Jordan's results

Scanned August 30, 2026

Overview

Jordan has a solid grasp of the overall *structure* of algebra: he consistently knows to undo the constant first, then divide by the coefficient, and he combines like terms accurately. Nearly every miss on these three pages traces back to one specific issue — handling negative signs, especially distributing a negative number across a subtraction (for example writing $-3(x-4) = -3x - 12$ instead of $-3x + 12$ ). There is also a smaller, intermittent slip where he adds instead of subtracts when isolating the variable term. These are narrow, very fixable issues rather than a broad misunderstanding of algebra.

What we read from the photos (14 problems)

Every answer below was re-checked by real computation. If we misread something, the gaps that relied on it were dropped.

1) 3x + 5 = 20
correct
Their work: 3x = 15, x = 5
2) 2x - 7 = 11
correct
Their work: 2x = 18, x = 9
3) 4x + 9 = 21
incorrect
Their work: 4x = 30, x = 7.5
4) 5x - 3 = 12
correct
Their work: 5x = 15, x = 3
5) 6x + 8 = 26
incorrect
Their work: 6x = 34, x = 34/6
6) -2x + 4 = 10
incorrect
Their work: -2x = 6, x = 3
Page 2, 1) 2(x + 3) =
correct
Their work: 2x + 6
Page 2, 2) -3(x - 4) =
incorrect
Their work: -3x - 12
Page 2, 3) 4(2x + 1) - 3x =
correct
Their work: 8x + 4 - 3x = 5x + 4
Page 2, 4) -(x - 5) =
incorrect
Their work: -x - 5
Page 2, 5) 3x + 2x - 7 =
correct
Their work: 5x - 7
Page 3, 1) 2(x + 3) = 14
correct
Their work: 2x + 6 = 14 2x = 8, x = 4
Page 3, 2) -2(x - 4) = 10
incorrect
Their work: -2x - 8 = 10 -2x = 18, x = -9
Page 3, 3) 3(x - 2) = 9
correct
Their work: 3x - 6 = 9 3x = 15, x = 5

What's going well

Reliable two-step equation procedure: on $3x + 5 = 20$ , $2x - 7 = 11$ , and $5x - 3 = 12$ he correctly undid the constant, then divided by the coefficient, showing both steps.
Distributes correctly when the multiplier is positive: $2(x+3) = 2x+6$ and $4(2x+1) - 3x = 8x + 4 - 3x = 5x + 4$ .
Combines like terms accurately and keeps unlike terms separate, as in $3x + 2x - 7 = 5x - 7$ and $8x - 3x = 5x$ .
Handles multi-step equations that require distributing first and then solving, e.g. $2(x+3)=14 \Rightarrow 2x+6=14 \Rightarrow x=4$ and $3(x-2)=9 \Rightarrow x=5$ .
Shows his intermediate steps clearly, which makes his thinking easy to follow and correct.

4 gaps to work on

Distributing a negative factor

significant

Usually taught in 7th-8th grade

When a negative number (or a bare minus sign) is multiplied across parentheses, Jordan applies the negative to the first term but not to the second — he treats the sign inside the parentheses as unchangeable. The rule that a negative times a negative gives a positive isn't being applied to the second term.

In their work: Page 2 #2: $-3(x-4)$ written as $-3x - 12$ instead of $-3x + 12$ ; Page 2 #4: $-(x-5)$ written as $-x - 5$ instead of $-x + 5$ ; Page 3 #2: $-2(x-4)=10$ expanded as $-2x - 8 = 10$ instead of $-2x + 8 = 10$ .

Guide

Jordan, you already know the hard part: you distribute, you combine like terms, and you undo operations in the right order. The one thing tripping you up is what happens to the SECOND term inside the parentheses when the factor outside is negative. Right now you multiply the negative into the first term and then copy the sign of the second term straight down. That copying step is the misconception. Here is the correct mental model. The factor outside the parentheses multiplies EVERY term inside, sign and all. And the sign that belongs to a term is the sign written in front of it. So in $(x - 4)$ the two terms are $+x$ and $-4$ . When you distribute $-3$ , you compute $(-3)(+x) = -3x$ and $(-3)(-4) = +12$ . Two negatives multiplied give a positive, so the $-4$ becomes $+12$ . Nothing is being "copied"; each term gets a fresh multiplication. A good way to feel why this must be true: $-3(x-4)$ means take $3(x-4) = 3x - 12$ and flip the sign of the whole thing, which gives $-3x + 12$ . Flipping the sign of a difference flips BOTH signs. If you only flipped the first sign you'd be changing the value of the expression. You can also check with a number: let $x = 4$ . Then $-3(4-4) = -3(0) = 0$ . Your version $-3x - 12$ gives $-12 - 12 = -24$ , which is wrong. The correct version $-3x + 12$ gives $-12 + 12 = 0$ . Correct. Substituting a number is your personal lie detector — use it whenever you are unsure. One more note: a bare minus sign in front of parentheses, like $-(x-5)$ , means $-1$ times the parentheses. So $(-1)(x) = -x$ and $(-1)(-5) = +5$ , giving $-x + 5$ . The invisible $-1$ is still a full multiplier. Worked Example 1: Expand $-4(x - 7)$ . Step 1: Name the terms inside. They are $+x$ and $-7$ . Step 2: Multiply the first term: $(-4)(+x) = -4x$ . Step 3: Multiply the second term, keeping its sign attached: $(-4)(-7) = +28$ . Step 4: Write the result: $-4x + 28$ . Step 5: Check with $x = 7$ . Original: $-4(7-7) = 0$ . Answer: $-4(7) + 28 = -28 + 28 = 0$ . Match, so the expansion is right. Worked Example 2: Solve $-2(x - 4) = 10$ . Step 1: Terms inside are $+x$ and $-4$ . Step 2: Distribute: $(-2)(+x) = -2x$ and $(-2)(-4) = +8$ , so the equation becomes $-2x + 8 = 10$ . Step 3: Undo the constant. Subtract $8$ from both sides: $-2x = 2$ . Step 4: Divide both sides by $-2$ : $x = -1$ . Step 5: Check in the ORIGINAL equation: $-2(-1 - 4) = -2(-5) = 10$ . True, so $x = -1$ . The recipe: 1. Write the two terms inside the parentheses WITH their signs (for example $+x$ and $-4$ ). 2. If there is a bare minus sign outside, rewrite it as $-1$ . 3. Multiply the outside factor by the first term; write the answer with its sign. 4. Multiply the outside factor by the second term; apply negative times negative equals positive. 5. Only now continue solving: subtract or add the constant, then divide by the coefficient. 6. Substitute your answer back into the ORIGINAL expression or equation to check. Most common mistake to watch for: copying the sign of the second term instead of multiplying it. If the factor outside is negative, the sign of the second term ALWAYS flips — a $-4$ becomes $+12$ under $-3$ , and a $+5$ becomes $-15$ . Before you move on, ask yourself out loud: "Did the second sign change?" If the outside factor was negative and the second sign looks the same as before, you skipped a multiplication.

