Students frequently misapply algebraic operations when isolating a variable that appears on both sides of a rational expression. Given the equation $\frac{a x + b}{c x + d} = k$, where $k \neq \frac{a}{c}$ and $c x + d \neq 0$, solve explicitly for $x$ by correctly collecting all like terms involving $x$ on one side and factoring.

Students frequently misapply algebraic operations when isolating a variable that appears on both sides of a rational expression. Given the equation $\frac{a x + b}{c x + d} = k$ , where $k \neq \frac{a}{c}$ and $c x + d \neq 0$ , solve explicitly for $x$ by correctly collecting all like terms involving $x$ on one side and factoring.

Final answer

$x = \frac{k d - b}{a - k c}$

Steps

Clear the denominator

To solve for $x$ in the literal equation $\frac{a x + b}{c x + d} = k$ , follow these systematic algebraic steps:

Multiply both sides of the equation by the non-zero denominator $(c x + d)$ : $a x + b = k(c x + d)$

Expand using the distributive property

Distribute $k$ to every term inside the parentheses on the right side: $a x + b = k c x + k d$

Common Pitfall: A frequent mistake is distributing $k$ to only the first term, writing $k c x + d$ . Ensure $k$ is multiplied by both $c x$ and $d$ .

Rearrange terms to group all terms containing $x$ together

To isolate $x$ , collect all terms involving $x$ on one side (typically the left) and all remaining terms without $x$ on the opposite side:

Subtract $k c x$ from both sides: $a x - k c x + b = k d$
Subtract $b$ from both sides: $a x - k c x = k d - b$

Factor out the common variable $x$

Notice that $x$ is a common factor in both terms on the left-hand side. Factor $x$ out: $x(a - k c) = k d - b$

Solve for $x$ by division

Divide both sides by the binomial coefficient $(a - k c)$ . Since the problem specifies that $k \neq \frac{a}{c}$ , it is guaranteed that $a - k c \neq 0$ , making division valid: $x = \frac{k d - b}{a - k c}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to distribute k to the second term d when multiplying both sides by (cx + d), resulting in an incorrect constant term.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$ax + b = k(cx + d)$
Step 2
$ax + b = kcx + d$
Correct
$ax + b = kcx + kd$
When expanding $k(cx + d)$ with the distributive property, the factor $k$ must multiply every term inside the parentheses. Distributing $k$ to $d$ gives $+ kd$ , not $+ d$ .
Step 3
$ax - kcx = d - b \implies x(a - kc) = d - b$
Correct
$ax - kcx = kd - b \implies x(a - kc) = kd - b$
Your algebraic logic for grouping all terms containing $x$ onto one side and factoring out $x$ was executed correctly, but the right-hand side inherited the missing factor $k$ from the previous distribution error.
Step 4
$x = \frac{d - b}{a - kc}$
Correct
$x = \frac{kd - b}{a - kc}$
You divided by the coefficient $(a - kc)$ correctly, but the final numerator is incorrect due to the propagated expansion error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by $(cx + d)$ .

You grouped all terms containing the target variable $x$ onto one side of the equation.

You correctly factored out the common factor $x$ from the binomial $(ax - kcx)$ .

You isolated $x$ by dividing both sides by $(a - kc)$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete Distributive Property

You multiplied $k$ by $cx$ to get $kcx$ , but forgot to multiply $k$ by $d$ , writing $d$ instead of $kd$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Partial distribution across parentheses: treating the multiplier outside parentheses as applying only to the immediately adjacent leading term rather than distributing to every term inside the grouping.

REASONING PATH

Your ACEpath teacher’s suggested approach

Full Parentheses Expansion: For any terms $k(A + B)$ , multiplication distributes over addition: $k(A + B) = k \cdot A + k \cdot B$ .
Draw distribution arcs from the external factor to each separate term inside parentheses before writing down the expanded expression.
Term-Count Verification: Multiplying a single factor by a binomial with two terms must always produce an expression with two terms, each carrying that factor.
Count the terms inside the parentheses and verify that every resulting term contains the distributed factor.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the distributive property fully to the right-hand side: $ax + b = kcx + kd$ .
Subtract $kcx$ and subtract $b$ on both sides to collect like terms: $ax - kcx = kd - b$ .
Factor out the common variable $x$ : $x(a - kc) = kd - b$ .
Divide both sides by $(a - kc)$ to isolate $x$ .

