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Arithmetic with Expressions: Free Response

5 questions in parts, 68 points in total. Work each one out on paper, taking a hint if you get stuck. When you have an answer, reveal the answer to check it, and the full solution only if you still want it. The rubric is there so you can mark your own work.

Free response · work it on paper Question 1 of 5

1. Read the expression before you rewrite it . Foundational, 12 points. Question 1 of 5.

Take the expression

$E = 7m^2 - m + 4mn - 3m^2 + 6m - 10nm.$

Simplifying safely starts with reading rather than writing: what the terms are, what number each one carries, and which of them are of the same kind. Do the reading, then the rewriting, then say why the rewriting was allowed.
1. Part A.
  List the terms of $E$ , each with the sign that belongs to it, and give the coefficient of every one.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Simplify $E$ completely by combining like terms.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  The terms $4mn$ and $-10nm$ collapse into a single term, while $7m^2$ and $6m$ do not. Explain what makes the first pair collapse, naming the law that licenses it, and say exactly where the same reasoning stops for the second pair.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  Everything asked here is decided by the variable part of a term, meaning whatever is left once the numerical factor has been set aside. Write that part down for all six terms before you gather anything at all.
2. Hint 2 of 3 · Part B
  Two of these terms look different but are of the same kind. Multiplication can be carried out in either order, so ask whether any two variable parts are the same product with its letters swapped.
3. Hint 3 of 3 · Part C
  Take the law $ab + ac = a(b + c)$ and read it from right to left. Ask what plays the part of $a$ in each of the two pairs, and what happens to the argument when nothing can.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

The terms are $7m^2$ , $-m$ , $4mn$ , $-3m^2$ , $6m$ and $-10nm$ , with coefficients $7$ , $-1$ , $4$ , $-3$ , $6$ and $-10$ .

Part B

$E = 4m^2 + 5m - 6mn$ .

$4m^2 - 6mn + 5m$ is the same expression: the terms may be written in any order
$-6nm$ and $-6mn$ name the same term

Part C

The first pair shares a variable part, since $nm$ and $mn$ name the same product, so the distributive law read from right to left pulls that part out as a common factor and leaves only the coefficients to add. The second pair has no shared variable part, so there is nothing the two coefficients could be collected in front of.

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

Cut the expression at every $+$ and every $-$ , and let each sign travel with the term that follows it. Written as a pure sum, with subtraction turned into adding the opposite,

$E = 7m^2 + (-m) + 4mn + (-3m^2) + 6m + (-10nm).$

So there are six terms: $7m^2$ , $-m$ , $4mn$ , $-3m^2$ , $6m$ and $-10nm$ .

The coefficient of a term is its numerical factor, and the variable part is whatever is left. Reading them off in order:

$7, \quad -1, \quad 4, \quad -3, \quad 6, \quad -10.$

The second one is the one to watch. A variable written with no number in front of it carries the invisible factor $1$ , so once the minus sign is attached the coefficient of $-m$ is $-1$ .

Part B

Sort the six terms into families that share a variable part. One family is easy to miss: because multiplication can be done in either order, $nm$ and $mn$ are the same product, so the last term belongs with $4mn$ .

$4mn - 10nm = 4mn - 10mn = (4 - 10)mn = -6mn.$

The other two families are collected the same way, by adding coefficients and leaving the variable part untouched:

$7m^2 - 3m^2 = 4m^2, \qquad -m + 6m = 5m.$

Remember that the lone $-m$ counts as $-1$ copies of $m$ , which is what makes the second sum $5m$ rather than $6m$ . Putting the three survivors together,

$E = 4m^2 + 5m - 6mn.$

No two of the variable parts $m^2$ , $m$ and $mn$ are the same, so nothing further will combine.

Part C

The rule for combining is one line of the distributive law, $ab + ac = a(b + c)$ , read from right to left.

For the first pair, start with the variable parts. Multiplication does not care about order, so $nm = mn$ and the two terms carry the very same variable part. With that shared part acting as the common factor,

$4mn - 10nm = 4mn - 10mn = (4 - 10)mn = -6mn.$

The shared $mn$ is exactly what the law takes outside. Once it is out, all that remains inside is $4 - 10$ , an ordinary subtraction of two numbers, and a single term comes back.

