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Consider the rational literal equation representing total output resistance in a feedback network: Rout=R1R2−kR1+R2R_{\text{out}} = \frac{R_1 R_2 - k}{R_1 + R_2}. Solve explicitly for R1R_1 in terms of RoutR_{\text{out}}, R2R_2, and kk. State the algebraic boundary condition (restriction) on the parameters for which the solution for R1R_1 is mathematically defined.

Answer and steps

Final answer

R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}, with boundary restriction R2≠RoutR_2 \neq R_{\text{out}}

Steps

01

Clear the fraction

To solve for the target variable R1R_1, we follow a standard algebraic process for rational literal equations: clear fractions, group like terms involving R1R_1, factor, and isolate R1R_1.

Multiply both sides of the equation by the denominator (R1+R2)(R_1 + R_2), assuming R1+R2≠0R_1 + R_2 \neq 0:

Rout(R1+R2)=R1R2−kR_{\text{out}}(R_1 + R_2) = R_1 R_2 - k

02

Expand and distribute

Expand the left-hand side:

RoutR1+RoutR2=R1R2−kR_{\text{out}} R_1 + R_{\text{out}} R_2 = R_1 R_2 - k

03

Collect terms containing R1R_1 on one side

Gather all terms containing R1R_1 on the right-hand side and all terms without R1R_1 on the left-hand side by subtracting RoutR1R_{\text{out}} R_1 and adding kk to both sides:

RoutR2+k=R1R2−RoutR1R_{\text{out}} R_2 + k = R_1 R_2 - R_{\text{out}} R_1

04

Factor out the variable R1R_1

Factor R1R_1 from both terms on the right-hand side:

RoutR2+k=R1(R2−Rout)R_{\text{out}} R_2 + k = R_1(R_2 - R_{\text{out}})

05

Isolate R1R_1

Divide both sides by (R2−Rout)(R_2 - R_{\text{out}}):

R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

(Note: Writing this as R1=−RoutR2−kRout−R2R_1 = \frac{-R_{\text{out}} R_2 - k}{R_{\text{out}} - R_2} is algebraically equivalent, but keeping the denominator positive or naturally ordered as R2−RoutR_2 - R_{\text{out}} avoids extra negative signs.)

Algebraic Boundary Restriction

Because division by zero is undefined in real arithmetic, the denominator in the solved expression cannot be zero:

R2−Rout≠0  ⟹  R2≠RoutR_2 - R_{\text{out}} \neq 0 \implies R_2 \neq R_{\text{out}}

Thus, the explicit solution is: R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}} with the restriction: R2≠RoutR_2 \neq R_{\text{out}}

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student fails to distribute R_out to both terms when multiplying by the denominator, omitting R_out from the R_2 term.

Sample Student Work

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

    Rout(R1+R2)=R1R2−kR_{\text{out}}(R_1 + R_2) = R_1 R_2 - k

  2. Step 2

    RoutR1+R2=R1R2−kR_{\text{out}} R_1 + R_2 = R_1 R_2 - k

    Correct

    RoutR1+RoutR2=R1R2−kR_{\text{out}} R_1 + R_{\text{out}} R_2 = R_1 R_2 - k

    When distributing RoutR_{\text{out}} across (R1+R2)(R_1 + R_2), you multiplied R1R_1 by RoutR_{\text{out}} but left R2R_2 unmultiplied instead of writing RoutR2R_{\text{out}} R_2.

  3. Step 3

    R2+k=R1R2−RoutR1  ⟹  R2+k=R1(R2−Rout)R_2 + k = R_1 R_2 - R_{\text{out}} R_1 \implies R_2 + k = R_1(R_2 - R_{\text{out}})

    Correct

    RoutR2+k=R1R2−RoutR1  ⟹  RoutR2+k=R1(R2−Rout)R_{\text{out}} R_2 + k = R_1 R_2 - R_{\text{out}} R_1 \implies R_{\text{out}} R_2 + k = R_1(R_2 - R_{\text{out}})

    Your algebraic technique for grouping R1R_1 terms on one side and factoring out R1R_1 was executed correctly, but the line carried forward the missing RoutR_{\text{out}} factor from the previous distribution error.