Practice problems

✓ answers independently checked

1. Expand $-5(x - 2)$ .
Hint
The terms inside are $+x$ and $-2$ . Multiply $-5$ by each of them separately, and remember what negative times negative gives.
Answer
$-5x + 10$ . The $-5$ times $-2$ is $+10$ , so the second sign flips to positive.
Worked solution
Identify the terms inside: $+x$ and $-2$ . Multiply the first term: $(-5)(+x) = -5x$ . Multiply the second term: $(-5)(-2) = +10$ . Combine: $-5x + 10$ . Check with $x = 2$ : original $-5(2-2) = 0$ ; answer $-5(2) + 10 = -10 + 10 = 0$ . Match.
2. Expand $-(x - 9)$ .
Hint
A bare minus sign means $-1$ times the parentheses. Multiply $-1$ by each term.
Answer
$-x + 9$ . Since $(-1)(-9) = +9$ , the second term becomes positive.
Worked solution
Rewrite the bare minus as a factor: $-1(x - 9)$ . Terms inside: $+x$ and $-9$ . Multiply the first term: $(-1)(+x) = -x$ . Multiply the second term: $(-1)(-9) = +9$ . Combine: $-x + 9$ . Check with $x = 9$ : original $-(9-9) = 0$ ; answer $-9 + 9 = 0$ . Match.
3. Expand $-6(2x - 3)$ .
Hint
Multiply $-6$ by $2x$ , then multiply $-6$ by $-3$ . Watch the sign on the second product.
Answer
$-12x + 18$ . The product $(-6)(-3) = +18$ .
Worked solution
Terms inside: $+2x$ and $-3$ . Multiply the first term: $(-6)(2x) = -12x$ . Multiply the second term: $(-6)(-3) = +18$ . Combine: $-12x + 18$ . Check with $x = 1$ : original $-6(2-3) = -6(-1) = 6$ ; answer $-12 + 18 = 6$ . Match.
4. Two students expanded $-4(x - 5)$ . Sam wrote $-4x - 20$ and Riley wrote $-4x + 20$ . Who is right, and prove it by substituting $x = 5$ .
Hint
Substitute $x = 5$ into the original expression first — the parentheses become something very simple.
Answer
Riley is right: $-4x + 20$ . At $x = 5$ the original gives $0$ , and only Riley's version gives $0$ .
Worked solution
Original at $x = 5$ : $-4(5 - 5) = -4(0) = 0$ . Sam's version at $x = 5$ : $-4(5) - 20 = -20 - 20 = -40$ . Not $0$ , so Sam is wrong. Riley's version at $x = 5$ : $-4(5) + 20 = -20 + 20 = 0$ . Matches. Why: the terms inside are $+x$ and $-5$ , and $(-4)(-5) = +20$ . So the correct expansion is $-4x + 20$ ; Riley is right.
5. Solve $-3(x - 6) = 21$ .
Hint
Distribute first and make sure the constant term comes out positive, then undo the constant and divide.
Answer
$x = -1$ . Distributing gives $-3x + 18 = 21$ , so $-3x = 3$ and $x = -1$ .
Worked solution
Distribute: $(-3)(x) = -3x$ and $(-3)(-6) = +18$ . Equation becomes $-3x + 18 = 21$ . Subtract $18$ from both sides: $-3x = 3$ . Divide both sides by $-3$ : $x = -1$ . Check in the original: $-3(-1 - 6) = -3(-7) = 21$ . True. So $x = -1$ .
6. Solve $-(x - 8) = 3$ .
Hint
Treat the bare minus as $-1$ and distribute to both terms before you solve.
Answer
$x = 5$ . Expanding gives $-x + 8 = 3$ , so $-x = -5$ and $x = 5$ .
Worked solution
Rewrite: $-1(x - 8) = 3$ . Distribute: $(-1)(x) = -x$ and $(-1)(-8) = +8$ . Equation becomes $-x + 8 = 3$ . Subtract $8$ from both sides: $-x = -5$ . Divide both sides by $-1$ : $x = 5$ . Check in the original: $-(5 - 8) = -(-3) = 3$ . True. So $x = 5$ .
7. Expand and simplify $-2(x - 5) - 3(x + 1)$ .
Hint
Do each distribution separately, being careful that the first one produces $+10$ and the second produces $-3$ , then combine like terms.
Answer
$-5x + 7$ . The two expansions are $-2x + 10$ and $-3x - 3$ , which combine to $-5x + 7$ .
Worked solution
First distribution: $(-2)(x) = -2x$ and $(-2)(-5) = +10$ , giving $-2x + 10$ . Second distribution: $(-3)(x) = -3x$ and $(-3)(+1) = -3$ , giving $-3x - 3$ . Write them together: $-2x + 10 - 3x - 3$ . Combine the $x$ terms: $-2x - 3x = -5x$ . Combine the constants: $10 - 3 = 7$ . Result: $-5x + 7$ . Check with $x = 1$ : original $-2(1-5) - 3(1+1) = -2(-4) - 3(2) = 8 - 6 = 2$ ; answer $-5 + 7 = 2$ . Match.
8. Solve $4 - 2(x - 3) = 14$ .
Hint
Distribute the $-2$ across both terms before touching the $4$ ; do not subtract the $2$ from the $4$ first.
Answer
$x = -2$ . Expanding gives $4 - 2x + 6 = 14$ , so $-2x + 10 = 14$ , $-2x = 4$ , $x = -2$ .
Worked solution
The factor multiplying the parentheses is $-2$ . Distribute: $(-2)(x) = -2x$ and $(-2)(-3) = +6$ . Equation becomes $4 - 2x + 6 = 14$ . Combine the constants on the left: $4 + 6 = 10$ , so $-2x + 10 = 14$ . Subtract $10$ from both sides: $-2x = 4$ . Divide both sides by $-2$ : $x = -2$ . Check in the original: $4 - 2(-2 - 3) = 4 - 2(-5) = 4 + 10 = 14$ . True. So $x = -2$ .