For the second pair there is no shared variable part to take out. The variable parts are $m^2$ and $m$ , which are different objects: $m^2$ means $m$ multiplied by itself, while $m$ is a single copy. Collecting needs the WHOLE variable part to be common, so that what is left inside the bracket is nothing but the two coefficients, and here it is not:

$7m^2 + 6m$

remains two separate terms. The difference between the pairs is not how similar they look; it is whether one and the same variable part sits in both.

Be careful what this does and does not say. These two terms do share the single letter $m$ , so something can be written outside a bracket, and the next lesson does exactly that. What no rearrangement can do is turn them into ONE term, because taking $m$ out leaves $7m + 6$ behind rather than a bare sum of coefficients.

In one line

The six terms of $E$ are $7m^2$ , $-m$ , $4mn$ , $-3m^2$ , $6m$ and $-10nm$ , with coefficients $7$ , $-1$ , $4$ , $-3$ , $6$ and $-10$ ; the expression simplifies to $4m^2 + 5m - 6mn$ ; and the two $mn$ terms collapse because they share a variable part that the distributive law, read from right to left, takes outside a bracket, which is precisely what an $m^2$ term and an $m$ term never offer.

Another way: Rewrite in gathered order before adding anything

Rather than scanning the line repeatedly and tallying one family at a time in your head, use the commutative and associative laws of addition to rewrite $E$ once with the families side by side:

$E = 7m^2 - 3m^2 - m + 6m + 4mn - 10nm.$

Nothing has been combined yet, but every combination that remains is now between neighbours, and the only work left is three subtractions of coefficients. The extra line buys protection against the two mistakes that cost most: a term counted twice, and a term whose sign was left behind when it moved.

When it is worth it When an expression runs to more than four or five terms, or mixes several variable parts, so that hunting for one family at a time means reading the whole line over and over.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Lists every term with the sign that belongs to it, rather than dropping a leading minus and reporting the bare term. . Worth 2 points.

Gives a coefficient for every term, including any term whose numerical factor is not written down. . Worth 1 point.

Part B 5 points

Sorts the terms into families that share a variable part before adding anything. . Worth 2 points.

Adds the coefficients within each family, keeping every sign, and leaves the variable part unchanged. . Worth 2 points.

Reports a form in which no two of the remaining terms are like. . Worth 1 point.

Part C 4 points

Names the law behind the combining step and shows it acting in the direction that takes a shared factor outside a bracket. . Worth 2 points. needs an explanation, not just an answer

Says what specifically is missing in the second pair, rather than only asserting that those two terms are unlike. . Worth 2 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Simplify $9c^2 - 4cd + c - 5c^2 + 7dc - 3c$ , and give the coefficient of each term of your answer.
2. The minus sign that has to reach every term . Foundational, 13 points. Question 2 of 5.

Let

$D = (4p^2 - 7p + 5) - (p^2 - 7p - 6).$

Subtracting a whole expression is where this lesson's arithmetic is most often lost, and a substitution is the cheapest way to find out whether it was. Simplify, check your own work, and then turn the same check on somebody else's.
1. Part A.
  Simplify $D$ completely.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Evaluate $D$ exactly as it is printed in the stem, and evaluate your simplified form, both at $p = -2$ . Then say what the outcome of that comparison does and does not settle.
  
  Carry your own answer forward Substitute into whichever simplified form you produced in part A. What earns the credit here is evaluating both expressions at one and the same value and setting the two results beside each other, whatever they turn out to be.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  A classmate hands in $3p^2 - 14p - 1$ for the same subtraction. Choose a value of $p$ , evaluate their expression and the printed $D$ at it, and then name the single step that would produce what they wrote.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  A minus sign written in front of a bracket is the number $-1$ waiting to be multiplied in. Work out what that factor does to each of the three terms it reaches before you combine anything.
2. Hint 2 of 3 · Part B
  Evaluate the printed expression exactly as it stands, brackets and all, and only afterwards evaluate your own version. Two computations that were kept apart and then meet are what makes a check worth anything.
3. Hint 3 of 3 · Part C
  Their answer differs from a correct one in the $p$ terms and in the constant, but not in the $p^2$ term. Ask which single step could leave one family untouched while changing the other two.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

$D = 3p^2 + 11$ .

