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Linear Equations in Disguise: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Three denominators, one multiplier

Solve $\dfrac{2x}{3} - \dfrac{x - 4}{2} = \dfrac{5}{6}$ , and check your value in the original equation.
1. Hint 1
  One multiplication can remove every denominator at once, as long as it reaches every term on both sides, including any term without a variable.
2. Hint 2
  The smallest number that $3$ , $2$ and $6$ all divide into is $6$ . Keep $x - 4$ in parentheses as the $6$ reaches it, because the minus sign in front of that fraction belongs to both of its terms.
3. Hint 3
  After clearing, the $-3$ in front of the parentheses multiplies the $-4$ as well. To check, work out each fraction of the original separately at your value, then combine them over a common denominator.
Answer

$x = -7$ ; check: both sides of the original equation equal $\tfrac{5}{6}$ .

Full solution

The denominators are $3$ , $2$ and $6$ , and the smallest number all three divide into is $6$ .
Multiply every term on both sides by $6$ .
It is a nonzero number, so the new equation has exactly the same solutions as the original.
Keep the numerator $x - 4$ in parentheses as the $6$ reaches it:
$\begin{aligned} 6 \cdot \frac{2x}{3} - 6 \cdot \frac{x - 4}{2} &= 6 \cdot \frac{5}{6} \\ 4x - 3(x - 4) &= 5. \end{aligned}$

Distribute the $-3$ across both terms inside the parentheses, so the $-4$ becomes $+12$ .
Writing $-3x - 12$ there would lose that sign, because the minus in front of the fraction belongs to its whole numerator.
Then collect the $x$ terms and subtract $12$ from both sides:
$\begin{aligned} 4x - 3x + 12 &= 5 \\ x + 12 &= 5 \\ x &= -7. \end{aligned}$

Check in the original equation.
At $x = -7$ the first fraction is $\dfrac{2(-7)}{3}$ , which is $-\dfrac{14}{3}$ , and the second is $\dfrac{-7 - 4}{2}$ , which is $-\dfrac{11}{2}$ .
Subtracting the second from the first over the common denominator $6$ gives
$\begin{aligned} -\frac{14}{3} + \frac{11}{2} &= \frac{-28 + 33}{6} \\ &= \frac{5}{6}, \end{aligned}$
which matches the right side.

Answer

$x = -7$ ; check: both sides of the original equation equal $\tfrac{5}{6}$ .

Key idea

Multiply every term on both sides by the LCD, keeping each numerator in parentheses, so that a minus sign in front of a fraction reaches its whole numerator.
Problem 1: Mark complete
Problem 2 Decimals that reach the hundredths

Solve $0.25x + 1.5 = 0.4x - 0.9$ , and check your value in the original equation.
1. Hint 1
  Every decimal here is a fraction with a power of ten underneath, so one multiplier that is a power of ten can clear all of them at once.
2. Hint 2
  Find the number that runs furthest past the decimal point and count its places. Each factor of ten moves every digit one place, so that count is how many factors of ten you need.
3. Hint 3
  Multiply every term on both sides, the constants included. Then collect the $x$ terms on the side where their coefficient is larger, so that it stays positive.
Answer

$x = 16$ ; check: both sides of the original equation equal $5.5$ .

Full solution

The coefficient $0.25$ runs to the hundredths place, so two factors of ten are needed and the multiplier is $100$ .
A multiplier of $10$ alone would leave $2.5x + 15 = 4x - 9$ , which still has a decimal coefficient.

Multiply every term on both sides by $100$ .
It is a nonzero number, so the solutions do not change:
$25x + 150 = 40x - 90.$

The larger coefficient of $x$ is on the right, so subtract $25x$ from both sides.
Then add $90$ to both sides and divide both sides by $15$ :
$\begin{aligned} 150 &= 15x - 90 \\ 240 &= 15x \\ x &= 16. \end{aligned}$

Check in the original equation.
The left side is $0.25(16) + 1.5$ , which is $4 + 1.5$ , or $5.5$ .
The right side is $0.4(16) - 0.9$ , which is $6.4 - 0.9$ , or $5.5$ .
The two sides agree.