Done practicing? Scan the finished worksheet → and see whether this gap is closing.

Using the inverse operation to isolate the variable term

moderate

Usually taught in 7th grade

Most of the time Jordan correctly subtracts the constant from both sides, but in two problems he added it instead. Notably, both errors happened where the subtraction required regrouping across a ten ( $21-9$ , $26-8$ ), which suggests he switched to the easier addition rather than a true misunderstanding of inverse operations. Checking the answer by substituting back would catch these immediately.

In their work: Problem 3: $4x+9=21$ became $4x = 30$ (should be $4x = 12$ , $x = 3$ ); Problem 5: $6x+8=26$ became $6x = 34$ (should be $6x = 18$ , $x = 3$ ). Contrast with Problems 1 and 4, where he subtracted correctly.

Guide

Jordan, here's the good news first: your algebra structure is solid. You know to deal with the constant first and then divide by the coefficient, and you combine like terms cleanly. The gap we're closing here is small and very fixable — it shows up in problems like $4x + 9 = 21$ and $6x + 8 = 26$ , where you wrote $4x = 30$ and $6x = 34$ instead of $4x = 12$ and $6x = 18$ . In both cases you added the constant instead of subtracting it. Interestingly, both were subtractions that needed regrouping across a ten ( $21 - 9$ and $26 - 8$ ). Adding is the easier mental move, and your brain quietly took the shortcut. Here is the correct mental model. An equation is a balance scale. $4x + 9 = 21$ says: "whatever $4x$ is, once you pile $9$ more onto it, the total is $21$ ." So $4x$ must be SMALLER than $21$ — the $9$ was part of the $21$ already. To find $4x$ alone, you take the $9$ back off: $4x = 21 - 9 = 12$ . If you add and get $4x = 30$ , that would mean $4x$ is bigger than the total, which is impossible when you're adding a positive number to it. That size check — "the variable term must be less than the total" — is your instant alarm bell for a $+$ constant. The operation rule behind it: you undo an operation with its inverse, and you do it to BOTH sides so the scale stays balanced. The equation has $+9$ , so you subtract $9$ from both sides. If the equation had $-9$ , you would add $9$ to both sides. Write the $-9$ under both sides so your eye can see the balance instead of relying on memory. And the safety net that would have caught both of your misses in about five seconds: substitute your answer back in. With $4x = 30$ you'd get $x = 7.5$ , and $4(7.5) + 9 = 39$ , not $21$ . Wrong. With $x = 3$ : $4(3) + 9 = 12 + 9 = 21$ . Correct. Always check. Worked Example 1: Solve $4x + 9 = 21$ . Step 1 — Name the constant and its sign: the constant on the variable side is $+9$ . Step 2 — The inverse of $+9$ is $-9$ , so subtract $9$ from both sides. Write it out: $4x + 9 - 9 = 21 - 9$ . Step 3 — Simplify. Left side: $4x$ . Right side: $21 - 9$ . This one needs regrouping, so slow down: $21 - 9 = 21 - 10 + 1 = 11 + 1 = 12$ . So $4x = 12$ . Step 4 — Size check: is $12$ less than $21$ ? Yes — good, that's what we expect when a positive constant was added. Step 5 — Divide both sides by $4$ : $x = \frac{12}{4} = 3$ . Step 6 — Check by substituting: $4(3) + 9 = 12 + 9 = 21$ . True. So $x = 3$ . Worked Example 2: Solve $6x - 8 = 22$ . Step 1 — The constant on the variable side is $-8$ . Step 2 — The inverse of $-8$ is $+8$ , so add $8$ to both sides: $6x - 8 + 8 = 22 + 8$ . Step 3 — Simplify: $6x = 30$ . Step 4 — Size check: here the constant was subtracted, so $6x$ should be BIGGER than $22$ . $30 > 22$ . Good. Step 5 — Divide both sides by $6$ : $x = \frac{30}{6} = 5$ . Step 6 — Check: $6(5) - 8 = 30 - 8 = 22$ . True. So $x = 5$ . The recipe: 1. Circle the constant on the variable side and keep its sign attached ( $+9$ , not $9$ ). 2. Apply the inverse to BOTH sides, written directly underneath: plus becomes minus, minus becomes plus. 3. Do the arithmetic slowly, especially if it crosses a ten. Break it up: $26 - 8 = 26 - 10 + 2 = 18$ . 4. Size check: if the constant was added, the variable term shrinks; if it was subtracted, the variable term grows. 5. Divide both sides by the coefficient. 6. Substitute your answer back into the ORIGINAL equation and confirm both sides match. Most common mistake to watch out for: switching to addition because the subtraction is harder — exactly what happened with $21 - 9$ and $26 - 8$ . Your rule from now on: whenever the subtraction crosses a ten, that is the signal to write the step out and use the make-a-ten trick, not the signal to change the operation. Then let the substitution check be the final judge.

