Linear Equations in Disguise

Intuition

You already solve equations like $2x + 3 = 11$ , where the variable sits in plain sight. Many equations hide that same simple core under a layer of clutter. The clutter may be a pile of fractions, a nest of parentheses, a scatter of decimals, or a variable stuck inside a denominator. None of these is a new kind of equation. Each one is an ordinary linear equation wearing a disguise. The whole job is to strip the disguise away until the familiar $ax + b = c$ is looking back at you. Clear the fractions, distribute the parentheses, clear the decimals, and every tool from the last lesson finishes the work. Along the way one genuinely new question turns up, and it is the most interesting part of the lesson. What does it mean when the variable disappears entirely as you simplify?

Learning goals

Clear fractions by multiplying every term by the LCD
Multiply through by a power of ten to clear decimals
Cross-multiply a proportion, which just scales both sides by $bd$
Name the excluded values before solving with a variable denominator
Read a cancelled variable as no solution or infinitely many
Ask which equation a verdict belongs to after clearing

Clearing the fractions

A fraction in an equation is intimidating only because of its denominator. Remove every denominator at once and what remains is an equation in whole-number pieces you already know how to solve. The tool is the multiplication property of equality from the last lesson: multiplying both sides of an equation by the same nonzero number produces an equivalent equation. Choose that number well and it wipes out all the denominators in a single step.

The number to choose is the least common denominator (LCD) of the fractions in the equation, the smallest number every denominator divides into evenly. Multiply both sides by the LCD, distribute it across every term, and each denominator cancels. Take $\dfrac{x}{2} + \dfrac{x}{3} = 5$ . The denominators are $2$ and $3$ , so the LCD is $6$ . Multiply both sides by $6$ and hand it to each term:

6 \cdot \frac{x}{2} + 6 \cdot \frac{x}{3} = 6 \cdot 5, \qquad 3x + 2x = 30.

The denominators are gone, and $5x = 30$ gives $x = 6$ . The one habit that makes this reliable is carrying the multiplier to every term, on both sides, not just to the fractions. The constant $5$ had no denominator, yet it still had to be multiplied by $6$ , because the property multiplies whole sides, not selected pieces.

Why multiplying by the LCD keeps the same solutions#

Multiplying both sides of an equation $L = R$ by a number $m$ gives $mL = mR$ . As long as $m \neq 0$ this is an equivalent equation, for the same reason every property of equality is reversible. That reversal is a division by that same nonzero $m$ . Dividing both sides of $mL = mR$ by $m$ returns you to $L = R$ , so neither equation has a solution the other lacks. That is exactly the multiplication property you proved last lesson, now used with $m$ equal to the LCD.

An LCD is a product of the denominators’ factors, and a denominator is never zero. So the LCD is a nonzero number, and the step is legal and reversible. Why does it clear the fractions? By construction the LCD is a multiple of every denominator, so for each term the denominator divides the LCD exactly and leaves a whole number behind. And because multiplication distributes over addition, multiplying an entire side by the LCD is the same as multiplying each of its terms by the LCD. Put those two facts together and every denominator cancels at once, with nothing left over. That is why one well-chosen multiplication turns a tangle of fractions into a clean whole-number equation.

When a numerator is itself a sum, such as $x + 1$ , keep it wrapped in parentheses as the LCD lands on it. That way the multiplication reaches both terms of the numerator, and any following subtraction reaches both as well.

Worked example 1 Solve $\dfrac{x+1}{2} - \dfrac{x-3}{5} = 2$

The denominators are $2$ and $5$ , so the LCD is $10$ . Multiply every term by $10$ , keeping each numerator in parentheses:

10 \cdot \frac{x+1}{2} - 10 \cdot \frac{x-3}{5} = 10 \cdot 2, \qquad 5(x+1) - 2(x-3) = 20.

Now distribute, and watch the minus sign in front of the second product reach both of its terms:

5x + 5 - 2x + 6 = 20, \qquad 3x + 11 = 20.

Subtract $11$ , then divide by $3$ :

3x = 9, \qquad x = 3.

Check in the original equation: $\dfrac{3+1}{2} - \dfrac{3-3}{5} = \dfrac{4}{2} - 0 = 2$ , which matches. The $-3$ turning into $+6$ was the step to guard: treating the subtraction as reaching only the $x$ would have thrown the constant off.

Multiplying both sides by the LCD 12 lands on every term, both sides, turning x over 3 plus x over 4 equals 7 into 4x plus 3x equals 84. Multiplying only some of the terms is the classic error.

