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From Arithmetic to Algebra: 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 One line for a list of facts

Each of these facts is true:
$\begin{aligned} 6 \cdot 9 &= 6 \cdot 10 - 6 \\ 7 \cdot 9 &= 7 \cdot 10 - 7 \\ 23 \cdot 9 &= 23 \cdot 10 - 23 \end{aligned}$
Write one equation, using the letter $n$ , that states this pattern for every number $n$ , and check it at $n = 4$ .
1. Hint 1
  Look for what stays the same from one fact to the next and what changes. A letter takes the place of the part that changes.
2. Hint 2
  In each fact, find every place where the changing number appears; there is more than one. Each of those places gets the letter, and the $9$ and the $10$ stay as they are.
Answer

$n \cdot 9 = n \cdot 10 - n$ (or $9n = 10n - n$ ); at $n = 4$ both sides equal $36$ .

Full solution

In each fact the numbers $9$ and $10$ stay the same, while one number changes from fact to fact.
That changing number appears three times: multiplied by $9$ , multiplied by $10$ , and subtracted at the end.

Replace each appearance of the changing number with $n$ :
$n \cdot 9 = n \cdot 10 - n.$
The same line can be written
$9n = 10n - n.$

Check at $n = 4$ .
The left side is $4 \cdot 9$ , which is $36$ , and the right side is $4 \cdot 10 - 4$ , which is $40 - 4$ , also $36$ .
The two sides agree.

One check is evidence, not proof.
The distributive law proves the line for every number: write $9$ as $10 + (-1)$ , and use the law with $a = n$ , $b = 10$ and $c = -1$ :
$\begin{aligned} n \cdot 9 &= n(10 + (-1)) \\ &= n \cdot 10 + n \cdot (-1) \\ &= n \cdot 10 - n. \end{aligned}$
The last step uses $n \cdot (-1) = -n$ : multiplying a number by $-1$ gives its opposite, and adding the opposite of $n$ is subtracting $n$ .

Answer

$n \cdot 9 = n \cdot 10 - n$ (or $9n = 10n - n$ ); at $n = 4$ both sides equal $36$ .

Key idea

One equation with a letter states a pattern for every number at once, where arithmetic would need an endless list of facts.
Problem 1: Mark complete
Problem 2 A negative number in two places

Evaluate $2a^2 - 3(a - 5)$ when $a = -2$ .
1. Hint 1
  Write the value in parentheses wherever $a$ appears, so that its minus sign stays attached. Then the order of operations decides what to work out first.
2. Hint 2
  The square comes before the multiplication by $2$ , and $a - 5$ is worked out inside its parentheses before the multiplication by $3$ . Subtracting a negative number is the same as adding its opposite.
Answer

$29$

Full solution

Substitute $-2$ for $a$ in both places, keeping it in parentheses:
$2(-2)^2 - 3(-2 - 5).$

Exponents come before multiplication, so square first.
The square of $-2$ is $(-2)(-2)$ , which is $4$ , positive because a negative times a negative is positive.
The first term is therefore $2 \cdot 4$ , which is $8$ .

Parentheses come first in the second term: $-2 - 5$ is $-7$ .
The second term is therefore $3(-7)$ , which is $-21$ .

Finally subtract the second term from the first.
Taking away $-21$ is the same as adding $21$ :
$\begin{aligned} 8 - (-21) &= 8 + 21 \\ &= 29. \end{aligned}$

Two slips to avoid: $2a^2$ means $2 \cdot a^2$ , so it is not $(2a)^2$ , which would give $16$ ; and writing $-2^2$ without parentheses would be read as $-(2^2)$ , which is $-4$ .

Answer

$29$

Key idea

Put a substituted value in parentheses, then let the order of operations decide the order of the arithmetic.
Problem 2: Mark complete
Problem 3 Where do the two sides agree?

Evaluate both sides of $m^2 + 6 = 5m$ at $m = 1$ , $m = 2$ , $m = 3$ and $m = 4$ , and say at which of these values the equation is true.
1. Hint 1
  An equation is true at a value exactly when its two sides give the same number there, so work out each side on its own.
2. Hint 2
  On the left, square $m$ first and then add $6$ ; on the right, multiply $m$ by $5$ . At $m = 1$ , for example, the left side is $1 + 6$ .
Answer

True at $m = 2$ (both sides $10$ ) and at $m = 3$ (both sides $15$ ); false at $m = 1$ ( $7$ against $5$ ) and at $m = 4$ ( $22$ against $20$ ).