Practice problems

✓ answers independently checked

1. Solve for $x$ : $5x + 3 = 18$ .
Hint
The constant is $+3$ . What is the inverse of adding $3$ ? Do it to both sides, and remember the variable term should end up smaller than $18$ .
Answer
$x = 3$ . Subtracting $3$ from both sides gives $5x = 15$ , then dividing by $5$ gives $x = 3$ .
Worked solution
Start: $5x + 3 = 18$ . The constant on the variable side is $+3$ , so subtract $3$ from both sides. $5x + 3 - 3 = 18 - 3$ $5x = 15$ Size check: $15 < 18$ , which is correct because a positive constant had been added. Divide both sides by $5$ : $x = \frac{15}{5}$ $x = 3$ Check: $5(3) + 3 = 15 + 3 = 18$ . True. Final answer: $x = 3$ .
2. Solve for $x$ : $4x + 9 = 21$ . Show the subtraction step and then check your answer by substituting.
Hint
$21 - 9$ crosses a ten, so slow down: $21 - 10 + 1$ . Do not switch to addition just because subtracting feels harder.
Answer
$x = 3$ . Subtracting $9$ gives $4x = 12$ , so $x = 3$ , and $4(3) + 9 = 21$ confirms it.
Worked solution
Start: $4x + 9 = 21$ . The constant is $+9$ , so subtract $9$ from both sides. $4x + 9 - 9 = 21 - 9$ Compute $21 - 9$ carefully: $21 - 10 = 11$ , then add back $1$ to get $12$ . $4x = 12$ Size check: $12 < 21$ . Good. Divide both sides by $4$ : $x = \frac{12}{4}$ $x = 3$ Check: $4(3) + 9 = 12 + 9 = 21$ . True. Final answer: $x = 3$ .
3. Solve for $x$ : $6x + 8 = 26$ . Then substitute your answer back to check.
Hint
$26 - 8$ needs regrouping. Break it up: $26 - 10 + 2$ . The variable term must come out smaller than $26$ .
Answer
$x = 3$ . Subtracting $8$ gives $6x = 18$ , so $x = 3$ , and $6(3) + 8 = 26$ .
Worked solution
Start: $6x + 8 = 26$ . The constant is $+8$ , so subtract $8$ from both sides. $6x + 8 - 8 = 26 - 8$ Compute $26 - 8$ : $26 - 10 = 16$ , then add back $2$ to get $18$ . $6x = 18$ Size check: $18 < 26$ . Good. Divide both sides by $6$ : $x = \frac{18}{6}$ $x = 3$ Check: $6(3) + 8 = 18 + 8 = 26$ . True. Final answer: $x = 3$ .
4. Solve for $y$ : $7y - 6 = 29$ .
Hint
Here the constant is negative. What is the inverse of subtracting $6$ ? After the step, should the variable term be bigger or smaller than $29$ ?
Answer
$y = 5$ . Adding $6$ to both sides gives $7y = 35$ , so $y = 5$ .
Worked solution
Start: $7y - 6 = 29$ . The constant on the variable side is $-6$ , so add $6$ to both sides. $7y - 6 + 6 = 29 + 6$ $7y = 35$ Size check: the constant was subtracted, so the variable term should be larger than $29$ . $35 > 29$ . Good. Divide both sides by $7$ : $y = \frac{35}{7}$ $y = 5$ Check: $7(5) - 6 = 35 - 6 = 29$ . True. Final answer: $y = 5$ .
5. A student solved $8x + 7 = 31$ and wrote $8x = 38$ , then $x = 4.75$ . Explain the error using a substitution check, and give the correct value of $x$ .
Hint
Plug $4.75$ back into the original equation. Also ask: can $8x$ be larger than $31$ if adding $7$ to it gives $31$ ?
Answer
The student added $7$ instead of subtracting it; correctly $8x = 24$ and $x = 3$ . The check $8(4.75) + 7 = 45\ne 31$ exposes the error.
Worked solution
Start: $8x + 7 = 31$ . The student's work: $8x = 31 + 7 = 38$ , giving $x = 4.75$ . Check the student's answer: $8(4.75) + 7 = 38 + 7 = 45$ , but the equation needs $31$ , so it is wrong. Size reasoning: since $7$ was added to $8x$ to reach $31$ , $8x$ must be less than $31$ , so $38$ is impossible. Correct step: the constant is $+7$ , so subtract $7$ from both sides. $8x + 7 - 7 = 31 - 7$ $8x = 24$ Divide both sides by $8$ : $x = \frac{24}{8}$ $x = 3$ Check: $8(3) + 7 = 24 + 7 = 31$ . True. Final answer: the error was adding instead of subtracting; $x = 3$ .
6. Solve for $m$ : $9m + 17 = 62$ .
Hint
$62 - 17$ crosses a ten twice. Try $62 - 20 + 3$ , or subtract in two chunks: $62 - 12 - 5$ .
Answer
$m = 5$ . Subtracting $17$ gives $9m = 45$ , so $m = 5$ .
Worked solution
Start: $9m + 17 = 62$ . The constant is $+17$ , so subtract $17$ from both sides. $9m + 17 - 17 = 62 - 17$ Compute $62 - 17$ : $62 - 20 = 42$ , then add back $3$ to get $45$ . $9m = 45$ Size check: $45 < 62$ . Good. Divide both sides by $9$ : $m = \frac{45}{9}$ $m = 5$ Check: $9(5) + 17 = 45 + 17 = 62$ . True. Final answer: $m = 5$ .
7. Solve for $x$ : $3x + 14 = 32 - 5$ . Simplify the right side first, then isolate the variable term.
Hint
Clean up the right side into a single number before you touch the $+14$ . Then use the inverse of adding $14$ .
Answer
$x = \frac{13}{3}$ . The right side is $27$ , so $3x = 13$ and $x = \frac{13}{3}$ .
Worked solution
Start: $3x + 14 = 32 - 5$ . Simplify the right side: $32 - 5 = 27$ . $3x + 14 = 27$ The constant on the variable side is $+14$ , so subtract $14$ from both sides. $3x + 14 - 14 = 27 - 14$ Compute $27 - 14$ : $27 - 10 = 17$ , then $17 - 4 = 13$ . $3x = 13$ Size check: $13 < 27$ . Good. Divide both sides by $3$ : $x = \frac{13}{3}$ Check: $3\left(\frac{13}{3}\right) + 14 = 13 + 14 = 27$ , and $32 - 5 = 27$ . True. Final answer: $x = \frac{13}{3}$ .
8. Solve for $t$ : $12t - 23 = 61$ . Then, without solving again, explain how a size check tells you your value for $12t$ is reasonable.
Hint
The constant is $-23$ , so use its inverse. Because a constant was subtracted from $12t$ to reach $61$ , the value of $12t$ must be on the larger side of $61$ .
Answer
$t = 7$ . Adding $23$ gives $12t = 84$ , and $84 > 61$ as expected since $23$ had been subtracted away.
Worked solution
Start: $12t - 23 = 61$ . The constant on the variable side is $-23$ , so add $23$ to both sides. $12t - 23 + 23 = 61 + 23$ $12t = 84$ Size check: $23$ had been taken away from $12t$ to leave $61$ , so $12t$ must be $23$ more than $61$ ; $84 > 61$ , which fits. Divide both sides by $12$ : $t = \frac{84}{12}$ $t = 7$ Check: $12(7) - 23 = 84 - 23 = 61$ . True. Final answer: $t = 7$ .