  4. Step 4

    R1=R2+kR2−RoutR_1 = \frac{R_2 + k}{R_2 - R_{\text{out}}}

    Correct

    R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

    You accurately divided both sides by the factored coefficient (R2−Rout)(R_2 - R_{\text{out}}), but the numerator remained missing the factor RoutR_{\text{out}}.

  5. Step 5

    R2−Rout≠0  ⟹  R2≠RoutR_2 - R_{\text{out}} \neq 0 \implies R_2 \neq R_{\text{out}}

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by the common denominator (R1+R2)(R_1 + R_2).

You collected all terms containing the target variable R1R_1 on one side and factored it out cleanly.

You identified the denominator restriction to ensure the expression is mathematically defined.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Incomplete distribution over parentheses

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WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the distributive property as applying primarily to the first term inside the parentheses, forgetting to multiply the outer factor RoutR_{\text{out}} by the second term R2R_2.

REASONING PATH
Your ACEpath teacher’s suggested approach
Distributive Property Check

  • Every single term inside the parentheses must be multiplied by the factor outside: a(b+c)=ab+aca(b + c) = ab + ac.

  • Draw distribution arcs from the outside multiplier to each individual term inside the parentheses before writing down the expanded line.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Distribute RoutR_{\text{out}} to both terms: Rout⋅R1+Rout⋅R2=R1R2−kR_{\text{out}} \cdot R_1 + R_{\text{out}} \cdot R_2 = R_1 R_2 - k

  2. Rearrange to isolate terms with R1R_1 on one side: RoutR2+k=R1R2−RoutR1R_{\text{out}} R_2 + k = R_1 R_2 - R_{\text{out}} R_1

  3. Factor out R1R_1: RoutR2+k=R1(R2−Rout)R_{\text{out}} R_2 + k = R_1(R_2 - R_{\text{out}})

  4. Divide by the coefficient (R2−Rout)(R_2 - R_{\text{out}}): R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

Final answer

R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}, with restriction R2≠RoutR_2 \neq R_{\text{out}}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the distributive property whenever expanding multi-term factors.

  • Practice checking each term after expanding by mentally dividing back the factor.

  • Work through similar literal equations that require clearing fractions and factoring the target variable.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when rearranging terms to collect R_1, subtracting k instead of adding it.

Sample Student Work

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

    Rout(R1+R2)=R1R2−k  ⟹  RoutR1+RoutR2=R1R2−kR_{\text{out}}(R_1 + R_2) = R_1 R_2 - k \implies R_{\text{out}} R_1 + R_{\text{out}} R_2 = R_1 R_2 - k

  2. Step 2

    RoutR1−R1R2=−RoutR2−kR_{\text{out}} R_1 - R_1 R_2 = -R_{\text{out}} R_2 - k

  3. Step 3

    R1(Rout−R2)=RoutR2−kR_1(R_{\text{out}} - R_2) = R_{\text{out}} R_2 - k

    Correct

    R1(Rout−R2)=−RoutR2−kR_1(R_{\text{out}} - R_2) = -R_{\text{out}} R_2 - k

    While you correctly factored R1R_1 from the left side, you dropped the leading negative sign on −RoutR2-R_{\text{out}} R_2 on the right side.

  4. Step 4

    R1=RoutR2−kRout−R2R_1 = \frac{R_{\text{out}} R_2 - k}{R_{\text{out}} - R_2}

    Correct

    R1=−RoutR2−kRout−R2=RoutR2+kR2−RoutR_1 = \frac{-R_{\text{out}} R_2 - k}{R_{\text{out}} - R_2} = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

    You correctly divided by the coefficient (Rout−R2)(R_{\text{out}} - R_2), but the expression inherited the dropped negative sign from the previous step.

