}{0} \implies \\lim_{a \to 3^+}xx\text{ does not exist (undefined)}$_

    Correct

    $\lim_{a →\to 3^+} x = \\lim_{a \to 3^+} \frac{13}{2(a-3)}=+ = +\infty$_

    When evaluating a limit of the form c0\frac{c}{0} with c>0c > 0, direct substitution gives an indeterminate asymptotic form rather than an undefined arithmetic value; as a→3+a \to 3^+, the factor a−3a - 3 approaches 00 through positive values (0+0^+), which drives the positive quotient to +∞+\infty.

WHAT YOU DID WELL
Compliments from Your Teacher

You cleared denominators and correctly distributed terms on both sides of the equation.

You factored and isolated the variable xx in terms of the parameter aa without any algebraic errors.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misinterpreting infinite one-sided limits as undefined arithmetic

You substituted a=3a = 3 directly into the expression to obtain 130\frac{13}{0} and concluded the limit is merely undefined or does not exist.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing arithmetic division by zero (which is undefined) with asymptotic limit behavior of the form k0+\frac{k}{0^+} (where k>0k > 0), which describes growth toward +∞+\infty.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the sign of approaching zero

  • For a non-zero numerator k>0k > 0, lim⁡kf(t)=+∞\lim \frac{k}{f(t)} = +\infty if f(t)→0+f(t) \to 0^+, and −∞-\infty if f(t)→0−f(t) \to 0^-.

  • Determine whether the denominator approaches zero through strictly positive or strictly negative numbers as the variable approaches the target from the specified side.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Observe that as a→3+a \to 3^+, a>3a > 3, meaning a−3>0a - 3 > 0.

  2. The denominator 2(a−3)2(a - 3) approaches 00 through positive values: 2(a−3)→0+2(a - 3) \to 0^+.

  3. Dividing the fixed positive numerator 1313 by increasingly tiny positive quantities yields values growing without bound.

Final answer

x = 132a−6\frac{13}{2a - 6};  $\lim_{a →\to 3^+}$ x = +\infty_

NEXT STEPS
Final Takeaways from Your Teacher

  • Review one-sided limits of rational functions where the denominator approaches zero while the numerator is non-zero.

  • Practice tracking the sign of approaching quantities (0+0^+ vs. 0−0^-) to distinguish between +∞+\infty, −∞-\infty, and bilateral two-sided non-existence.

OverviewACEpath Teacher's Diagnosis

03 The student forgets to distribute the factor of 2 to the constant term on the left-hand side, leading to an incorrect expression for x.

Sample Student Work

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

    2(ax+1)=3(2x+5)2(ax + 1) = 3(2x + 5)

  2. Step 2

    2ax+1=6x+15  ⟹  2(a−3)x=142ax + 1 = 6x + 15 \implies 2(a - 3)x = 14

    Correct

    2ax+2=6x+15  ⟹  x(2a−6)=132ax + 2 = 6x + 15 \implies x(2a - 6) = 13

    When distributing 22 across (ax+1)(ax + 1), you multiplied 2⋅ax=2ax2 \cdot ax = 2ax but did not multiply 2⋅12 \cdot 1, writing +1+1 instead of +2+2. This resulted in 15−1=1415 - 1 = 14 rather than 15−2=1315 - 2 = 13 on the constant side.

  3. Step 3

    x=142(a−3)=7a−3x = \frac{14}{2(a - 3)} = \frac{7}{a - 3}

    Correct

    x=132a−6x = \frac{13}{2a - 6}

    You solved for xx correctly using your previous equation, but the result inherited the distribution error from the previous step.

  4. Step 4

    lim⁡a→3+x=+∞\lim_{a \to 3^+} x = +\infty

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly cleared the fractions by cross-multiplying both sides.

You correctly isolated the variable term xx by factoring out (a−3)(a - 3).