Check your understanding

Solve $\dfrac{x}{2} + \dfrac{x}{5} = 7$ .

x = 10

x = 7

x = 35

x = 70

Clearing parentheses and decimals

Parentheses are handled by the skill from the expanding lesson: distribute to remove each one, then combine like terms. When groups are nested inside groups, work from the innermost outward, clearing one layer at a time before touching the next. Only after a side is fully simplified do you start moving terms across the equals sign.

Worked example 2 Solve $2\big(3x - (x - 4)\big) = 28$

Start with the inner group. Subtracting $(x - 4)$ flips both of its terms, so $3x - (x - 4) = 3x - x + 4 = 2x + 4$ . The equation becomes

2(2x + 4) = 28.

Distribute the outer $2$ , then solve the two-step equation that appears:

4x + 8 = 28, \qquad 4x = 20, \qquad x = 5.

Check in the original: $2\big(3(5) - (5 - 4)\big) = 2\big(15 - 1\big) = 2 \cdot 14 = 28$ . Peeling the inner parentheses first is what kept the signs straight.

Decimals clear the same way, only now the useful multiplier is a power of ten. Multiplying by $10$ shifts every digit one place to the left, so $10 \cdot 0.1x = x$ and $10 \cdot 0.5 = 5$ ; one decimal place disappears for each factor of ten. Pick the smallest power of ten that turns every coefficient into a whole number, $10$ when the numbers reach tenths and $100$ when any reaches hundredths. Multiplying both sides by it is again the multiplication property with a nonzero multiplier, so the solution is untouched, and a whole-number equation is left in its place.

Worked example 3 Solve $0.5x - 0.2 = 0.1x + 1$

Every number reaches the tenths place, so multiply both sides by $10$ to clear the decimals:

10(0.5x - 0.2) = 10(0.1x + 1), \qquad 5x - 2 = x + 10.

Collect the variable on the left by subtracting $x$ , then finish:

4x - 2 = 10, \qquad 4x = 12, \qquad x = 3.

Check in the original: the left is $0.5(3) - 0.2 = 1.5 - 0.2 = 1.3$ and the right is $0.1(3) + 1 = 0.3

1 = 1.3 $, so$ x = 3$.

Proportions and cross-multiplication

A proportion is an equation stating that two ratios are equal, such as $\dfrac{a}{b} = \dfrac{c}{d}$ . You met these in pre-algebra and solved them with a shortcut called cross-multiplication: multiply each numerator by the other side’s denominator to get $ad = bc$ . That shortcut is not a rule to trust on faith; it is a single application of the moves you already own.

Why cross-multiplication is legal#

A proportion $\dfrac{a}{b} = \dfrac{c}{d}$ is an equation like any other, with $b \neq 0$ and $d \neq 0$ so that both fractions are defined. To clear both denominators in one stroke, multiply both sides by the product $bd$ . That product is nonzero, since neither factor is zero, so the multiplication property of equality applies and the step can be undone.

On the left, the $b$ in the denominator cancels the $b$ in $bd$ , leaving

bd \cdot \frac{a}{b} = \frac{bd}{b} \cdot a = d \cdot a = ad.

On the right, the $d$ cancels in the same way, giving $bd \cdot \dfrac{c}{d} = bc$ . So the proportion becomes

ad = bc.

That is the whole of cross-multiplication. The name comes from the picture of each numerator reaching across to the far denominator, but the justification is nothing more exotic than multiplying both sides by $bd$ . Because the step reverses (divide both sides by $bd$ to return), the proportion and $ad = bc$ are equivalent, so solving one solves the other.

Worked example 4 Solve $\dfrac{x}{6} = \dfrac{x+4}{9}$

Cross-multiply, sending each numerator across to the opposite denominator:

9 \cdot x = 6 \cdot (x + 4), \qquad 9x = 6x + 24.

The result is an ordinary linear equation with the variable on both sides. Subtract $6x$ , then divide by $3$ :

3x = 24, \qquad x = 8.

Check the proportion at $x = 8$ : the left ratio is $\dfrac{8}{6} = \dfrac{4}{3}$ and the right ratio is $\dfrac{8 + 4}{9} = \dfrac{12}{9} = \dfrac{4}{3}$ . The two ratios agree, so $x = 8$ .

When the variable sits in the denominator

An equation can hide a variable in a denominator, as in $\dfrac{6}{x} = 2$ or $\dfrac{x+1}{x-2} = 3$ . The plan is unchanged: clear the denominator by multiplying both sides by it, which leaves a linear equation. One extra habit comes with it. Because a denominator can never be zero, first find the excluded values, the values of the variable that would make a denominator zero, and set them aside. Then solve, and at the end confirm your answer is not one of those excluded values.