Answer

$x = 16$ ; check: both sides of the original equation equal $5.5$ .

Key idea

To clear decimals, multiply every term by the smallest power of ten that makes each number whole: $100$ when any of them reaches the hundredths.
Problem 2: Mark complete
Problem 3 Two ratios set equal

Solve the proportion $\dfrac{x - 5}{6} = \dfrac{x + 3}{10}$ , and check your value in the original equation.
1. Hint 1
  Each side is a single fraction, so one multiplication of both sides can clear both denominators at once and leave an ordinary linear equation.
2. Hint 2
  Cross-multiplying sends each numerator across to the other side's denominator. Keep each numerator in parentheses so that the multiplier reaches both of its terms.
3. Hint 3
  After expanding, collect the $x$ terms on the side where their coefficient is larger. To check, work out each ratio on its own at your value.
Answer

$x = 17$ ; check: both ratios equal $2$ .

Full solution

Each side is one fraction, so cross-multiplication applies.
It is the same as multiplying both sides by the product of the denominators, $60$ , which is not zero.
On the left the $6$ cancels and leaves $10(x - 5)$ ; on the right the $10$ cancels and leaves $6(x + 3)$ .

Expand both sides.
Subtract $6x$ from both sides, add $50$ to both sides, and divide both sides by $4$ :
$\begin{aligned} 10(x - 5) &= 6(x + 3) \\ 10x - 50 &= 6x + 18 \\ 4x - 50 &= 18 \\ 4x &= 68 \\ x &= 17. \end{aligned}$

Check in the original proportion.
At $x = 17$ the left ratio is $\dfrac{12}{6}$ , which is $2$ , and the right ratio is $\dfrac{20}{10}$ , which is also $2$ .

The smaller multiplier $30$ , the LCD, works too.
It gives $5(x - 5) = 3(x + 3)$ , each side half of the cross-multiplied one, and it leads to the same $x = 17$ .

Answer

$x = 17$ ; check: both ratios equal $2$ .

Key idea

Cross-multiplying a proportion is multiplying both sides by the product of its denominators; when those are nonzero numbers, the solutions are unchanged.
Problem 3: Mark complete
Problem 4 Two equations to unwrap

For each equation below, find every value of $x$ that makes it true.
$\begin{aligned} 0.3(x + 4) - 0.5x &= 1.5 - 0.2x \\ 4x - \big(x - 2(5 - x)\big) &= x + 10 \end{aligned}$
1. Hint 1
  Strip each disguise first, then gather the $x$ terms on one side. Look at whether any $x$ is left at all, and at what the remaining statement says.
2. Hint 2
  Multiply every term of the first equation by $10$ . In the second, simplify the inner group $x - 2(5 - x)$ before removing the outer parentheses, and remember that the minus sign in front of them reaches every term inside.
3. Hint 3
  If the $x$ terms cancel, no choice of $x$ can change the statement that is left. Neither equation has $x$ in a denominator, so a false one means no number works, and a true one means every number does.
Answer

The first equation has no solution. Every number satisfies the second, so it has infinitely many solutions.

Full solution

In the first equation every number reaches the tenths place, so multiply every term on both sides by $10$ , a nonzero number.
The $10$ turns $0.3(x + 4)$ into $3(x + 4)$ .
Then expand and combine:
$\begin{aligned} 3(x + 4) - 5x &= 15 - 2x \\ 3x + 12 - 5x &= 15 - 2x \\ 12 - 2x &= 15 - 2x. \end{aligned}$

Adding $2x$ to both sides removes the variable from both sides at once and leaves $12 = 15$ , which is false.
No value of $x$ can change a statement with no $x$ in it, so the first equation has no solution.
Testing agrees: at $x = 0$ the sides are $1.2$ and $1.5$ , and at $x = 10$ they are $-0.8$ and $-0.5$ .