Final answer

$x = \frac{kd - b}{a - kc}$

NEXT STEPS

Final Takeaways from Your Teacher

Draw distribution arrows from the outside factor to each term inside parentheses before writing the expanded form.
Practice distributing constants and variables over binomials and trinomials with literal symbols.
Plug a simple set of numbers (such as $a=1, b=0, c=1, d=1, k=2$ ) into both the original equation and your final isolated expression to check consistency.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when transposing terms across the equals sign, adding b instead of subtracting it.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$ax + b = kcx + kd$
Step 2
$ax - kcx = kd + b$
Correct
$ax - kcx = kd - b$
When moving $+b$ from the left-hand side to the right-hand side, you must subtract $b$ from both sides, which yields $kd - b$ rather than $kd + b$ .
Step 3
$x(a - kc) = kd + b$
Correct
$x(a - kc) = kd - b$
You correctly factored out the common variable $x$ , but this step carries forward the sign error from the previous line.
Step 4
$x = \frac{kd + b}{a - kc}$
Correct
$x = \frac{kd - b}{a - kc}$
You isolated $x$ by dividing both sides by $(a - kc)$ , but the final numerator remains affected by the earlier sign slip.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by $(cx + d)$ .

You distributed $k$ across both terms on the right side without dropping any factors.

You correctly factored out $x$ and divided by its coefficient binomial to isolate the variable.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign Error in Transposition

You wrote $+b$ instead of $-b$ on the right side when moving the constant term across the equals sign.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dropping the inverse operation during multi-term rearrangement: terms moved across the equals sign keep their original operation instead of applying subtraction to cancel an addition.

REASONING PATH

Your ACEpath teacher’s suggested approach

Inverse Operation Check: To eliminate $+b$ from one side of an equation, apply its inverse operation (subtract $b$ ) to both sides.
Write out the explicit balance operation ' $- b$ ' directly under both sides before simplifying to the next line.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $kcx$ and subtract $b$ from both sides: $ax - kcx = kd - b$
Factor out $x$ on the left side: $x(a - kc) = kd - b$
Divide both sides by $(a - kc)$ : $x = \frac{kd - b}{a - kc}$

Final answer

$x = \frac{kd - b}{a - kc}$

NEXT STEPS

Final Takeaways from Your Teacher

Always write out the intermediate balancing step (such as ' $-b$ ') beneath both sides when transposing multiple terms simultaneously.
Substitute simple test values (for example $a=2, b=3, c=1, d=1, k=1$ ) into both your original expression and final formula to verify that the signs stay consistent.

OverviewACEpath Teacher's Diagnosis

03 The student attempts to divide by the coefficient of x before collecting all x-terms on one side, failing to isolate x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$ax + b = kcx + kd$
Step 2
$ax = kcx + kd - b$
Step 3
$x = \frac{kcx + kd - b}{a}$
Correct
$ax - kcx = kd - b \implies x(a - kc) = kd - b \implies x = \frac{kd - b}{a - kc}$
You divided by $a$ while the variable $x$ was still present on the right-hand side in $kcx$ . To solve explicitly for $x$ , all terms containing $x$ must be grouped on one side and factored out before dividing.

WHAT YOU DID WELL

Compliments from Your Teacher

You cleared the rational denominator correctly by multiplying both sides by $(cx + d)$ .

You applied the distributive property accurately to obtain $kcx + kd$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Premature division without isolating variable

Dividing by $a$ to isolate $x$ while another term with $x$ ( $kcx$ ) remains on the other side of the equation.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating isolation as simply dividing by the visible leading coefficient of one term, rather than recognizing that all variable terms must be collected together and factored out before dividing by a single numerical or variable coefficient.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Variable Presence: An explicit solution for $x$ cannot contain $x$ in the expression on the other side.
Scan both sides of your final expression to ensure $x$ appears only once, by itself, on one side of the equals sign.
Collect Before Factoring: When a target variable appears in multiple terms across an equation, move all terms containing that variable to one side and all other terms to the opposite side.
Subtract $kcx$ from both sides to get $ax - kcx = kd - b$ , then factor $x(a - kc)$ before dividing.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract $kcx$ from both sides to collect all terms with $x$ on the left side: $ax - kcx = kd - b$
Factor the common factor $x$ out of the left-hand side: $x(a - kc) = kd - b$
Divide both sides by the coefficient binomial $(a - kc)$ to isolate $x$ : $x = \frac{kd - b}{a - kc}$

Final answer

$x = \frac{kd - b}{a - kc}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the requirement that an explicit solution for a variable must not express that variable in terms of itself.
Practice moving all variable terms to one side of the equals sign before dividing.
Drill factoring out the common variable $x(A - B)$ from expressions of the form $Ax - Bx$ .

Lock it in

Question

What does the distributive property state for an expression of the form $k(A + B)$ ?

Answer

$k(A + B) = kA + kB$ . The factor outside the parentheses must multiply every term inside.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to distribute k to the second term d when multiplying both sides by (cx + d), resulting in an incorrect constant term.

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 makes a sign error when transposing terms across the equals sign, adding b instead of subtracting it.

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 attempts to divide by the coefficient of x before collecting all x-terms on one side, failing to isolate x.

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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