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An engineer reorganizes the linear thermal expansion relation for a composite structural strut, Lf=L0(1+αΔT)L_f = L_0 (1 + \alpha \Delta T), into an explicit expression for the temperature change ΔT=Lf−L0αL0\Delta T = \frac{L_f - L_0}{\alpha L_0}. Reverse engineer this formulation to solve for the initial baseline length L0L_0 entirely in terms of the final length LfL_f, linear expansion coefficient α\alpha, and temperature change ΔT\Delta T.

Answer and steps

Final answer

L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

Steps

01

Clear the denominator

To solve for the initial baseline length L0L_0 in terms of LfL_f, α\alpha, and ΔT\Delta T, we can reverse engineer the expression ΔT=Lf−L0αL0\Delta T = \frac{L_f - L_0}{\alpha L_0}.

Multiply both sides of the equation by αL0\alpha L_0 to eliminate the fraction: ΔT⋅(αL0)=Lf−L0\Delta T \cdot (\alpha L_0) = L_f - L_0 αL0ΔT=Lf−L0\alpha L_0 \Delta T = L_f - L_0

02

Collect all terms containing L0L_0 on one side

Add L0L_0 to both sides so that every term with the variable L0L_0 is gathered on the left side: L0+αL0ΔT=LfL_0 + \alpha L_0 \Delta T = L_f

03

Factor out the target variable L0L_0

Since L0L_0 is a common factor in both terms on the left side, factor it out using the distributive property: L0(1+αΔT)=LfL_0(1 + \alpha \Delta T) = L_f

04

Isolate L0L_0

Divide both sides by the binomial expression (1+αΔT)(1 + \alpha \Delta T): L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

Verification: Starting directly from the original relation Lf=L0(1+αΔT)L_f = L_0(1 + \alpha \Delta T), dividing both sides directly by (1+αΔT)(1 + \alpha \Delta T) immediately yields: L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

Common Pitfall: When rearranging αL0ΔT=Lf−L0\alpha L_0 \Delta T = L_f - L_0, learners sometimes forget to factor out L0L_0 and instead try to divide by terms still containing L0L_0, leading to an incomplete isolation like L0=Lf−αL0ΔT1L_0 = \frac{L_f - \alpha L_0 \Delta T}{1}. Remember to gather all terms involving the unknown on one side before factoring.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student subtracts the term containing L_0 rather than factoring, leaving the target variable L_0 on both sides of the final expression.

Sample Student Work

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

    L_f = L_0 + α\alpha L_0 Δ\Delta T

  2. Step 2

    L_0 = L_f - α\alpha L_0 Δ\Delta T

    Correct

    L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

    Subtracting αL0ΔT\alpha L_0 \Delta T from both sides does not solve for L0L_0 because L0L_0 still appears on the right side. When the unknown appears in multiple terms, factor it out as a common factor instead of moving one term over.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly recognized that L0L_0 distributes across the binomial (1+αΔT)(1 + \alpha \Delta T) to yield Lf=L0+αL0ΔTL_f = L_0 + \alpha L_0 \Delta T.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Incomplete variable isolation

You isolated one instance of L0L_0 on the left side while leaving another instance of L0L_0 on the right side.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Treating a literal equation as solved simply because the target variable is isolated on one side, without ensuring that the target variable does not appear on the opposite side. When a variable appears in multiple terms, you must group them and factor out the common variable.

REASONING PATH
Your ACEpath teacher’s suggested approach
Definition of Explicit Isolation

  • A variable is only solved for if it appears exactly once, on one side of the equation, and nowhere in the other side's expression.

  • Always scan the entire right side after isolating your variable to verify that the target letter does not appear there.

Factoring Shared Variables

  • When the variable to isolate appears in more than one term (ax+bxa x + b x), combine them by factoring: x(a+b)x(a + b).

  • Gather all terms containing the target variable on one side, factor the variable out, and divide by the remaining expression.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Factor the common variable L0L_0 from the terms on the right: Lf=L0(1+αΔT)L_f = L_0(1 + \alpha \Delta T).

  2. Divide both sides by the entire factor (1+αΔT)(1 + \alpha \Delta T) to isolate L0L_0: L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}.

Final answer

L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the rule that a solved variable cannot appear on both sides of the equals sign.

  • Practice factoring out a common variable when solving literal equations.

OverviewACEpath Teacher's Diagnosis

02 The student treats the compound denominator 1 + αΔT as separate additive terms and divides by αΔT before subtracting 1, producing an algebraically invalid isolation.

Sample Student Work

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

    L_f = L_0(1 + α\alpha Δ\Delta T)

  2. Step 2

    LfαΔT=L0+1\frac{L_f}{\alpha \Delta T} = L_0 + 1

    Correct

    L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

    The quantity (1+αΔT)(1 + \alpha \Delta T) is multiplied by L0L_0 as a single composite factor. Dividing only by αΔT\alpha \Delta T does not isolate L0+1L_0 + 1; to eliminate the entire multiplying factor, you must divide the whole equation by (1+αΔT)(1 + \alpha \Delta T).

  3. Step 3

    L_0 = LfαΔT\frac{L_f}{\alpha \Delta T} - 1

    Correct

    L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

    You subtracted 11 logically from your previous expression, but the result is incorrect due to the invalid division in the previous step.