For $\dfrac{6}{x} = 2$ the only excluded value is $x = 0$ . Multiplying both sides by $x$ gives $6 = 2x$ , so $x = 3$ , and since $3$ is not the excluded $0$ , it is a genuine solution.

Worked example 5 Solve $\dfrac{x+1}{x-2} = 3$

The denominator $x - 2$ is zero when $x = 2$ , so record the restriction $x \neq 2$ before doing any algebra. This is already a proportion, $\dfrac{x+1}{x-2} = \dfrac{3}{1}$ , so cross-multiply:

x + 1 = 3(x - 2), \qquad x + 1 = 3x - 6.

Collect the variable and solve:

7 = 2x, \qquad x = \frac{7}{2}.

The excluded value was $x = 2$ , and $\dfrac{7}{2} = 3.5$ is not $2$ , so the answer survives the check. Substituting confirms it: $\dfrac{\frac{7}{2} + 1}{\frac{7}{2} - 2} = \dfrac{\frac{9}{2}}{\frac{3}{2}} = \dfrac{9}{3} = 3$ .

The excluded-value check is not a formality. Once in a while the value you solve for is exactly one of the values that was ruled out, and then it has to be thrown away. Consider $\dfrac{x}{x-3} = \dfrac{3}{x-3}$ , where $x = 3$ is excluded. Multiplying both sides by $x - 3$ gives $x = 3$ , but that is the forbidden value, so it cannot count as a solution, and the equation in fact has no solution. Rejecting a candidate that lands on an excluded value is a first taste of an idea a later chapter develops in full. For now, simply run the check every time a variable appears in a denominator.

Check your understanding

Solve $\dfrac{10}{x + 1} = 2$ , given $x \neq -1$ .

x = 4

x = 1

x = 6

D No solution

One solution, no solution, or infinitely many

Almost every equation so far has settled on exactly one solution. The exception was the last one, whose only candidate turned out to be an excluded value. The candidate had to be thrown away, leaving the equation with no solution at all. That failure came from the denominator. There is a second way for an equation to miss having exactly one solution, and it has nothing to do with denominators. Once you clear a linear equation’s disguise and gather the variable on one side and the constants on the other, the variable can cancel completely. What is left is a bare statement about numbers, and that statement tells you which of three cases you are in.

Watch it happen. In $2x + 1 = 2x + 3$ , subtracting $2x$ from both sides removes the variable and leaves

1 = 3,

which is false. In $2(x + 1) = 2x + 2$ , the left side is $2x + 2$ , and subtracting $2x$ leaves

2 = 2,

which is true. The first equation demands something impossible, so no value of $x$ can satisfy it. The second says something that holds no matter what, so every value of $x$ satisfies it. The next proof explains exactly why the leftover statement is the whole story.

Why a vanished variable settles the number of solutions#

Clear the disguise from a linear equation, multiplying through by nonzero numbers only. Then gather every variable term on one side and every constant on the other, just as in the last lesson. Whatever you began with, the result has the shape

ax = b,

where $a$ is the net coefficient of the variable and $b$ is the net constant, both definite numbers. Everything now hinges on whether $a$ is zero.

If $a \neq 0$ , you may divide both sides by $a$ , and the cleared equation has the single solution $x = \dfrac{b}{a}$ . This is the ordinary case: one value and no other.

If $a = 0$ , the variable has vanished, because $0 \cdot x$ equals $0$ for every value of $x$ . The equation now reads $0 = b$ , with no $x$ left in it to adjust, so the constant $b$ decides. When $b \neq 0$ , the statement $0 = b$ is simply false, and no choice of $x$ can rescue a numeric falsehood, so the equation has no solution. When $b = 0$ , the statement $0 = 0$ is true for every $x$ , so every number is a solution and there are infinitely many.

This is why you can read the verdict straight off the leftover numbers the instant the variable cancels. A false line like $1 = 3$ means the disguise hid a contradiction, true for nothing; a true line like $0 = 0$ means it hid an identity, true for everything. And if the variable never cancels, the cleared equation names one value and no other. These are the same three kinds of equation you first met in this subject as conditional equations, contradictions, and identities.