$11 + 3p^2$ is the same expression written in the other order

Part B

Both come to $23$ at $p = -2$ . Agreement at one value is evidence that the simplification is sound, not proof of it; a disagreement would have been proof of a mistake.

Part C

At $p = 1$ the printed expression gives $14$ while theirs gives $-12$ , so the two are not equivalent. Their form is what comes out if the minus sign in front of the second bracket is spent on that bracket's first term only, leaving the other two signs exactly as they were written.

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

To subtract an expression is to add its opposite, so the minus sign in front of the second bracket is a factor of $-1$ waiting to be distributed, and it reaches all three terms inside:

$D = 4p^2 - 7p + 5 - p^2 + 7p + 6.$

The first bracket is untouched. In the second, the $p^2$ arrives as $-p^2$ , the $-7p$ comes out as $+7p$ , and the $-6$ comes out as $+6$ . Both of those last two sign flips are the step that gets skipped.

Now collect the three families:

$4p^2 - p^2 = 3p^2, \qquad -7p + 7p = 0, \qquad 5 + 6 = 11.$

The two $p$ terms cancel exactly, so nothing of that family survives:

$D = 3p^2 + 11.$

Part B

Substitute $p = -2$ into the printed expression, finishing each bracket before the subtraction between them. In the first bracket, $(-2)^2 = 4$ and $-7(-2) = +14$ :

$4(4) - 7(-2) + 5 = 16 + 14 + 5 = 35.$

In the second bracket,

$(-2)^2 - 7(-2) - 6 = 4 + 14 - 6 = 12,$

so the printed expression comes to $35 - 12 = 23$ .

Now the simplified form at the same value:

$3(-2)^2 + 11 = 3(4) + 11 = 23.$

The two agree, which is worth having, but be exact about what it buys. Only one value has been tested, and an expression that is wrong somewhere can still be right at a particular number, so agreement raises confidence without settling the matter. A disagreement would have been decisive in the other direction: simplifying is supposed to preserve the value at every input, so any single mismatch proves a slip. That asymmetry is what makes the check worth the thirty seconds it costs.

Part C

Any value will do; $p = 1$ keeps the arithmetic light. The printed expression gives

$(4 - 7 + 5) - (1 - 7 - 6) = 2 - (-12) = 14,$

and the classmate's expression gives

$3(1)^2 - 14(1) - 1 = 3 - 14 - 1 = -12.$

One value at which two expressions disagree is enough to say they are not equivalent, so their answer cannot be a simplification of $D$ . Notice that this verdict was reached without reading a single line of their algebra.

Now name the step. Their $p$ coefficient is $-14$ , which is $-7 - 7$ : both $p$ terms arrived carrying a minus. Their constant is $-1$ , which is $5 - 6$ : the second bracket's constant also arrived unchanged. Both are what you get from

$4p^2 - 7p + 5 - p^2 - 7p - 6 = 3p^2 - 14p - 1,$

where the minus sign was applied to $p^2$ alone. So it is one slip, not three: the subtraction was treated as reaching only the first term of the bracket, when subtracting an expression means adding the opposite of every term in it.

In one line

$D = 3p^2 + 11$ . At $p = -2$ both the printed expression and the simplified form come to $23$ , which is evidence of a sound simplification rather than proof of one. The classmate's $3p^2 - 14p - 1$ disagrees with $D$ at $p = 1$ , giving $-12$ against $14$ , because the minus sign in front of the second bracket was spent on that bracket's first term alone.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Adds the opposite of the second expression, so that every one of its terms is accounted for and not just the one written first. . Worth 2 points.

Collects each family of like terms and reports a form in which no two terms are like. . Worth 2 points.

Part B 4 points

Evaluates the printed expression as it stands, respecting both brackets and the order of operations, including the square of a negative number. . Worth 2 points.

Evaluates the simplified form at the SAME value and sets the two results beside each other. . Worth 1 point.