Done practicing? Scan the finished worksheet → and see whether this gap is closing.

Dividing by a negative coefficient

minor

Usually taught in 7th grade

When the final step is to divide by a negative number, the sign of the answer is sometimes dropped. Dividing a positive by a negative must give a negative result.

In their work: Problem 6: from the correct line $-2x = 6$ he wrote $x = 3$ instead of $x = -3$ . (He did handle this correctly on Page 3 #2, going from $-2x = 18$ to $x = -9$ , so this is inconsistent rather than absent.)

Guide

Jordan's misconception happens at the very last step of solving an equation, and only on some problems, not all: when he divides a positive number by a negative number, he sometimes forgets to carry the negative sign into the answer. For example, from $-2x = 6$ he wrote $x = 3$ instead of $x = -3$ . Notice that on a similar problem, $-2x = 18$ , he correctly got $x = -9$ . This tells us the issue isn't that he doesn't know the rule — it's that the rule isn't yet automatic, so under pressure or with certain sign combinations he skips it. Here is the mental model to make it automatic: dividing is really just multiplying by the reciprocal, and the sign rules for division are exactly the same as for multiplication. A positive divided by a positive is positive. A negative divided by a negative is positive. But whenever the two numbers have different signs — one positive, one negative — the answer is always negative. So a positive number divided by a negative number must come out negative, no exceptions. A good habit is to handle the sign as its own separate step, out loud: 'positive six divided by negative two — different signs, so the answer is negative — six divided by two is three, so the answer is negative three.' Doing this every single time, even on 'easy' problems, is what turns the rule into a reflex instead of something he has to remember under pressure. Worked Example 1: Solve $-3x + 7 = 1$ . Step 1: Undo the constant. Subtract 7 from both sides: $-3x = 1 - 7$ , which gives $-3x = -6$ . Step 2: Divide both sides by the coefficient, $-3$ : $x = \dfrac{-6}{-3}$ . Step 3: Check the signs before doing the arithmetic. Negative divided by negative — same signs — so the answer is positive. Step 4: $6 \div 3 = 2$ , and since the sign is positive, $x = 2$ . Worked Example 2: Solve $9 - 5x = 24$ . Step 1: Undo the constant. Subtract 9 from both sides: $-5x = 24 - 9$ , which gives $-5x = 15$ . Step 2: Divide both sides by the coefficient, $-5$ : $x = \dfrac{15}{-5}$ . Step 3: Check the signs first. Positive divided by negative — different signs — so the answer is negative. Step 4: $15 \div 5 = 3$ , and since the sign is negative, $x = -3$ . Recipe to follow every time: 1. Undo the constant term (add or subtract) so the equation looks like (number) $x$ = (number). 2. Look at the sign of the coefficient and the sign of the number on the other side before dividing. 3. Decide the sign of the answer first, using the rule: same signs give positive, different signs give negative. 4. Divide the digits (ignore the signs) to get the size of the answer. 5. Attach the sign you decided in step 3 to that number — this is your final answer. 6. Plug the answer back into the original equation to check it makes both sides equal. The most common mistake to watch for: when a positive number is divided by a negative coefficient (like $\dfrac{6}{-2}$ ), it is easy to only think about the '6 divided by 2 is 3' part and forget the leftover negative sign. Always say the sign rule out loud before writing the digits down, especially when the number being divided is positive and the coefficient is negative.