In the second equation, work from the innermost parentheses outward.
The product $-2(5 - x)$ is $-10 + 2x$ , so the inner group $x - 2(5 - x)$ simplifies to $3x - 10$ .
Subtracting that whole group from $4x$ flips both of its terms:
$\begin{aligned} 4x - (3x - 10) &= x + 10 \\ x + 10 &= x + 10. \end{aligned}$

The two sides are the same expression.
Subtracting $x$ from both sides leaves $10 = 10$ , which is true whatever $x$ is, so every number satisfies the second equation.
Testing agrees: at $x = 0$ both sides are $10$ , and at $x = 1$ both sides are $11$ .

Answer

The first equation has no solution. Every number satisfies the second, so it has infinitely many solutions.

Key idea

When the $x$ terms cancel after clearing with nonzero numbers only, the statement left behind decides: false means no solution, and true means every number is a solution.
Problem 4: Mark complete
Problem 5 Two equations over $x + 3$

Each equation below rules out one value of $x$ . For each, name that value before doing any algebra, then solve the equation and say what becomes of the value you find.
$\begin{aligned} \frac{1}{x + 3} + 2 &= \frac{x + 5}{x + 3} \\ \frac{2}{x + 3} + 2 &= \frac{x + 5}{x + 3} \end{aligned}$
1. Hint 1
  A denominator can never be zero, so a value of $x$ that makes $x + 3$ zero cannot be a solution. Find it first, and keep it in view until the end.
2. Hint 2
  The left side is a sum, so cross-multiplication is not available. Multiply every term on both sides by $x + 3$ instead, the constant $2$ included.
3. Hint 3
  Each cleared equation is an ordinary linear equation. Compare the value it gives with the one you ruled out at the start, and ask whether the original equation says anything at that value.
Answer

Both rule out $x = -3$ . The first has the solution $x = -2$ . The second clears to the candidate $x = -3$ , the ruled-out value, so the candidate is rejected and that equation has no solution.

Full solution

Both equations have the denominator $x + 3$ , which is zero at $x = -3$ .
So $x = -3$ is excluded from both before any algebra starts.

In the first equation the left side is a sum, so multiply every term on both sides by $x + 3$ .
The fractions lose their denominators and the $2$ becomes $2(x + 3)$ :
$\begin{aligned} 1 + 2(x + 3) &= x + 5 \\ 2x + 7 &= x + 5 \\ x + 7 &= 5 \\ x &= -2. \end{aligned}$

The value $-2$ is not the excluded $-3$ , so it stands.
Checking in the original, $x + 3$ is $1$ at $x = -2$ , so the left side is $\tfrac{1}{1} + 2$ , which is $3$ , and the right side is $\tfrac{3}{1}$ , which is also $3$ .

In the second equation the same multiplication gives
$\begin{aligned} 2 + 2(x + 3) &= x + 5 \\ 2x + 8 &= x + 5 \\ x + 8 &= 5 \\ x &= -3. \end{aligned}$

The only candidate is $-3$ , exactly the value ruled out at the start.
Putting it into the original asks for $\tfrac{2}{0}$ , which names no number, so the candidate is rejected.
Any genuine solution of the original would also satisfy the cleared equation, because $x + 3$ is a nonzero number at every allowed value.
The cleared equation has no other solution, so the second equation has no solution at all.

Answer

Both rule out $x = -3$ . The first has the solution $x = -2$ . The second clears to the candidate $x = -3$ , the ruled-out value, so the candidate is rejected and that equation has no solution.

Key idea

With a variable in a denominator, name the excluded value first; a candidate that lands on it is thrown away, and if no other candidate remains, the equation has no solution.
Problem 5: Mark complete
Problem 6 A map drawn to scale

A hiking map is drawn to a fixed scale: $3$ centimeters on the map stands for $8$ kilometers of trail, and the same ratio holds everywhere on the sheet. Write $c$ for a length in centimeters on the map, and $d$ for the number of kilometers of trail it stands for.