  5. Step 5

    Rout−R2≠0  ⟹  Rout≠R2R_{\text{out}} - R_2 \neq 0 \implies R_{\text{out}} \neq R_2

WHAT YOU DID WELL
Compliments from Your Teacher

You cleared the rational denominator and expanded the parentheses accurately.

You correctly isolated all terms containing the target variable on one side of the equation.

You identified the non-zero denominator condition to establish the correct parameter restriction.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Dropped sign during intermediate line copy

You accidentally omitted the negative sign in front of RoutR2R_{\text{out}} R_2 when writing down the right-hand side of line 4.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Losing visual track of leading signs across multi-symbol expressions when focusing attention primarily on factoring the other side.

REASONING PATH
Your ACEpath teacher’s suggested approach
Side-by-Side Term Audit

  • Every term on the unaltered side of an equation must retain its exact sign and value when performing an operation on the opposite side.

  • Before proceeding to divide, pause and point your pen at each term on the non-factored side to verify its sign matches the preceding line.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. RoutR1−R1R2=−RoutR2−kR_{\text{out}} R_1 - R_1 R_2 = -R_{\text{out}} R_2 - k

  2. R1(Rout−R2)=−(RoutR2+k)R_1(R_{\text{out}} - R_2) = -(R_{\text{out}} R_2 + k)

  3. R1=−(RoutR2+k)Rout−R2=RoutR2+kR2−RoutR_1 = \frac{-(R_{\text{out}} R_2 + k)}{R_{\text{out}} - R_2} = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

Final answer

R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}, with restriction R2≠RoutR_2 \neq R_{\text{out}}

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice checking signs by factoring out a negative from the entire side: write −A−B-A - B as −(A+B)-(A + B) immediately to prevent dropped minus signs.

  • Alternatively, gather terms with the target variable on the side that keeps positive coefficients to avoid dealing with multiple leading negative signs.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly identifies the boundary condition from the original denominator R_1 + R_2 ≠ 0 instead of the solved parameter constraint R_2 - R_out ≠ 0.

Sample Student Work

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

    Rout(R1+R2)=R1R2−k  ⟹  RoutR1+RoutR2=R1R2−kR_{\text{out}}(R_1 + R_2) = R_1 R_2 - k \implies R_{\text{out}} R_1 + R_{\text{out}} R_2 = R_1 R_2 - k

  2. Step 2

    RoutR2+k=R1R2−RoutR1  ⟹  RoutR2+k=R1(R2−Rout)R_{\text{out}} R_2 + k = R_1 R_2 - R_{\text{out}} R_1 \implies R_{\text{out}} R_2 + k = R_1(R_2 - R_{\text{out}})

  3. Step 3

    R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}

  4. Step 4

    R1+R2≠0  ⟹  R1≠−R2R_1 + R_2 \neq 0 \implies R_1 \neq -R_2

    Correct

    R2−Rout≠0  ⟹  R2≠RoutR_2 - R_{\text{out}} \neq 0 \implies R_2 \neq R_{\text{out}}

    A restriction on the parameters for an explicit solution R1=f(Rout,R2,k)R_1 = f(R_{\text{out}}, R_2, k) must restrict the input parameters themselves, not the isolated variable R1R_1. The explicit solution is defined when the denominator of the solved expression is non-zero, meaning R2−Rout≠0R_2 - R_{\text{out}} \neq 0.

WHAT YOU DID WELL
Compliments from Your Teacher

You cleared the rational denominator and expanded parentheses accurately.

You collected like terms containing the target variable on one side.

You factored out R1R_1 and isolated it to achieve the correct explicit formula.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Parameter Domain vs. Original Domain

You checked the denominator of the initial equation instead of the denominator of your final solved formula.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing the domain restriction of the initial implicit relation with the parameter constraint required for the isolated solution to exist. When expressing y=f(x,a)y = f(x, a), the solution is mathematically defined only when the formula for yy does not divide by zero, requiring the denominator in terms of the parameters to be non-zero.