Says what a check of this kind establishes and what it leaves open, rather than treating the outcome as the end of the matter. . Worth 1 point.

Part C 5 points

Evaluates the classmate's expression and the printed expression at one and the same value, showing the arithmetic for both. . Worth 2 points.

Names one step that accounts for the whole of the classmate's expression and says what that step failed to do, rather than only reporting that the two forms differ. . Worth 3 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Simplify $(6q^2 - 5q - 2) - (2q^2 - 5q + 7)$ , then check the result by substituting $q = 3$ into both forms.
3. How much more fence . Application, 14 points. Question 3 of 5.

A rectangular vegetable bed measures $x$ metres across and $2x + 3$ metres along, and is to be fenced on all four sides. The gardener then wants a second bed, rectangular as well, whose width and whose length are each $3$ metres greater than the first bed's. Fencing is sold by the metre, so the extra has to be worked out before either bed is dug.
1. Part A.
  Write an expression in simplest form for the length of fencing the first bed needs.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 4 points
2. Part B.
  Write an expression in simplest form for the length of fencing the second bed needs.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Using your two expressions, work out how much more fencing the second bed needs than the first. Give the amount with its unit.
  
  Carry your own answer forward Work from the two expressions you wrote in parts A and B, whatever they came out as. What is credited here is subtracting one whole expression from the other and labelling what comes out of it.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
4. Part D.
  Say what your part C answer implies for two gardeners whose first beds are of different widths, and name the feature of the two fencing expressions that is responsible for it.
  
  Carry your own answer forward Read this off the comparison you produced in part C, in whatever form it came out. The question is what that form tells you about the two beds, so answer it about your own result.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 4
  Fencing runs all the way round, so the quantity in play is the perimeter. Build each one from the four sides, and simplify each expression fully before it goes anywhere near the other.
2. Hint 2 of 4 · Part B
  The second bed's sides are not given to you directly. Work out its width and its length from the first bed's, write both of them down, and only then assemble the fence.
3. Hint 3 of 4 · Part C
  Taking one fencing length away from another is subtracting a whole expression, so put brackets round both before the minus sign moves. Every term of the one being taken away changes sign.
4. Hint 4 of 4 · Part D
  Look at what survived the subtraction and, just as importantly, at what did not. Then ask a gardener with a far wider bed the same question and see whether your answer would have to change.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

$6x + 6$ metres.

$6 + 6x$ metres is the same expression written in the other order

Part B

$6x + 18$ metres.

$18 + 6x$ metres is the same expression written in the other order

Part C

$12$ metres.

Part D

It carries no $x$ , so the extra fencing is the same whatever the width: a gardener with a much wider bed orders exactly the same extra length. That happens because the two expressions have identical $x$ terms, which therefore cancel in the subtraction and leave only the constants behind.

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

The fence runs right round the bed, so the quantity wanted is the perimeter: two widths and two lengths.

$P_1 = 2(x) + 2(2x + 3).$

Scale the bracket by the $2$ in front of it, reaching both terms inside:

$2(2x + 3) = 4x + 6.$

The two widths contribute $2x$ , so

$P_1 = 2x + 4x + 6 = 6x + 6,$

and the unit is metres. The two $x$ terms were like, so they combined; the $6$ had nothing to pair with.

Part B

The second bed's sides are not handed to you; build them from the first bed's. Its width is $x + 3$ metres and its length is $(2x + 3) + 3 = 2x + 6$ metres. Its fence is again two widths and two lengths:

$P_2 = 2(x + 3) + 2(2x + 6).$

Scale each bracket, reaching every term:

$2(x + 3) = 2x + 6, \qquad 2(2x + 6) = 4x + 12.$

Collecting like terms,

$P_2 = 2x + 6 + 4x + 12 = 6x + 18$

metres.

Part C

Take the first fencing length away from the second. Both are whole expressions, so bracket them before the minus sign goes anywhere, and let that sign reach both terms of the one being subtracted:

$P_2 - P_1 = (6x + 18) - (6x + 6) = 6x + 18 - 6x - 6.$

The two $6x$ terms cancel, and the constants leave

$18 - 6 = 12.$

So the second bed needs $12$ metres more fencing than the first.