Full solution

An equation is true at a value when its two sides come out equal there.
Work out each side on its own, squaring before adding on the left.

At $m = 1$ the left side is $1 + 6$ , which is $7$ , and the right side is $5 \cdot 1$ , which is $5$ .
The sides differ, so the equation is false at $m = 1$ .

At $m = 2$ the left side is $4 + 6$ , which is $10$ , and the right side is $5 \cdot 2$ , which is $10$ .
The sides agree, so the equation is true at $m = 2$ .

At $m = 3$ the left side is $9 + 6$ , which is $15$ , and the right side is $5 \cdot 3$ , which is $15$ .
The sides agree, so the equation is true at $m = 3$ .

At $m = 4$ the left side is $16 + 6$ , which is $22$ , and the right side is $5 \cdot 4$ , which is $20$ .
The sides differ, so the equation is false at $m = 4$ .

Of these four values, then, the equation is true at $m = 2$ and $m = 3$ and false at $m = 1$ and $m = 4$ .
Because it fails at some values, it is not an identity, even though it holds at two of them.

Answer

True at $m = 2$ (both sides $10$ ) and at $m = 3$ (both sides $15$ ); false at $m = 1$ ( $7$ against $5$ ) and at $m = 4$ ( $22$ against $20$ ).

Key idea

An equation with a letter can be true at some values and false at others, and each value is tested by working out both sides there.
Problem 3: Mark complete
Problem 4 One law for each step

Each line of this chain follows from the line before it by exactly one law of addition:
$\begin{aligned} (-6 + y) + 6 &= (y + (-6)) + 6 \\ &= y + ((-6) + 6) \\ &= y + 0 \\ &= y. \end{aligned}$
For each of the four steps, name the law, and say what each letter of that law stands for in the step.
1. Hint 1
  A law holds for every number, so its letters can stand for a letter such as $y$ or a negative number such as $-6$ . For each step, ask what changed: the order, the grouping, or a part replaced by something simpler.
2. Hint 2
  Write out the four laws of addition with letters: $a + b = b + a$ , $(a + b) + c = a + (b + c)$ , $a + 0 = a$ and $a + (-a) = 0$ . Then match each step to the law whose two sides look like its before and after.
3. Hint 3
  In the third step, $(-6) + 6$ has the shape $a + (-a)$ . Decide what $a$ must be for $-a$ to be $6$ .
Answer

Step 1: commutative law, $a = -6$ , $b = y$ . Step 2: associative law, $a = y$ , $b = -6$ , $c = 6$ . Step 3: inverse law, $a = -6$ (so $-a = 6$ ). Step 4: identity law, $a = y$ . All four are laws of addition.

Full solution

Each law is a claim about every number, so its letters may stand for any numbers, a letter such as $y$ and a negative number such as $-6$ included.
Matching a step to a law means finding what each letter of the law stands for.

Step 1 swaps the order of the two numbers inside the parentheses and leaves the $+ 6$ outside alone.
That is the commutative law $a + b = b + a$ with $a = -6$ and $b = y$ , which turns $-6 + y$ into $y + (-6)$ .

Step 2 keeps the order $y$ , $-6$ , $6$ and moves the parentheses.
That is the associative law $(a + b) + c = a + (b + c)$ with $a = y$ , $b = -6$ and $c = 6$ .

Step 3 replaces $(-6) + 6$ with $0$ .
That is the inverse law $a + (-a) = 0$ with $a = -6$ : the opposite of $-6$ is $6$ , so $-a$ is $6$ .

Step 4 replaces $y + 0$ with $y$ .
That is the identity law $a + 0 = a$ with $a = y$ .

Read as a whole, the chain shows that adding $6$ undoes adding $-6$ .
Every step used a law that holds for every number, so this is true whatever number $y$ is.