Find the trail distance that $10.5$ centimeters on the map stands for, and the length on the map of a $20$ kilometer trail. Then decide which of the proportions $\dfrac{c}{3} = \dfrac{d}{8}$ and $\dfrac{c}{8} = \dfrac{d}{3}$ describes this map.
1. Hint 1
  A fixed scale is one ratio that holds everywhere, so set two ratios equal, matching map length with map length and trail distance with trail distance.
2. Hint 2
  Write the scale as $\dfrac{c}{d} = \dfrac{3}{8}$ . Put the length you know into the position that matches what it measures, then cross-multiply to reach an ordinary linear equation in the other letter.
3. Hint 3
  To judge an arrangement, cross-multiply it and compare the equation it gives with the one that $\dfrac{c}{d} = \dfrac{3}{8}$ gives. Testing the scale's own pair, $c = 3$ with $d = 8$ , is a quick check.
Answer

$10.5$ centimeters stands for $28$ kilometers of trail, and a $20$ kilometer trail is $7.5$ centimeters long on the map. The proportion $\dfrac{c}{3} = \dfrac{d}{8}$ describes this map; $\dfrac{c}{8} = \dfrac{d}{3}$ does not.

Full solution

The scale matches $3$ centimeters of map with $8$ kilometers of trail, and every other pair is in the same ratio.
So
$\frac{c}{d} = \frac{3}{8}.$

For $10.5$ centimeters of map, put $c = 10.5$ .
The distance $d$ is not zero, so cross-multiplying is legal, and it gives
$\begin{aligned} 3d &= 8(10.5) \\ 3d &= 84 \\ d &= 28. \end{aligned}$
So $10.5$ centimeters stands for $28$ kilometers of trail.
As a check, $10.5$ is $3.5$ times $3$ , and $28$ is $3.5$ times $8$ .

For a trail of $20$ kilometers, put $d = 20$ and cross-multiply:
$\begin{aligned} 8c &= 3(20) \\ 8c &= 60 \\ c &= 7.5. \end{aligned}$
So the trail is $7.5$ centimeters long on the map.
As a check, $20$ is $2.5$ times $8$ , and $7.5$ is $2.5$ times $3$ .

To judge an arrangement, cross-multiply it.
The map's proportion $\dfrac{c}{d} = \dfrac{3}{8}$ gives $8c = 3d$ .
The arrangement $\dfrac{c}{3} = \dfrac{d}{8}$ also gives $8c = 3d$ , the same equation, so it describes the same map: it compares each length with its own entry in the scale.

The arrangement $\dfrac{c}{8} = \dfrac{d}{3}$ gives $3c = 8d$ , a different equation.
On the scale's own pair, $c = 3$ with $d = 8$ , it would need $3(3) = 8(8)$ , that is $9 = 64$ , which is false.
So it does not describe this map; it pairs $8$ centimeters of map with $3$ kilometers of trail.

Answer

$10.5$ centimeters stands for $28$ kilometers of trail, and a $20$ kilometer trail is $7.5$ centimeters long on the map. The proportion $\dfrac{c}{3} = \dfrac{d}{8}$ describes this map; $\dfrac{c}{8} = \dfrac{d}{3}$ does not.

Key idea

Cross-multiply two arrangements of a proportion to compare them: if both give the same equation, they describe the same relation wherever the denominators are nonzero.
Problem 6: Mark complete
Problem 7 Checking Priya's work

Priya solves $\dfrac{x}{4} + \dfrac{1}{2} = \dfrac{x + 6}{3}$ by writing these lines.
$\begin{aligned} 3x &= 4(x + 6) \\ 3x &= 4x + 24 \\ x &= -24 \end{aligned}$
Find the first of her lines that does not follow from the equation before it, and say what the move that produced it requires. Then solve the equation correctly, and check your value in the equation as given.
1. Hint 1
  Test each line against the one before it, starting with the given equation. One step treats the equation as though it had a shape it does not have.
2. Hint 2
  Her first line is what cross-multiplying would give. Cross-multiplying is multiplying both sides by the product of two denominators, so ask what happened to the $\tfrac{1}{2}$ .
3. Hint 3
  To solve correctly, multiply every term on both sides by the smallest number that $4$ , $2$ and $3$ all divide into, the $\tfrac{1}{2}$ included.
Answer

Her first line, $3x = 4(x + 6)$ : it cross-multiplies, which needs a single fraction on each side, and the left side is a sum. The solution is $x = -18$ , where both sides of the original equal $-4$ .