You correctly analyzed the behavior of the one-sided limit as the parameter approached the vertical asymptote.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Distributive Property Omission

You distributed the coefficient 22 to the variable term axax but forgot to multiply it by the constant term 11.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Incomplete distribution across parentheses often occurs when focusing primarily on grouping the variable term, leaving trailing constants unmultiplied.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distribute to Every Term

  • For any expression c(u+v)c(u + v), the factor cc must multiply every term inside the parentheses: c(u+v)=cu+cvc(u + v) = cu + cv.

  • Draw distribution arrows or mentally check that every term inside the parentheses is multiplied by the outside factor before dropping brackets.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Distribute 22 to both terms on the left side: 2(ax+1)=2ax+22(ax + 1) = 2ax + 2.

  2. Set the expanded sides equal: 2ax+2=6x+152ax + 2 = 6x + 15.

  3. Subtract 6x6x and 22 from both sides: 2ax−6x=15−2=132ax - 6x = 15 - 2 = 13.

  4. Factor out xx: x(2a−6)=13x(2a - 6) = 13, which gives x=132a−6x = \frac{13}{2a - 6}.

Final answer

x=132a−6x = \frac{13}{2a - 6}; lim⁡a→3+x=+∞\lim_{a \to 3^+} x = +\infty

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice expanding linear expressions with parentheses to ensure every term receives the outside factor.

  • Verify your algebraic steps by substituting a simple test value for aa (such as a=4a = 4) into both the original and simplified equations.

Lock it in

Question

When transposing a positive constant +c+c from the left side of an equation to the right side, what operation must be applied?

Answer

Subtract cc from both sides (+c+c becomes −c-c). Remember: moving a term across an equals sign always inverts its sign.

Similar question

01

Solve the equation for xx in terms of kk: kx+3x−2=4\frac{k x + 3}{x - 2} = 4.

Answer

x=−11k−4x = \frac{-11}{k - 4} (or 114−k\frac{11}{4 - k}). Steps: Cross-multiply to get kx+3=4(x−2)=4x−8kx + 3 = 4(x - 2) = 4x - 8. Subtract 4x4x and subtract 33 from both sides: kx−4x=−8−3=−11kx - 4x = -8 - 3 = -11. Factor and divide: x(k−4)=−11x(k - 4) = -11.

02

Given the equation mx−43x+1=2\frac{m x - 4}{3x + 1} = 2, express xx in terms of mm and determine lim⁡m→6+x\lim_{m \to 6^+} x.

Answer

x=6m−6x = \frac{6}{m - 6} and lim⁡m→6+x=+∞\lim_{m \to 6^+} x = +\infty. Steps: Cross-multiply: mx−4=2(3x+1)=6x+2mx - 4 = 2(3x + 1) = 6x + 2. Add 44 to both sides and subtract 6x6x: (m−6)x=2+4=6(m - 6)x = 2 + 4 = 6, giving x=6m−6x = \frac{6}{m - 6}. As m→6+m \to 6^+, the denominator approaches 0+0^+ while the numerator is positive 66, so the limit is +∞+\infty.

03

Solve the rational equation bx+5x+1=3\frac{b x + 5}{x + 1} = 3 for xx. Then, evaluate both one-sided limits as bb approaches the critical value 33: lim⁡b→3+x\lim_{b \to 3^+} x and lim⁡b→3−x\lim_{b \to 3^-} x.

Answer

x=−2b−3x = \frac{-2}{b - 3} (or 23−b\frac{2}{3 - b}). Limits: lim⁡b→3+x=−∞\lim_{b \to 3^+} x = -\infty and lim⁡b→3−x=+∞\lim_{b \to 3^-} x = +\infty. Steps: Cross-multiply to obtain bx+5=3x+3bx + 5 = 3x + 3. Subtract 55 and 3x3x from both sides: (b−3)x=3−5=−2(b - 3)x = 3 - 5 = -2, so x=−2b−3x = \frac{-2}{b - 3}. As b→3+b \to 3^+, −20+=−∞\frac{-2}{0^+} = -\infty. As b→3−b \to 3^-, −20−=+∞\frac{-2}{0^-} = +\infty.

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