Answer

Step 1: commutative law, $a = -6$ , $b = y$ . Step 2: associative law, $a = y$ , $b = -6$ , $c = 6$ . Step 3: inverse law, $a = -6$ (so $-a = 6$ ). Step 4: identity law, $a = y$ . All four are laws of addition.

Key idea

A law holds for every number, so a letter or a negative number can stand in for its letters, and that is what lets a law justify a step.
Problem 4: Mark complete
Problem 5 Every value, one value, or none?

For each equation below, decide whether it is true for every value of $y$ , for exactly one value of $y$ , or for no value of $y$ , and say how you know. Where it is true for exactly one value, give that value.

Equation A: $6(y + 2) = 6y + 12$

Equation B: $y + 11 = 4$

Equation C: $y - 2 = y + 5$
1. Hint 1
  An equation that is true for every value can be shown to be so with a law. One that is not an identity may still be true for exactly one value, or for none at all.
2. Hint 2
  For A, think of the distributive law. For B, ask which number gives $4$ when $11$ is added to it. For C, compare $y - 2$ and $y + 5$ for the same $y$ : how far apart are they?
Answer

A: true for every value of $y$ (an identity). B: true for exactly one value, $y = -7$ . C: true for no value of $y$ .

Full solution

Equation A: the distributive law $a(b + c) = ab + ac$ , with $a = 6$ , $b = y$ and $c = 2$ , says that $6(y + 2)$ equals $6y + 12$ for every number $y$ .
So equation A is true for every value of $y$ ; it is an identity.

Equation B: its left side is $11$ more than $y$ .
For that to be $4$ , $y$ must be $11$ less than $4$ , which is $-7$ .
Check:
$-7 + 11 = 4.$
A different value of $y$ gives a different sum, so equation B is true for exactly one value, $y = -7$ .

Equation C: for every number $y$ , the right side $y + 5$ is $7$ more than the left side $y - 2$ , since climbing from $y - 2$ to $y$ adds $2$ and climbing on to $y + 5$ adds $5$ more.
Two numbers that differ by $7$ are never equal, so equation C is true for no value of $y$ .
At $y = 0$ , for instance, the sides are $-2$ and $5$ .

So equation A holds whatever $y$ is, equation B singles out one number to be found, and equation C is satisfied by no number at all.

Answer

A: true for every value of $y$ (an identity). B: true for exactly one value, $y = -7$ . C: true for no value of $y$ .

Key idea

An identity holds for every value of its letter, while another equation may hold for one value, for several, or for none, so decide which kind an equation is before working on it.
Problem 5: Mark complete
Problem 6 Equivalent or not?

For each pair below, decide whether the two expressions are equivalent. If they are, name the law or laws that make them equal for every value of $k$ ; if they are not, give a value of $k$ at which they differ.

Pair A: $3(k \cdot 4)$ and $12k$

Pair B: $4(k + 3)$ and $k + 12$

Pair C: $-2(k - 3)$ and $6 - 2k$
1. Hint 1
  Equivalent means equal for every value of $k$ . A law that holds for every number can guarantee that, and one value where the two expressions differ rules it out.
2. Hint 2
  For a pair you suspect is not equivalent, substitute a simple value into both. Zero can make two different expressions agree, so do not rely on it alone.
3. Hint 3
  For a pair you suspect is equivalent, look for laws that turn one expression into the other: reordering and regrouping factors, or spreading a factor over both terms inside parentheses.
Answer

A: equivalent (commutative and associative laws of multiplication). B: not equivalent; at $k = 1$ they give $16$ and $13$ (any $k \neq 0$ works). C: equivalent (distributive law, then commutative law of addition).

Full solution

Equivalent means equal for every value of $k$ .
A law that holds for every number can guarantee that; a single value where the two expressions differ rules it out.

Pair A: by the commutative law of multiplication, $k \cdot 4$ equals $4k$ , so $3(k \cdot 4)$ equals $3(4k)$ .
By the associative law of multiplication, $3(4k)$ equals $(3 \cdot 4)k$ , which is $12k$ .
Both laws hold for every number, so the pair is equivalent.