Full solution

Her first line $3x = 4(x + 6)$ is what cross-multiplying $\dfrac{x}{4} = \dfrac{x + 6}{3}$ would give.
Cross-multiplication turns one fraction equal to one fraction into a product equal to a product, and it is nothing more than multiplying both sides by the product of the two denominators.

That move needs each side to be a single fraction.
Here the left side is the sum $\dfrac{x}{4} + \dfrac{1}{2}$ , so the move does not apply, and the $\dfrac{1}{2}$ simply disappeared.
Multiplying the whole left side by $12$ gives $3x + 6$ , not $3x$ .
Her next two lines do follow from her first, which is why the rest of the page looks sound.

To solve correctly, multiply every term on both sides by $12$ , the smallest number that $4$ , $2$ and $3$ all divide into.
Then expand, subtract $3x$ from both sides, and subtract $24$ from both sides:
$\begin{aligned} 3x + 6 &= 4(x + 6) \\ 3x + 6 &= 4x + 24 \\ 6 &= x + 24 \\ x &= -18. \end{aligned}$

Check in the original equation.
The left side is $-\tfrac{18}{4} + \tfrac{1}{2}$ , which is $-4.5 + 0.5$ , or $-4$ .
The right side is $\tfrac{-18 + 6}{3}$ , which is $\tfrac{-12}{3}$ , or $-4$ .
The sides agree.

Priya's value fails the same test.
At $x = -24$ the left side is $-6 + 0.5$ , or $-5.5$ , while the right side is $\tfrac{-18}{3}$ , or $-6$ .
So $-24$ solves her first line but not the given equation.

A second correct route combines the left side into one fraction first: $\tfrac{x}{4} + \tfrac{2}{4}$ is $\tfrac{x + 2}{4}$ .
Now each side is a single fraction, and cross-multiplying $\dfrac{x + 2}{4} = \dfrac{x + 6}{3}$ gives $3(x + 2) = 4(x + 6)$ , which leads to $x = -18$ again.

Answer

Her first line, $3x = 4(x + 6)$ : it cross-multiplies, which needs a single fraction on each side, and the left side is a sum. The solution is $x = -18$ , where both sides of the original equal $-4$ .

Key idea

Cross-multiplication needs exactly one fraction on each side; when a side is a sum, combine it into one fraction first or clear every denominator with the LCD.
Problem 7: Mark complete
Problem 8 Sam's two verdicts

Sam clears the fractions in each equation below. Both times the $x$ terms cancel and leave a true statement, so under each equation he writes that every number is a solution.
$\begin{aligned} \frac{x}{3} + \frac{x + 6}{2} &= \frac{5x}{6} + 3 \\ \frac{x + 1}{x + 4} + \frac{x + 7}{x + 4} &= 2 \end{aligned}$
For each equation, say whether Sam's verdict is right for the equation as given, and describe every number that satisfies it. Then explain what your decision on each verdict depended on.
1. Hint 1
  A verdict read off a cleared equation belongs first to the cleared equation. For each one, ask what Sam multiplied by, and whether that step can be undone at every number.
2. Hint 2
  The first equation is cleared with $6$ . The second is cleared with $x + 4$ : find the value of $x$ at which that multiplier is zero, and ask whether the original equation makes sense there.
3. Hint 3
  Clear each equation yourself to confirm the true statement. Then keep the numbers the original equation allows, and remove any number it excludes.
Answer

First: right, every number is a solution. Second: wrong; it excludes $x = -4$ , and its solutions are every number except $-4$ . The first was cleared with the number $6$ , the second with $x + 4$ , which is zero at $x = -4$ .