Part D

The comparison came out with no $x$ in it. An expression with no variable in it takes the same value whatever the variable is, so the extra fencing does not depend on the width of the first bed at all: a gardener whose bed is $2$ metres across and a gardener whose bed is $50$ metres across order exactly the same extra length.

The reason is visible in the two expressions side by side:

$P_1 = 6x + 6, \qquad P_2 = 6x + 18.$

They carry identical $x$ terms. Subtracting one from the other cancels those terms exactly, and only the constants survive the subtraction. That is the algebra saying something the situation already suggests: widening the first bed lengthens both fences by the same amount, so it cannot change the gap between them.

In one line

The first bed needs $6x + 6$ metres of fencing and the second needs $6x + 18$ metres, so the second needs $12$ metres more. That comparison carries no $x$ , so it is the same for every width: the two expressions have identical $x$ terms, which cancel when one is subtracted from the other.

Another way: Count the extra metres side by side

The comparison can be had without building either perimeter. Making the width $3$ metres greater lengthens each of the two width sides by $3$ ; making the length $3$ metres greater lengthens each of the two length sides by $3$ . That is four sides, each gaining $3$ metres:

$4 \times 3 = 12.$

So the extra fencing is $12$ metres. No expression in $x$ was ever written down, which is why the answer could not have depended on $x$ in the first place.

When it is worth it When only the comparison is wanted and neither perimeter is needed for its own sake. It also makes an independent check on the algebra, since the two routes share no steps at all.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Builds the fencing length from all four sides of the rectangle, rather than from one width and one length. . Worth 2 points.

Scales each side length by its factor so that every term inside a bracket is reached, then collects the like terms. . Worth 2 points.

Part B 3 points

States both of the second bed's side lengths before any perimeter is assembled. . Worth 1 point.

Builds and simplifies the second fencing length by the same route used for the first. . Worth 2 points.

Part C 3 points

Subtracts one whole fencing expression from the other, with the sign of every term of the subtracted expression accounted for. . Worth 2 points.

Reports the comparison as a length in metres rather than as a bare number. . Worth 1 point.

Part D 4 points

Says what the form of the comparison means for beds of different widths, rather than restating the comparison itself. . Worth 2 points. needs an explanation, not just an answer

Points at the feature of the two fencing expressions that is responsible for the comparison taking that form. . Worth 2 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

A rectangular pen is $y$ metres wide and $3y + 1$ metres long. Write its fencing length in simplest form, then find how much more fencing a pen whose width and length are each $2$ metres greater would need.
4. Where the work stopped being equivalent . Reasoning, 15 points. Question 4 of 5.

A student is asked to simplify

$6 - 2x[3x - 5(x - 1)]$

and hands in these five lines, numbered $1$ to $5$ reading downwards.

$6 - 2x[3x - 5(x - 1)]$

$= 6 - 2x[3x - 5x - 5]$

$= 6 - 2x[-2x - 5]$

$= 6 + 4x^2 + 10x$

$= 4x^2 + 10x + 6$

The last line is not a simplification of the expression they were given. Exactly one line, however, fails to follow from the line above it; every other line is honest work on whatever it inherited.
1. Part A.
  Name the first line that does not follow from the line above it, say what went wrong in it, and write that line as it should have read.
  
  Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 5 points
2. Part B.
  Carry the corrected work through and simplify the expression the student was given, completely.
  
  Carry your own answer forward Continue from the corrected line you wrote in part A and finish from there. What is credited is the order in which you clear the grouping symbols and the signs you carry out of each one.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 5 points
3. Part C.
  Choose a value of $x$ and evaluate both the expression the student was given and their last line at it. Say what the comparison establishes, and why a check of this kind is worth running on your own work when nobody has told you that anything is wrong.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 4
  Nothing is settled by staring at the last line. Read the work as a chain and ask of each line in turn whether it follows from the one directly above it, because everything after a failure can be perfectly good work on the wrong expression.
2. Hint 2 of 4 · Part A
  The innermost parentheses are cleared by a factor carrying a minus sign, and such a factor lands on both terms inside. Work out that one little product on its own, away from the rest of the line.
3. Hint 3 of 4 · Part B
  Grouping symbols come apart from the inside out. Finish the innermost bracket completely, then treat the whole factor in front of the square brackets, sign and letter together, as the thing being distributed.
4. Hint 4 of 4 · Part C
  Pick a small value and evaluate the printed expression exactly as it stands, innermost bracket first. Then ask what a single disagreement is enough to prove, and what it would take to prove the opposite.
That is every hint for this question.
Answer and solution
Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.
The answer