Pair B: try $k = 1$ .
The first expression is $4(1 + 3)$ , which is $16$ , and the second is $1 + 12$ , which is $13$ .
They differ, so the pair is not equivalent.
The distributive law makes the $4$ multiply the $k$ as well as the $3$ , giving $4k + 12$ ; the second expression is what you get by multiplying only the $3$ .

Testing $k = 0$ alone would have misled you in pair B, because both expressions give $12$ there.
Every other value of $k$ exposes the difference.
Both $4k + 12$ and $k + 12$ add $12$ , so they agree only where $4k$ equals $k$ , and four times a number equals the number only when the number is $0$ .

Pair C: the distributive law, with $a = -2$ , $b = k$ and $c = -3$ , gives $-2k + 6$ , since $(-2)(-3)$ is $6$ .
The commutative law of addition turns $-2k + 6$ into $6 + (-2k)$ , which is $6 - 2k$ .
So the pair is equivalent.

Answer

A: equivalent (commutative and associative laws of multiplication). B: not equivalent; at $k = 1$ they give $16$ and $13$ (any $k \neq 0$ works). C: equivalent (distributive law, then commutative law of addition).

Key idea

A law settles that two expressions are equivalent, and one value where they differ settles that they are not; a value where they agree, such as $0$ in pair B, settles neither.
Problem 6: Mark complete
Problem 7 Stickers in symbols

Leo has $n$ stickers. Ava has $6$ fewer than three times as many stickers as Leo, and together they have $58$ stickers.

Write Ava's number of stickers as an expression in $n$ , and use it to write an equation that says they have $58$ stickers together. Then decide, by substitution, whether Leo can have $16$ stickers.
1. Hint 1
  Build Ava's amount in the order the words set it up: first three times Leo's amount, then six fewer than that. A sentence saying that two amounts are equal becomes an equation.
2. Hint 2
  Six fewer than a quantity takes $6$ away from that quantity, so the $6$ is written after it, not before. The total is Leo's amount added to Ava's.
3. Hint 3
  To test whether Leo can have $16$ , put $16$ in place of every $n$ in your equation and see whether its two sides agree.
Answer

Ava has $3n - 6$ stickers; the equation is $n + (3n - 6) = 58$ (or an equivalent form such as $4n - 6 = 58$ ). Leo can have $16$ : then Ava has $42$ , and $16 + 42 = 58$ .

Full solution

Three times as many stickers as Leo is $3n$ .
Six fewer than that takes $6$ away from $3n$ , so Ava has $3n - 6$ stickers.
The order matters: $6 - 3n$ would take $3n$ away from $6$ , the reverse of what the words say.

The phrase for Ava's stickers names an amount and claims nothing, so $3n - 6$ is an expression.
The sentence saying that together they have $58$ stickers claims that two amounts are equal, so it becomes an equation:
$n + (3n - 6) = 58.$

Substitute $16$ for $n$ .
Ava then has $3(16) - 6$ stickers, which is $48 - 6$ , or $42$ .
Together they have $16 + 42$ , which is $58$ , so the two sides of the equation agree and Leo can have $16$ stickers.

Here the letter $n$ names one particular number, the number of stickers Leo actually has, and the equation is a question about it.
That is a different job from the one a letter does in a law such as $a + b = b + a$ , where it stands for every number.

Answer

Ava has $3n - 6$ stickers; the equation is $n + (3n - 6) = 58$ (or an equivalent form such as $4n - 6 = 58$ ). Leo can have $16$ : then Ava has $42$ , and $16 + 42 = 58$ .

Key idea

A phrase becomes an expression and a sentence that says two amounts are equal becomes an equation, which a substitution can then test.
Problem 7: Mark complete
Problem 8 A number trick

A number trick goes like this: pick a number, double it, add $10$ , halve the result, and then subtract the number you picked.

Omar tries the trick with $1$ , $7$ and $20$ , and ends on the same number each time. Explain why his three tries do not settle whether every starting number ends on that number. Then settle it, by following the trick with a letter $n$ for the number picked.
1. Hint 1
  A try tells you what happens to the number tried and nothing about the others. A claim about every starting number needs an argument that works whatever the number is.
2. Hint 2
  Run the trick on a letter: picking $n$ and doubling gives $2n$ . Write each later instruction as an operation on the expression you have so far.
3. Hint 3
  Halving a sum halves each part, by the distributive law with the factor $\tfrac{1}{2}$ . Then see what subtracting $n$ does to the result.
Answer

Three tries leave every other starting number untested. Every starting number ends on $5$ : the trick gives $\tfrac{1}{2}(2n + 10) - n$ , which equals $5$ for every $n$ .