Practice problems

✓ answers independently checked

1. Solve for $x$ : $-2x = 6$
Hint
You have a positive number divided by a negative coefficient. Decide the sign of the answer before you divide the digits.
Answer
$x = -3$ , because $6 \div (-2)$ has different signs, so the result is negative.
Worked solution
Start with $-2x = 6$ . Divide both sides by $-2$ : $x = \dfrac{6}{-2}$ . Check signs: positive divided by negative — different signs — answer is negative. Divide the numbers: $6 \div 2 = 3$ . Attach the negative sign: $x = -3$ . Check: $-2(-3) = 6$ . Correct.
2. Solve for $x$ : $-5x = -20$
Hint
Both numbers are negative here — think about what sign a negative divided by a negative gives.
Answer
$x = 4$ , because $-20 \div (-5)$ has the same signs, so the result is positive.
Worked solution
Start with $-5x = -20$ . Divide both sides by $-5$ : $x = \dfrac{-20}{-5}$ . Check signs: negative divided by negative — same signs — answer is positive. Divide the numbers: $20 \div 5 = 4$ . Attach the positive sign: $x = 4$ . Check: $-5(4) = -20$ . Correct.
3. Solve for $x$ : $-3x + 7 = 1$
Hint
First undo the +7 by subtracting 7 from both sides, then handle the sign of the division carefully.
Answer
$x = 2$ , because after subtracting 7 you get $-3x = -6$ , and negative divided by negative is positive.
Worked solution
Start with $-3x + 7 = 1$ . Subtract 7 from both sides: $-3x = 1 - 7$ . Simplify the right side: $-3x = -6$ . Divide both sides by $-3$ : $x = \dfrac{-6}{-3}$ . Check signs: negative divided by negative — same signs — answer is positive. Divide the numbers: $6 \div 3 = 2$ , so $x = 2$ . Check: $-3(2) + 7 = -6 + 7 = 1$ . Correct.
4. Solve for $x$ : $-4x - 5 = 11$
Hint
Undo the subtraction of 5 first, then look carefully at whether the remaining division gives a positive or negative answer.
Answer
$x = -4$ , because after adding 5 to both sides you get $-4x = 16$ , and positive divided by negative is negative.
Worked solution
Start with $-4x - 5 = 11$ . Add 5 to both sides: $-4x = 11 + 5$ . Simplify the right side: $-4x = 16$ . Divide both sides by $-4$ : $x = \dfrac{16}{-4}$ . Check signs: positive divided by negative — different signs — answer is negative. Divide the numbers: $16 \div 4 = 4$ , so $x = -4$ . Check: $-4(-4) - 5 = 16 - 5 = 11$ . Correct.
5. Solve for $x$ : $2x - 6x = 16$
Hint
Combine the like terms on the left first to get a single negative coefficient, then divide carefully.
Answer
$x = -4$ , because combining terms gives $-4x = 16$ , and positive divided by negative is negative.
Worked solution
Start with $2x - 6x = 16$ . Combine like terms: $2x - 6x = -4x$ , so $-4x = 16$ . Divide both sides by $-4$ : $x = \dfrac{16}{-4}$ . Check signs: positive divided by negative — different signs — answer is negative. Divide the numbers: $16 \div 4 = 4$ , so $x = -4$ . Check: $2(-4) - 6(-4) = -8 + 24 = 16$ . Correct.
6. Solve for $x$ : $9 - 5x = 24$
Hint
Subtract 9 from both sides first, then decide the sign of the answer before doing the division.
Answer
$x = -3$ , because after subtracting 9 you get $-5x = 15$ , and positive divided by negative is negative.
Worked solution
Start with $9 - 5x = 24$ . Subtract 9 from both sides: $-5x = 24 - 9$ . Simplify the right side: $-5x = 15$ . Divide both sides by $-5$ : $x = \dfrac{15}{-5}$ . Check signs: positive divided by negative — different signs — answer is negative. Divide the numbers: $15 \div 5 = 3$ , so $x = -3$ . Check: $9 - 5(-3) = 9 + 15 = 24$ . Correct.
7. Solve for $x$ : $-7x + 2x - 3 = 27$
Hint
Combine the $x$ terms first, then undo the constant, and finally check the sign rule before dividing.
Answer
$x = -6$ , because combining and simplifying gives $-5x = 30$ , and positive divided by negative is negative.
Worked solution
Start with $-7x + 2x - 3 = 27$ . Combine like terms: $-7x + 2x = -5x$ , so $-5x - 3 = 27$ . Add 3 to both sides: $-5x = 27 + 3$ . Simplify: $-5x = 30$ . Divide both sides by $-5$ : $x = \dfrac{30}{-5}$ . Check signs: positive divided by negative — different signs — answer is negative. Divide the numbers: $30 \div 5 = 6$ , so $x = -6$ . Check: $-7(-6) + 2(-6) - 3 = 42 - 12 - 3 = 27$ . Correct.
8. Solve for $x$ : $20 - 8x = -4$
Hint
Subtract 20 from both sides first — this will make the right side negative too, so think carefully about the sign of negative divided by negative.
Answer
$x = 3$ , because after subtracting 20 you get $-8x = -24$ , and negative divided by negative is positive.
Worked solution
Start with $20 - 8x = -4$ . Subtract 20 from both sides: $-8x = -4 - 20$ . Simplify the right side: $-8x = -24$ . Divide both sides by $-8$ : $x = \dfrac{-24}{-8}$ . Check signs: negative divided by negative — same signs — answer is positive. Divide the numbers: $24 \div 8 = 3$ , so $x = 3$ . Check: $20 - 8(3) = 20 - 24 = -4$ . Correct.