Full solution

Clear the first equation by multiplying every term on both sides by $6$ , the LCD of $3$ , $2$ and $6$ :
$\begin{aligned} 2x + 3(x + 6) &= 5x + 18 \\ 5x + 18 &= 5x + 18. \end{aligned}$
Subtracting $5x$ from both sides leaves $18 = 18$ , which is true.

The multiplier $6$ is a nonzero number, so dividing by $6$ undoes the step, and the original has exactly the same solutions as the cleared equation.
Sam's verdict is right: every number satisfies the first equation.
Testing agrees: at $x = 0$ both sides are $3$ , and at $x = 6$ both sides are $8$ .

In the second equation both denominators are $x + 4$ , which is zero at $x = -4$ , so the equation as given excludes $x = -4$ .
Multiplying every term on both sides by $x + 4$ gives
$\begin{aligned} (x + 1) + (x + 7) &= 2(x + 4) \\ 2x + 8 &= 2x + 8. \end{aligned}$
The $x$ terms cancel and leave $8 = 8$ , which is true, so every number satisfies the cleared equation.

That verdict belongs to the cleared equation, not yet to the given one.
At $x = -4$ the given equation divides by zero and says nothing, so $-4$ cannot be a solution of it.
At every other number the multiplier $x + 4$ is not zero, so the step undoes there and the given equation holds.
So Sam's verdict is wrong for the second equation: its solutions are every number except $-4$ .
At $x = 0$ , for example, the left side is $\tfrac{1}{4} + \tfrac{7}{4}$ , which is $2$ .

Each decision depended on the multiplier.
Clearing with a nonzero number can be undone at every value, so the verdict carries back to the given equation.
Clearing with an expression that holds the variable cannot be undone where that expression is zero, so the verdict has to face the excluded values first.

Answer

First: right, every number is a solution. Second: wrong; it excludes $x = -4$ , and its solutions are every number except $-4$ . The first was cleared with the number $6$ , the second with $x + 4$ , which is zero at $x = -4$ .

Key idea

A verdict read off a cleared equation is the original's too when every multiplier was a nonzero number; after clearing with an expression that holds the variable, remove the excluded values from it.
Problem 8: Mark complete
Problem 9 Two values that both work

Kai substitutes $x = 1$ and then $x = 5$ into the equation below, and both values make its two sides equal.
$\frac{3x - 1}{4} - \frac{x - 3}{6} = \frac{7x + 3}{12}$
Before clearing any fractions, use those two values to decide how many solutions the equation has, and explain your reasoning. Then clear the fractions to confirm your answer.
1. Hint 1
  Clearing the fractions with a number and gathering the terms turns the equation into the form $ax = b$ without changing its solutions. Ask which of the three possible outcomes for $ax = b$ allows two different solutions.
2. Hint 2
  If $a$ were not zero, $ax = b$ would have the single solution $\dfrac{b}{a}$ , which cannot equal both $1$ and $5$ . So what must $a$ be, and then what does $x = 1$ working say about $b$ ?
3. Hint 3
  To confirm, multiply every term by $12$ , keeping $x - 3$ in parentheses so that the $-2$ reaches both of its terms, and see what is left.
Answer

Infinitely many: every number is a solution. In the cleared form $ax = b$ , both $a$ and $b$ are $0$ . Clearing with $12$ gives $7x + 3 = 7x + 3$ , which confirms it.

Full solution

Clearing the fractions multiplies both sides by a nonzero number, and gathering terms adds or subtracts the same thing on both sides.
Neither changes the solutions, so the equation can be rewritten as $ax = b$ , for some numbers $a$ and $b$ , with exactly the same solutions as the original.
Both $1$ and $5$ therefore satisfy $ax = b$ .

If $a$ were not zero, dividing both sides by $a$ would give the single solution $x = \dfrac{b}{a}$ , one value and no other.
That value cannot be both $1$ and $5$ , so $a = 0$ .

With $a = 0$ the equation reads $0 = b$ , with no $x$ left in it.
Because $x = 1$ satisfies it, the statement $0 = b$ must be true, so $b = 0$ .
The statement $0 = 0$ holds whatever $x$ is, so every number is a solution and there are infinitely many.