Full solution

Omar's tries: $1$ becomes $2$ , then $12$ , then $6$ , then $5$ ; $7$ becomes $14$ , $24$ , $12$ , then $5$ ; and $20$ becomes $40$ , $50$ , $25$ , then $5$ .
Each try ends on $5$ .

Those tries show what happens to $1$ , $7$ and $20$ , and nothing more.
There are infinitely many other starting numbers, and agreement on three of them cannot rule out one that ends somewhere else.
Settling a claim about every starting number takes an argument that does not depend on which number was picked.

Follow the trick with a letter.
Pick $n$ .
Doubling gives $2n$ , adding $10$ gives $2n + 10$ , and halving gives $\tfrac{1}{2}(2n + 10)$ .
Subtracting the number picked leaves
$\tfrac{1}{2}(2n + 10) - n.$

Halve with the distributive law, multiplying each part of the sum by $\tfrac{1}{2}$ :
$\begin{aligned} \tfrac{1}{2}(2n + 10) &= \tfrac{1}{2} \cdot 2n + \tfrac{1}{2} \cdot 10 \\ &= n + 5. \end{aligned}$
Here $\tfrac{1}{2} \cdot 2n$ is $(\tfrac{1}{2} \cdot 2)n$ by the associative law, which is $1 \cdot n$ because $\tfrac{1}{2}$ and $2$ multiply to $1$ (the inverse law).
That is $n \cdot 1$ by the commutative law, and $n$ by the identity law.

Now subtract the number picked, writing it as adding $-n$ :
$\begin{aligned} (n + 5) - n &= (5 + n) + (-n) \\ &= 5 + (n + (-n)) \\ &= 5 + 0 \\ &= 5. \end{aligned}$
The four steps use the commutative, associative, inverse and identity laws of addition, in that order.

Every step used a law that holds for every number, so every starting number ends on $5$ , the number Omar kept reaching.

Answer

Three tries leave every other starting number untested. Every starting number ends on $5$ : the trick gives $\tfrac{1}{2}(2n + 10) - n$ , which equals $5$ for every $n$ .

Key idea

Tries check the numbers tried; following the steps with a letter, using laws that hold for every number, shows what happens to every starting number at once.
Problem 8: Mark complete
Problem 9 Rui's check

Rui wants to know whether $4(x - 5) + 20 = 4x$ is true for every value of $x$ . He substitutes $x = 6$ and writes
$4 \cdot 6 - 5 + 20 = 39.$
Since $39$ is not $4 \cdot 6$ , which is $24$ , he concludes that the equation is not an identity.

Find the mistake in Rui's check and redo the check correctly. Then decide whether the equation is an identity, and justify your decision.
1. Hint 1
  A counterexample settles a claim only when its arithmetic is right. Compare what Rui wrote with the left side of the equation he was testing.
2. Hint 2
  Substitution puts $6$ where $x$ was and keeps the parentheses, so the $4$ multiplies all of $6 - 5$ . Work out the parentheses first.
3. Hint 3
  Agreement at one value does not prove an identity. Rewrite $4(x - 5)$ with the distributive law and see what adding $20$ then does.
Answer

Rui dropped the parentheses; correctly, $4(6 - 5) + 20 = 24$ , which matches $4 \cdot 6$ . The equation is an identity: its left side equals $4x$ for every $x$ .

Full solution

Substituting replaces $x$ with $6$ and keeps everything else, the parentheses included.
Rui wrote $4 \cdot 6 - 5$ , which multiplies only the $6$ by $4$ .
The correct substitution is $4(6 - 5) + 20$ , where the $4$ multiplies the whole of $6 - 5$ .

Redo the check, working out the parentheses first:
$\begin{aligned} 4(6 - 5) + 20 &= 4(1) + 20 \\ &= 24. \end{aligned}$
The right side is $4 \cdot 6$ , which is also $24$ , so the two sides agree at $x = 6$ .
Rui's value was never a counterexample; it came from a slip in the substitution.