Part A

Line $2$ . The factor $-5$ reached the $x$ but its product with the $-1$ was recorded as $-5$ rather than $+5$ . The line should have read $6 - 2x[3x - 5x + 5]$ .

Part B

$4x^2 - 10x + 6$ .

$6 - 10x + 4x^2$ is the same expression written in the other order

Part C

At $x = 2$ the given expression comes to $2$ while the student's last line comes to $42$ , so the two are not equivalent and that line cannot be a simplification of it. The check costs one substitution, needs no suspicion to start it, and any lost sign shows up as a disagreement.

Not what you got? Look for the slip on your own paper before you open the solution. Finding it yourself is worth more than reading it.
Worked solution

Part A

Test each line against the one directly above it, on its own.

Line $2$ comes from clearing the innermost parentheses, which means distributing $-5$ across $x - 1$ . That factor lands on both terms, and a negative times a negative is positive:

$-5(x - 1) = -5x + 5.$

The student wrote $-5x - 5$ , so the second product came out with the wrong sign and line $2$ should have read

$6 - 2x[3x - 5x + 5].$

Everything after that is careful work on a corrupted bracket, which is why nothing later looks suspicious. Line $3$ collects $3x - 5x = -2x$ correctly. Line $4$ distributes $-2x$ over $[-2x - 5]$ correctly, since $(-2x)(-2x) = 4x^2$ and $(-2x)(-5) = +10x$ . Line $5$ only reorders the three terms of line $4$ , which the commutative law permits. One step failed, and it was line $2$ .

Part B

Nested grouping symbols come apart from the inside out, so finish the innermost parentheses first:

$3x - 5(x - 1) = 3x - 5x + 5 = -2x + 5.$

Now scale that bracket by the factor sitting in front of it. The factor is a single term, $-2x$ , and both its sign and its letter go to both terms inside:

$-2x(-2x + 5) = 4x^2 - 10x,$

since $(-2x)(-2x) = 4x^2$ and $(-2x)(5) = -10x$ . Finally bring down the $6$ that has been waiting outside all along:

$6 + 4x^2 - 10x = 4x^2 - 10x + 6.$

Only the middle term differs from what the student handed in, which is the whole cost of one mishandled sign inside a bracket.

Part C

Take $x = 2$ . In the given expression, clear the innermost parentheses first:

$6 - 2(2)[3(2) - 5(2 - 1)] = 6 - 4[6 - 5] = 6 - 4 = 2.$

The student's last line at the same value is

$4(2)^2 + 10(2) + 6 = 16 + 20 + 6 = 42.$

Two expressions that disagree at even one value are not equivalent, so that line cannot be a simplification of the expression they were given. The verdict is reached by arithmetic alone, without reading a single algebraic step.

That is exactly why the check is worth running unprompted. Simplifying is supposed to leave the value untouched at every input, so a slip that changes the value announces itself the moment any one value is tested. It costs one substitution, it needs no suspicion to set it going, and the errors it catches, a sign left behind or a term never reached, are the two this lesson makes easiest to commit. What it cannot do is certify a correct answer: agreement at one value is evidence, since only one number out of infinitely many has been tried.

In one line

Line $2$ is the first that does not follow: $-5(x - 1)$ is $-5x + 5$ , so it should have read $6 - 2x[3x - 5x + 5]$ . Carried through, the expression simplifies to $4x^2 - 10x + 6$ . At $x = 2$ the given expression comes to $2$ while the student's last line comes to $42$ , which settles that the two are not equivalent.