REASONING PATH
Your ACEpath teacher’s suggested approach
Inspect the final explicit expression

  • For a formula A=BCA = \frac{B}{C}, the solution exists if and only if C≠0C \neq 0.

  • Set the denominator of the solved expression not equal to zero to identify the parameter boundary condition.

Check parameter variables

  • A parameter restriction for R1R_1 must not contain R1R_1 itself.

  • Verify that your stated restriction contains only the given parameters (RoutR_{\text{out}}, R2R_2, kk).

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Observe the denominator of the explicit formula for R1R_1: the denominator is (R2−Rout)(R_2 - R_{\text{out}}).

  2. Apply the non-zero denominator condition for real division: R2−Rout≠0R_2 - R_{\text{out}} \neq 0.

  3. Isolate the parameter relation: R2≠RoutR_2 \neq R_{\text{out}}.

Final answer

R1=RoutR2+kR2−RoutR_1 = \frac{R_{\text{out}} R_2 + k}{R_2 - R_{\text{out}}}, with boundary condition R2≠RoutR_2 \neq R_{\text{out}}

NEXT STEPS
Final Takeaways from Your Teacher

  • Review the distinction between original rational domain restrictions and parameter solvability conditions.

  • When asked for a condition on parameters, ensure the variable being solved for does not appear in your condition.

  • Practice identifying undefined values directly from the denominator of the final isolated expression.

Lock it in

Question

State the distributive property for multiplying a single factor across a binomial sum: a(b+c)=?a(b + c) = \text{?}

Answer

ab+acab + ac. Every term inside the parentheses must be multiplied by the external factor.

Similar question

01

Solve explicitly for xx in terms of aa, bb, and yy: y=x−3x+ay = \frac{x - 3}{x + a}. State any restriction on the parameters.

Answer

x=ay+31−yx = \frac{ay + 3}{1 - y} (or x=−ay−3y−1x = \frac{-ay - 3}{y - 1}) with restriction y≠1y \neq 1. Key steps: multiply both sides by (x+a)(x + a) to get y(x+a)=x−3y(x + a) = x - 3, distribute fully to obtain yx+ya=x−3yx + ya = x - 3, collect terms ya+3=x−yx=x(1−y)ya + 3 = x - yx = x(1 - y), and divide by (1−y)(1 - y).

02

Solve explicitly for m1m_1 in the gravitational-like formula: F=G(m1m2+c)m1+dF = \frac{G(m_1 m_2 + c)}{m_1 + d}. State the boundary restriction.

Answer

m1=Gc−FdF−Gm2m_1 = \frac{Gc - Fd}{F - Gm_2} (or m1=Fd−GcGm2−Fm_1 = \frac{Fd - Gc}{Gm_2 - F}) with restriction F≠Gm2F \neq Gm_2. Key steps: clear the denominator to get F(m1+d)=G(m1m2+c)F(m_1 + d) = G(m_1 m_2 + c), distribute both sides to yield Fm1+Fd=Gm1m2+GcFm_1 + Fd = Gm_1 m_2 + Gc, group terms containing m1m_1 on one side as Fd−Gc=Gm1m2−Fm1=m1(Gm2−F)Fd - Gc = Gm_1 m_2 - Fm_1 = m_1(Gm_2 - F), and divide.

03

In an optical system, the focal relation is given by M=fs−Ls+fM = \frac{f s - L}{s + f}. Solve explicitly for ff in terms of MM, ss, and LL, and state the parameter restriction.

Answer

f=Ms+Ls−Mf = \frac{Ms + L}{s - M} with restriction s≠Ms \neq M. Key steps: multiply both sides by (s+f)(s + f) to get M(s+f)=fs−LM(s + f) = fs - L, expand to obtain Ms+Mf=fs−LMs + Mf = fs - L, collect terms with ff on one side as Ms+L=fs−Mf=f(s−M)Ms + L = fs - Mf = f(s - M), and divide by (s−M)(s - M).

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