One agreeing value does not prove an identity either, so turn to the laws.
The distributive law, with $a = 4$ , $b = x$ and $c = -5$ , says that $4(x - 5)$ equals $4x - 20$ for every $x$ .
The left side is therefore $4x - 20 + 20$ , and the laws of addition finish it:
$\begin{aligned} 4x - 20 + 20 &= 4x + (-20 + 20) \\ &= 4x + 0 \\ &= 4x. \end{aligned}$
These steps use the associative, inverse and identity laws, in that order.

Every step holds for every number, so the left side equals $4x$ whatever $x$ is, and the equation is an identity.

Answer

Rui dropped the parentheses; correctly, $4(6 - 5) + 20 = 24$ , which matches $4 \cdot 6$ . The equation is an identity: its left side equals $4x$ for every $x$ .

Key idea

A counterexample disproves a claimed identity when its arithmetic is right, so substitute with the parentheses kept and follow the order of operations.
Problem 9: Mark complete
Problem 10 Does subtraction regroup?

The associative law of addition says that $(a + b) + c = a + (b + c)$ for every choice of numbers $a$ , $b$ and $c$ . Jess claims that subtraction regroups in the same way:
$(a - b) - c = a - (b - c)$
for every choice of numbers $a$ , $b$ and $c$ .

Decide whether Jess is right. Then find every value of $c$ for which the two sides of her equation are equal, whatever the numbers $a$ and $b$ are, and explain why.
1. Hint 1
  A claim about every choice of $a$ , $b$ and $c$ fails as soon as one choice makes the two sides differ, so start by trying some numbers.
2. Hint 2
  Compare both sides with $a - b$ . The left side takes $c$ away from $a - b$ . The right side takes away $b - c$ instead of $b$ , and $b - c$ is $c$ less than $b$ .
3. Hint 3
  So one side is $c$ less than $a - b$ and the other is $c$ more. Ask which values of $c$ make subtracting $c$ and adding $c$ give the same result.
Answer

Jess is wrong: for example, $a = 9$ , $b = 5$ and $c = 2$ give $2$ on the left and $6$ on the right. The two sides are equal exactly when $c = 0$ , whatever $a$ and $b$ are.

Full solution

Try numbers first.
With $a = 9$ , $b = 5$ and $c = 2$ , the left side is $(9 - 5) - 2$ , which is $2$ , and the right side is $9 - (5 - 2)$ , which is $9 - 3$ , or $6$ .
The sides differ, so this one choice is a counterexample, and Jess's claim about every choice is false.

To find which values of $c$ work, compare each side with $a - b$ .
The left side, $(a - b) - c$ , takes $c$ away from $a - b$ , so it is $c$ less than $a - b$ .

The right side takes away $b - c$ instead of $b$ .
Since $b - c$ is $c$ less than $b$ , taking it away removes $c$ less than taking away $b$ would, so the right side is $c$ more than $a - b$ :
$a - (b - c) = (a - b) + c.$
With the numbers above, $5 - 2$ is $2$ less than $5$ , so $9 - 3$ is $2$ more than $9 - 5$ .

The two sides are therefore $(a - b) - c$ and $(a - b) + c$ .
Going from the left side to $a - b$ adds $c$ , and going on to the right side adds $c$ again, so the right side is the left side plus $2c$ .
Adding $2c$ leaves a number unchanged exactly when $2c$ is $0$ , that is, when $c = 0$ .
So the sides are equal when $c = 0$ , whatever $a$ and $b$ are, and differ for every other value of $c$ .

An equation that holds for some choices and fails for others is not a law.
Subtraction has no associative law, even though regrouping happens to work whenever $c = 0$ .

Answer

Jess is wrong: for example, $a = 9$ , $b = 5$ and $c = 2$ give $2$ on the left and $6$ on the right. The two sides are equal exactly when $c = 0$ , whatever $a$ and $b$ are.

Key idea

A claim about every choice of numbers is false once one choice fails, even when many choices, such as every choice with $c = 0$ , make it true.
Problem 10: Mark complete

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