Notice what the opening line of this proof was doing. Every step here rewrites one equation as another with exactly the same solutions, and multiplying through by a nonzero number is what buys that. A number like $12$ is not zero once and for all, whatever $x$ turns out to be. So the multiplication undoes itself by division, and nothing can be gained or lost. Clear a denominator that holds the variable and you have multiplied by something you cannot certify. The reason is that $x - 3$ is a perfectly good nonzero number at almost every value of $x$ and is zero at one of them. There the step does not undo, and the verdict above is then a verdict about the equation you cleared to, not yet about the one you started with. That gap is exactly what the excluded-value check closes. The same gap is why the equation $\dfrac{x}{x-3} = \dfrac{3}{x-3}$ earlier in this lesson has no solution, even though its variable survived the clearing and named the single value $x = 3$ .

A linear equation has one of three fates. Left: the variable survives, giving a single value. Middle: it cancels to a false statement, so no value works. Right: it cancels to a true statement, so every value works.

The disguise can make these cases easy to miss, because the cancellation only shows up after you simplify. The next two examples look like ordinary equations until the variable slips away.

Worked example 6 Solve $6x - 2(x - 1) = 4x + 5$

Distribute the $-2$ on the left and combine like terms:

6x - 2x + 2 = 4x + 5, \qquad 4x + 2 = 4x + 5.

Both sides carry $4x$ , so subtract $4x$ from both. The variable cancels and leaves

2 = 5.

This statement is false, and no value of $x$ can make $2$ equal $5$ , so the equation has no solution. Note the trap: the answer is not $x = 0$ or $x = 2$ . A cancelled variable with a false remainder means there is no solution at all.

Worked example 7 Solve $5x - (x - 4) = 4(x + 1)$

Clear the parentheses on each side, letting the minus sign reach both terms of $x - 4$ :

5x - x + 4 = 4x + 4, \qquad 4x + 4 = 4x + 4.

The two sides are identical. Subtract $4x$ from both and the variable disappears, leaving

4 = 4,

which is true for every value of $x$ . The equation is an identity, so every real number is a solution, and there are infinitely many. Written honestly, the solution is “all real numbers,” not a single value.

Check your understanding

How many solutions does $5x + 3 = 5x - 2$ have?

A Exactly one solution B No solution C Infinitely many solutions D

x = 0

Common mistakes

Multiplying only some terms by the LCD. The multiplier lands on every term of both sides, not just the fractions. Clearing $\tfrac{x}{2} + \tfrac{x}{3} = 5$ with the LCD $6$ gives $3x + 2x = 30$ ; forgetting to multiply the $5$ leaves the wrong $3x + 2x = 5$ .
Dropping a sign when a numerator is a sum. A fraction bar groups its numerator, so $-\tfrac{x-3}{5}$ multiplied by the LCD becomes $-2(x - 3) = -2x + 6$ , not $-2x - 6$ . Keep each numerator in parentheses as the LCD arrives.
Cross-multiplying something that is not a single ratio equal to a single ratio. Cross-multiplication applies to $\tfrac{a}{b} = \tfrac{c}{d}$ . If a side is a sum of fractions or has extra terms, clear the denominators with the LCD first, then solve.
Forgetting a solution can be an excluded value. When the variable sits in a denominator, list the excluded values first and check your answer against them. If the value you find is one that was excluded, reject it; the equation has no solution there.
Calling a contradiction " $x = 0$ ". When the variable cancels and leaves a false statement such as $2 = 5$ , the answer is “no solution,” never $x = 0$ . Likewise a true statement such as $0 = 0$ means every value works, not “no solution.”

Key takeaways

Equations with fractions, parentheses, decimals, or a variable in a denominator are ordinary linear equations in disguise. Simplify each side to reach the form $ax + b = c$ and solve as before.
Clear fractions by multiplying both sides by the LCD, which is legal because the LCD is a nonzero number. Distribute it to every term so no denominator is left behind.
Clear decimals by multiplying both sides by the power of ten ( $10$ , $100$ , and so on) that makes every coefficient a whole number.
A proportion $\tfrac{a}{b} = \tfrac{c}{d}$ becomes $ad = bc$ by cross-multiplication, which is just multiplying both sides by $bd$ .
With a variable in a denominator, name the excluded values first, then solve, then confirm the answer is not excluded.
A linear equation has one solution when the variable survives ( $ax = b$ with $a \neq 0$ ). It has no solution when the variable cancels to a false statement (a contradiction), and infinitely many when it cancels to a true statement (an identity).
That verdict belongs to the equation you cleared to. It is the original’s verdict as well when every multiplier you cleared with was a number, because such a step undoes itself. Clear with something holding the variable and the value that survives still has to face the excluded-value check.

Linear Equations in Disguise

Learning goals

Clearing the fractions

Clearing parentheses and decimals

Proportions and cross-multiplication

When the variable sits in the denominator

One solution, no solution, or infinitely many

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)