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Simplifying or evaluating the final answer

minor

Usually taught in 5th-6th grade

Answers are sometimes left in an unsimplified fractional form rather than reduced or expressed as a decimal/mixed number, which can hide whether the answer is reasonable.

In their work: Problem 5: final answer written as $x = 34/6$ rather than reduced to $\frac{17}{3}$ .

Guide

Here is the pattern we noticed: Jordan solves the equation correctly all the way to the end, but then stops one step too soon. He gets a fraction like $x = \frac{34}{6}$ and writes that down as the final answer instead of simplifying it to $\frac{17}{3}$ . The algebra is fine — the finishing touch is missing. Think of a fraction like a photo that hasn't been cropped yet. $\frac{34}{6}$ and $\frac{17}{3}$ are the exact same number, the same point on the number line, but $\frac{34}{6}$ still has extra 'zoom' in it that both the numerator and denominator share. Simplifying means zooming out until there's nothing left to crop — until the top and bottom number share no common factor except 1. This matters for two reasons: first, it's the expected final form in math (just like you wouldn't leave a bill as \$3.00 written as \$300/100); second, an unreduced fraction hides whether an answer is reasonable. $\frac{34}{6}$ doesn't obviously look like 'a little more than 5', but $\frac{17}{3}$ does once you notice $17 \div 3 \approx 5.7$ . The correct mental model: solving the equation gets you a number. Simplifying is a separate, mandatory last step, exactly like remembering to write units on a word problem. Every single time you land on a fraction, ask 'can I still divide the top and bottom by the same whole number?' If yes, do it, and keep asking until the answer is no. Worked example 1: Solve $6x + 4 = 19$ . Step 1 (undo the constant): subtract 4 from both sides: $6x = 15$ . Step 2 (undo the coefficient): divide both sides by 6: $x = \frac{15}{6}$ . Step 3 (the step Jordan skips): look at $15$ and $6$ . Both are divisible by 3. Divide top and bottom by 3: $x = \frac{5}{2}$ . Step 4 (check): $5$ and $2$ share no common factor, so $\frac{5}{2}$ is fully simplified. As a check, $\frac{5}{2} = 2.5$ , and $6(2.5) + 4 = 15 + 4 = 19$ . It works. Worked example 2: Solve $3(2x + 4) = 46$ . Step 1 (distribute): $6x + 12 = 46$ . Step 2 (undo the constant): subtract 12 from both sides: $6x = 34$ . Step 3 (undo the coefficient): divide both sides by 6: $x = \frac{34}{6}$ . Step 4 (simplify — the missing step): both 34 and 6 are even, so divide top and bottom by their greatest common factor, 2: $x = \frac{17}{3}$ . Step 5 (double check it's fully reduced): 17 is prime and does not divide 3, so $\frac{17}{3}$ cannot be reduced further. This is the final answer. Here is the short recipe to run every time you finish solving an equation and land on a fraction: 1. Write the fraction you got from dividing. 2. Find the greatest common factor (GCF) of the numerator and the denominator. (List a few small factors — 2, 3, 5 — and see which one divides both evenly.) 3. Divide both the numerator and the denominator by that GCF. 4. Ask again: is there still a common factor? If yes, repeat step 3. If no, you're done. 5. Optionally, turn the fraction into a decimal or mixed number to sanity-check that the size of the answer makes sense. The most common mistake to watch for: stopping the instant you see a fraction and treating that as automatically 'the answer,' without checking whether the top and bottom share a factor. A fraction is not finished just because it's a fraction — it's finished only when it can't be reduced any further.