Another way: Let a substitution find the line for you

Instead of auditing the algebra step by step, evaluate every line at one convenient value and look for the place where the value jumps. At $x = 2$ the printed expression is $2$ , while the student's lines $2$ through $5$ give

$42, \quad 42, \quad 42, \quad 42.$

The value changed between line $1$ and line $2$ , so line $2$ is where the work parted company with the expression, and every line after it is faithful to the wrong one. The algebra then has to be examined in a single line rather than in five.

When it is worth it On long work, or on somebody else's, where reading every step is slower than testing every step. It locates the failure but does not say what the failure was, so you still have to look hard at the line it points to.
Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Names one specific line as the first that does not follow, and leaves the lines above it standing rather than faulting a step that is sound. . Worth 2 points.

Says what the offending step did, in terms of the factor and the particular term it was multiplying. . Worth 2 points. needs an explanation, not just an answer

Rewrites that one line as it should have read, changing nothing else about it. . Worth 1 point.

Part B 5 points

Clears the innermost grouping symbol completely before touching the one outside it. . Worth 2 points.

Scales the remaining bracket by the whole factor in front of it, sign and letter together, and reaches both of its terms. . Worth 2 points.

Reports a form in which no two of the remaining terms are like. . Worth 1 point.

Part C 5 points

Evaluates the given expression and the student's last line at one and the same value, showing the arithmetic for both. . Worth 2 points.

States what the comparison settles, and gives a reason for running such a check on work nobody has flagged, rather than only asserting that it is a good habit. . Worth 3 points. needs an explanation, not just an answer

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Simplify $4 - 3y[2y - 6(y - 1)]$ , then check your result by substituting $y = 1$ into both the original and your simplified form.
5. One line, and everything that follows from it . Reasoning, 14 points. Question 5 of 5.

Everything you may and may not do when gathering terms comes from a single line of algebra, which is the distributive law read from right to left. State that line, prove it, and then let it settle three pairs of terms that are easy to misjudge.
1. Part A.
  Let $v$ stand for a variable part, and let $m$ and $n$ be any numbers. Prove that $mv + nv = (m + n)v$ , naming the law behind each rewriting, and say which step is the one that needs both terms to carry the same $v$ .
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
2. Part B.
  Decide, for each of these three pairs, whether the two terms gather into a single term, and gather the ones that do: $6x^2y$ and $-4yx^2$ ; then $3ab$ and $3a^2b$ ; then $\dfrac{2}{5}k$ and $\dfrac{7}{10}k$ .
  
  Justify your claim State the claim, then give the reason it has to be true. 6 points
3. Part C.
  A classmate reasons: since $3x + 5x = 8x$ , it must be that $3x + 5y = 8xy$ . Name the requirement their second step drops, explain why the gathering cannot begin without it, and say how far $3x + 5y$ can honestly be taken.
  
  Explain why it works A sentence or two. Reasons, not steps. 3 points
Hints
One at a time, each one a step further than the last. Take only as many as you need.
1. Hint 1 of 3
  One identity does all the work in this question. Get it written in letters first, with a name attached to every rewriting, and then treat each later part as a question about whether that identity is available.
2. Hint 2 of 3 · Part B
  Set the coefficients aside completely and look only at what is left of each term. Two of these pairs hide their verdict in the order the letters are written, or in what kind of number a coefficient is allowed to be.
3. Hint 3 of 3 · Part C
  The classmate's first calculation is sound. Put the two lines side by side and find the one condition that holds in the first and fails in the second, then trace what that condition was needed for.
That is every hint for this question.
Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 5 points

Derives the identity by attaching a named law to each rewriting, rather than asserting the result or testing it on particular numbers. . Worth 3 points. needs an explanation, not just an answer

Identifies the single step that consumes the shared quantity, and says what becomes unavailable without it. . Worth 2 points. needs an explanation, not just an answer

Part B 6 points

Reaches a stated verdict on each of the three pairs, leaving none of them unaddressed. . Worth 2 points.

Decides each pair by comparing variable parts, rather than by how similar the two written terms happen to look. . Worth 2 points. needs an explanation, not just an answer

Carries out the coefficient arithmetic for every pair that admits it, leaving the variable part unchanged. . Worth 2 points.