To confirm, multiply every term on both sides by $12$ , the LCD of $4$ , $6$ and $12$ , keeping $x - 3$ in parentheses:
$\begin{aligned} 3(3x - 1) - 2(x - 3) &= 7x + 3 \\ 9x - 3 - 2x + 6 &= 7x + 3 \\ 7x + 3 &= 7x + 3. \end{aligned}$
The two sides are the same expression, so subtracting $7x + 3$ from both leaves $0 = 0$ , exactly as the reasoning predicted.
At $x = 0$ , for example, both sides of the original are $\tfrac{1}{4}$ .

Answer

Infinitely many: every number is a solution. In the cleared form $ax = b$ , both $a$ and $b$ are $0$ . Clearing with $12$ gives $7x + 3 = 7x + 3$ , which confirms it.

Key idea

A linear equation cleared with numbers has exactly one solution, no solution, or every number as a solution, so if two different values satisfy it, every number does.
Problem 9: Mark complete
Problem 10 Lena's claim

Lena claims that multiplying both sides of an equation by the same expression always gives an equation with exactly the same solutions.

Test her claim on $\dfrac{2x - 5}{x - 6} = \dfrac{7}{x - 6}$ : name the value of $x$ that the original rules out, multiply both sides by $x - 6$ and solve the new equation, and compare the solutions of the two equations. Then state a condition on the multiplier under which Lena's claim is true.
1. Hint 1
  Compare the two equations value by value. They can disagree only at a value where the multiplier does something that a nonzero number cannot do.
2. Hint 2
  The original equation says nothing at a value that makes its denominator zero. Solve the new equation, and ask whether its solution is such a value.
3. Hint 3
  For the condition, think about which multipliers can be undone by dividing both sides by them, at every value of $x$ .
Answer

The original rules out $x = 6$ and has no solution. The new equation $2x - 5 = 7$ has the solution $x = 6$ , so the solutions differ and the claim is false. It holds when the multiplier is nonzero at every value of $x$ .

Full solution

The denominator $x - 6$ is zero at $x = 6$ , so the original equation rules out $x = 6$ .
Multiplying both sides by $x - 6$ cancels both denominators and leaves the new equation, which solves as follows:
$\begin{aligned} 2x - 5 &= 7 \\ 2x &= 12 \\ x &= 6. \end{aligned}$

So $x = 6$ satisfies the new equation, since
$2(6) - 5 = 7.$
In the original, $x = 6$ makes both denominators zero, so each side asks for a division by zero and names no number.
The value $6$ is a solution of the new equation and not of the original, so the two equations have different solutions and Lena's claim is false.
The original in fact has no solution, because its only candidate is the excluded value.

The failure happened exactly where the multiplier is zero.
Canceling $x - 6$ against each denominator is valid only where $x - 6$ is not zero.
The new equation has no denominator, so it makes a statement at $x = 6$ , where the original makes none, and there $2(6) - 5 = 7$ is true.
Undoing the step would mean dividing both sides by $x - 6$ , which is impossible at $x = 6$ , so there the step cannot be reversed.
At any other value, $x - 6$ is a nonzero number, and dividing both sides of the new equation by it returns the original.

The step also cannot lose a solution.
A number that solves the original makes every denominator nonzero, so there $x - 6$ is a nonzero number, and multiplying both sides of a true equality by a nonzero number keeps it true.
So the only change the step can make is to add a value where the multiplier is zero.

Lena's claim is therefore true whenever the multiplier is nonzero at every value of $x$ .
A nonzero number, such as the LCD used to clear numerical fractions, meets that condition.
A multiplier that holds the variable need not, which is why its excluded values have to be checked.

Answer

The original rules out $x = 6$ and has no solution. The new equation $2x - 5 = 7$ has the solution $x = 6$ , so the solutions differ and the claim is false. It holds when the multiplier is nonzero at every value of $x$ .

Key idea

Clearing a denominator that holds the variable keeps every solution of the original but can add the value that makes the denominator zero, so each candidate must be checked against the excluded values.
Problem 10: Mark complete

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