Practice problems

✓ answers independently checked

1. Solve for $x$ : $9x = 21$
Hint
Divide both sides by 9 to get a fraction, then look for a number that divides both the top and bottom evenly.
Answer
$x = \frac{7}{3}$ — divide $\frac{21}{9}$ by their GCF of 3.
Worked solution
$9x = 21$ Divide both sides by 9: $x = \frac{21}{9}$ Find the GCF of 21 and 9: it is 3 Divide top and bottom by 3: $x = \frac{7}{3}$ Check: 7 and 3 share no common factor, so this is fully simplified Final answer: $x = \frac{7}{3}$
2. Solve for $x$ : $6x + 4 = 19$
Hint
Undo the +4 first, then divide by 6 — and don't stop at the raw fraction, check if it reduces.
Answer
$x = \frac{5}{2}$ — $\frac{15}{6}$ reduces since both share a factor of 3.
Worked solution
$6x + 4 = 19$ Subtract 4 from both sides: $6x = 15$ Divide both sides by 6: $x = \frac{15}{6}$ Find the GCF of 15 and 6: it is 3 Divide top and bottom by 3: $x = \frac{5}{2}$ Check: 5 and 2 share no common factor, so this is fully simplified Final answer: $x = \frac{5}{2}$
3. Solve for $x$ : $-12x + 5 = -13$
Hint
Watch your signs when moving the 5, then divide — a negative divided by a negative gives a positive fraction, which you should still check for reducing.
Answer
$x = \frac{3}{2}$ — $\frac{18}{12}$ reduces to $\frac{3}{2}$ using GCF 6.
Worked solution
$-12x + 5 = -13$ Subtract 5 from both sides: $-12x = -18$ Divide both sides by -12: $x = \frac{-18}{-12} = \frac{18}{12}$ Find the GCF of 18 and 12: it is 6 Divide top and bottom by 6: $x = \frac{3}{2}$ Check: 3 and 2 share no common factor, so this is fully simplified Final answer: $x = \frac{3}{2}$
4. Solve for $x$ : $-4(x - 3) = 2x + 8$
Hint
Distribute the -4 carefully across both terms first, then collect the x-terms before dividing — and remember to reduce your final fraction.
Answer
$x = \frac{2}{3}$ — after solving, $\frac{4}{6}$ reduces to $\frac{2}{3}$ .
Worked solution
$-4(x - 3) = 2x + 8$ Distribute: $-4x + 12 = 2x + 8$ Add $4x$ to both sides: $12 = 6x + 8$ Subtract 8 from both sides: $4 = 6x$ Divide both sides by 6: $x = \frac{4}{6}$ Find the GCF of 4 and 6: it is 2 Divide top and bottom by 2: $x = \frac{2}{3}$ Check: 2 and 3 share no common factor, so this is fully simplified Final answer: $x = \frac{2}{3}$
5. Solve for $x$ : $2(3x + 1) = 2x + 12$
Hint
Distribute first, then move all the x-terms to one side before dividing. Once you have a fraction, check if it can be reduced.
Answer
$x = \frac{5}{2}$ — after solving, $\frac{10}{4}$ reduces to $\frac{5}{2}$ .
Worked solution
$2(3x + 1) = 2x + 12$ Distribute: $6x + 2 = 2x + 12$ Subtract $2x$ from both sides: $4x + 2 = 12$ Subtract 2 from both sides: $4x = 10$ Divide both sides by 4: $x = \frac{10}{4}$ Find the GCF of 10 and 4: it is 2 Divide top and bottom by 2: $x = \frac{5}{2}$ Check: 5 and 2 share no common factor, so this is fully simplified Final answer: $x = \frac{5}{2}$
6. Solve for $x$ : $-3(x - 2) = x + 4$
Hint
Distribute the negative across the subtraction, gather the x-terms, then simplify the resulting fraction all the way down.
Answer
$x = \frac{1}{2}$ — after solving, $\frac{2}{4}$ reduces to $\frac{1}{2}$ .
Worked solution
$-3(x - 2) = x + 4$ Distribute: $-3x + 6 = x + 4$ Add $3x$ to both sides: $6 = 4x + 4$ Subtract 4 from both sides: $2 = 4x$ Divide both sides by 4: $x = \frac{2}{4}$ Find the GCF of 2 and 4: it is 2 Divide top and bottom by 2: $x = \frac{1}{2}$ Check: 1 and 2 share no common factor, so this is fully simplified Final answer: $x = \frac{1}{2}$
7. Solve for $x$ : $3(2x + 4) = 46$
Hint
Distribute, undo the constant, divide by the coefficient — then look closely at whether the fraction you get can still be reduced.
Answer
$x = \frac{17}{3}$ — after solving, $\frac{34}{6}$ reduces to $\frac{17}{3}$ using GCF 2.
Worked solution
$3(2x + 4) = 46$ Distribute: $6x + 12 = 46$ Subtract 12 from both sides: $6x = 34$ Divide both sides by 6: $x = \frac{34}{6}$ Find the GCF of 34 and 6: it is 2 Divide top and bottom by 2: $x = \frac{17}{3}$ Check: 17 is prime and does not divide 3, so this is fully simplified Final answer: $x = \frac{17}{3}$
8. Solve for $x$ : $-2(3x - 5) = 4x + 8$
Hint
This one combines a negative distribution with variables on both sides — take it slow, and don't forget to reduce the final fraction fully.
Answer
$x = \frac{1}{5}$ — after solving, $\frac{2}{10}$ reduces to $\frac{1}{5}$ .
Worked solution
$-2(3x - 5) = 4x + 8$ Distribute: $-6x + 10 = 4x + 8$ Add $6x$ to both sides: $10 = 10x + 8$ Subtract 8 from both sides: $2 = 10x$ Divide both sides by 10: $x = \frac{2}{10}$ Find the GCF of 2 and 10: it is 2 Divide top and bottom by 2: $x = \frac{1}{5}$ Check: 1 and 5 share no common factor, so this is fully simplified Final answer: $x = \frac{1}{5}$

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