Part C 3 points

Names the requirement the second step drops, in terms of the parts of the two terms, and says what that requirement was needed for. . Worth 2 points. needs an explanation, not just an answer

States how far the second expression can be taken, and why that is as far as it goes. . Worth 1 point.

Try a similar problem (Optional)

Same idea, different numbers. Work it on paper, then check yourself the same way.

Decide which of these pairs gather into a single term, and gather the ones that do: $8u^3w$ and $-3wu^3$ ; then $5r^2$ and $5r$ ; then $\dfrac{3}{4}h$ and $\dfrac{5}{6}h$ .

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. Read the expression before you rewrite it 12 points

Part A 3

Lists every term with the sign that belongs to it, rather than dropping a leading minus and reporting the bare term. 2
Gives a coefficient for every term, including any term whose numerical factor is not written down. 1

Part B 5

Sorts the terms into families that share a variable part before adding anything. 2
Adds the coefficients within each family, keeping every sign, and leaves the variable part unchanged. 2
Reports a form in which no two of the remaining terms are like. 1

Part C 4

Names the law behind the combining step and shows it acting in the direction that takes a shared factor outside a bracket. 2
Says what specifically is missing in the second pair, rather than only asserting that those two terms are unlike. 2

2. The minus sign that has to reach every term 13 points

Part A 4

Adds the opposite of the second expression, so that every one of its terms is accounted for and not just the one written first. 2
Collects each family of like terms and reports a form in which no two terms are like. 2

Part B 4

Evaluates the printed expression as it stands, respecting both brackets and the order of operations, including the square of a negative number. 2
Evaluates the simplified form at the SAME value and sets the two results beside each other. 1
Says what a check of this kind establishes and what it leaves open, rather than treating the outcome as the end of the matter. 1

Part C 5

Evaluates the classmate's expression and the printed expression at one and the same value, showing the arithmetic for both. 2
Names one step that accounts for the whole of the classmate's expression and says what that step failed to do, rather than only reporting that the two forms differ. 3

3. How much more fence 14 points

Part A 4

Builds the fencing length from all four sides of the rectangle, rather than from one width and one length. 2
Scales each side length by its factor so that every term inside a bracket is reached, then collects the like terms. 2

Part B 3

States both of the second bed's side lengths before any perimeter is assembled. 1
Builds and simplifies the second fencing length by the same route used for the first. 2

Part C 3

Subtracts one whole fencing expression from the other, with the sign of every term of the subtracted expression accounted for. 2
Reports the comparison as a length in metres rather than as a bare number. 1

Part D 4

Says what the form of the comparison means for beds of different widths, rather than restating the comparison itself. 2
Points at the feature of the two fencing expressions that is responsible for the comparison taking that form. 2

4. Where the work stopped being equivalent 15 points

Part A 5

Names one specific line as the first that does not follow, and leaves the lines above it standing rather than faulting a step that is sound. 2
Says what the offending step did, in terms of the factor and the particular term it was multiplying. 2
Rewrites that one line as it should have read, changing nothing else about it. 1

Part B 5

Clears the innermost grouping symbol completely before touching the one outside it. 2
Scales the remaining bracket by the whole factor in front of it, sign and letter together, and reaches both of its terms. 2
Reports a form in which no two of the remaining terms are like. 1

Part C 5

Evaluates the given expression and the student's last line at one and the same value, showing the arithmetic for both. 2
States what the comparison settles, and gives a reason for running such a check on work nobody has flagged, rather than only asserting that it is a good habit. 3

5. One line, and everything that follows from it 14 points

Part A 5

Derives the identity by attaching a named law to each rewriting, rather than asserting the result or testing it on particular numbers. 3
Identifies the single step that consumes the shared quantity, and says what becomes unavailable without it. 2

Part B 6

Reaches a stated verdict on each of the three pairs, leaving none of them unaddressed. 2
Decides each pair by comparing variable parts, rather than by how similar the two written terms happen to look. 2
Carries out the coefficient arithmetic for every pair that admits it, leaving the variable part unchanged. 2

Part C 3

Names the requirement the second step drops, in terms of the parts of the two terms, and says what that requirement was needed for. 2
States how far the second expression can be taken, and why that is as far as it goes. 1

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