WHAT YOU DID WELL
Compliments from Your Teacher

You identified the correct governing formula connecting all four physical variables.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Invalid isolation of grouped factor

You treated terms inside the parentheses (1+αΔT)(1 + \alpha \Delta T) as independently detachable rather than as a single composite multiplying factor.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Misunderstanding how to undo operations on grouped factors. In A=B(C+D)A = B(C + D), the entire binomial (C+D)(C + D) acts as a single factor multiplying BB, so isolating BB requires dividing both sides by (C+D)(C + D), not dividing by DD and adding or subtracting CC.

REASONING PATH
Your ACEpath teacher’s suggested approach
Treat Parentheses as a Single Unit

  • When an unknown variable is multiplied by a grouped expression B⋅(… )B \cdot (\dots), the entire group is a single multiplier.

  • Divide both sides by the complete grouped factor (1+αΔT)(1 + \alpha \Delta T) in one step: L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Recognize that L0L_0 is multiplied by the complete binomial expression (1+αΔT)(1 + \alpha \Delta T).

  2. Divide both sides of the equation by (1+αΔT)(1 + \alpha \Delta T) to isolate L0L_0 in a single operation: L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

Final answer

L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice identifying composite multiplying factors in literal equations before carrying out inverse operations.

  • Check your rearrangements by substituting simple test numbers (for example, let L0=10L_0 = 10, αΔT=2\alpha \Delta T = 2) to verify that the rearranged equation remains true.

OverviewACEpath Teacher's Diagnosis

03 When expanding or rearranging L_f - L_0 = αL_0ΔT, the student incorrectly factors out L_0 as L_0(αΔT - 1) due to a sign error.

Sample Student Work

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

    ΔT(αL0)=Lf−L0\Delta T (\alpha L_0) = L_f - L_0

  2. Step 2

    αL0ΔT−L0=Lf\alpha L_0 \Delta T - L_0 = L_f

    Correct

    αL0ΔT+L0=Lf\alpha L_0 \Delta T + L_0 = L_f

    To move the term −L0-L_0 from the right side to the left side, you must add L0L_0 to both sides, which gives +L0+L_0 rather than −L0-L_0.

  3. Step 3

    L_0(α\alpha Δ\Delta T - 1) = L_f

    Correct

    L_0(1 + α\alpha Δ\Delta T) = L_f

    Factoring L0L_0 out of your previous expression was algebraically consistent with your line 2, but the binomial factor should be (1+αΔT)(1 + \alpha \Delta T).

  4. Step 4

    L_0 = LfαΔT−1\frac{L_f}{\alpha \Delta T - 1}

    Correct

    L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

    Dividing by the factor was mathematically consistent with your prior lines, but it carries forward the earlier sign error.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by the denominator αL0\alpha L_0.

You recognized that factoring out the common variable L0L_0 is the key strategy to isolate it.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign error during term transposition

When moving −L0-L_0 from the right-hand side to the left-hand side, you kept the minus sign instead of changing it to an addition.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Applying an additive inverse to transpose terms across the equals sign without reversing the negative sign.

REASONING PATH
Your ACEpath teacher’s suggested approach
Additive Inverse Check

  • Adding or subtracting a term across the equals sign flips its sign (A−B=C  ⟹  A=C+BA - B = C \implies A = C + B).

  • Always explicitly write out +L0+ L_0 on both sides when canceling a subtracted term.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Add L0L_0 to both sides: αL0ΔT+L0=Lf\alpha L_0 \Delta T + L_0 = L_f

  2. Factor out L0L_0: L0(αΔT+1)=LfL_0(\alpha \Delta T + 1) = L_f

  3. Divide both sides by (1+αΔT)(1 + \alpha \Delta T): L0=Lf1+αΔTL_0 = \frac{L_f}{1 + \alpha \Delta T}

Final answer

L_0 = Lf1+αΔT\frac{L_f}{1 + \alpha \Delta T}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice moving subtracted terms across an equation by explicitly showing the addition step.

  • Double-check your rearranged equation by expanding it back out to ensure it matches the original line.

Lock it in

Question

What does it mean to solve a literal equation explicitly for a variable xx?

Answer

xx must be completely isolated on one side and cannot appear anywhere in the expression on the other side.

Similar question

01

Solve the equation y=4x+kxy = 4x + kx for xx in terms of yy and kk.

Answer

x=y4+kx = \frac{y}{4 + k}. Factor out xx on the right side: y=x(4+k)y = x(4 + k), then divide both sides by (4+k)(4 + k).

02

The total resistance formula for a branch circuit is given by V=IR1+IR2V = I R_1 + I R_2. Solve this equation for the current II in terms of VV, R1R_1, and R2R_2.

Answer

I=VR1+R2I = \frac{V}{R_1 + R_2}. Factor out II on the right side: V=I(R1+R2)V = I(R_1 + R_2), then divide both sides by (R1+R2)(R_1 + R_2).

03

Solve the rational literal equation m=x−kx+cm = \frac{x - k}{x + c} for xx in terms of mm, kk, and cc.

Answer

x=mc+k1−mx = \frac{mc + k}{1 - m}. Multiply both sides by (x+c)(x + c) to get m(x+c)=x−km(x + c) = x - k, expand to mx+mc=x−kmx + mc = x - k, group terms with xx on one side: mc+k=x−mxmc + k = x - mx, factor out xx: mc+k=x(1−m)mc + k = x(1 - m), and divide by (1−m)(1 - m).

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