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Solving for a Variable: 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 A cylinder's volume, aimed at its height

The volume of a cylinder with radius $r$ and height $h$ is $V = \pi r^2 h$ . Solve this formula for $h$ , and state the condition your final step requires.
1. Hint 1
  Treat every letter except $h$ as a fixed number. Then the whole product $\pi r^2$ is one fixed quantity multiplying $h$ , the same shape as $12 = 4h$ .
2. Hint 2
  Divide both sides by all of $\pi r^2$ , not by one factor of it. A division by an expression containing a letter needs that expression to be nonzero, so ask when $\pi r^2$ is zero.
Answer

$h = \dfrac{V}{\pi r^2}$ , valid when $r \neq 0$ (equivalently, $\pi r^2 \neq 0$ ).

Full solution

Every letter except the target counts as a fixed number, so read the formula as one fixed quantity, $\pi r^2$ , multiplying $h$ .
One division undoes that multiplication.

Divide both sides by the whole coefficient $\pi r^2$ .
On the right it cancels and leaves $h$ alone, so
$h = \frac{V}{\pi r^2}.$

The divisor $\pi r^2$ contains a letter, so the division assumes it is not zero.
The number $\pi$ is not zero, and $r^2 = r \cdot r$ is zero exactly when $r$ is zero.
So the rearranged formula holds provided $r \neq 0$ .

For an actual cylinder the radius is positive, so the condition is met, but it still belongs with the formula.

Answer

$h = \dfrac{V}{\pi r^2}$ , valid when $r \neq 0$ (equivalently, $\pi r^2 \neq 0$ ).

Key idea

To solve for a letter multiplied by a product, divide both sides by the whole product, and when that product contains a letter, state the condition that it is not zero.
Problem 1: Mark complete
Problem 2 Counting sides from an angle sum

The interior angles of a polygon with $n$ sides add up to $S = 180(n - 2)$ degrees. Solve this formula for $n$ , and check your rearranged formula with a hexagon, whose six interior angles add up to $720$ degrees.
1. Hint 1
  Two operations are wrapped around $n$ : first $2$ is subtracted from it, then the result is multiplied by $180$ . Solving undoes them in the reverse order.
2. Hint 2
  Divide both sides by $180$ first, since the multiplication came last, and then undo the subtraction. For the check, put $S = 720$ into your formula and see whether it returns $6$ .
Answer

$n = \dfrac{S}{180} + 2$ (or $n = \dfrac{S + 360}{180}$ ). At $S = 720$ it gives $n = 6$ , the hexagon's number of sides.

Full solution

Building $180(n - 2)$ from $n$ subtracts $2$ and then multiplies by $180$ .
Undo those operations in reverse order, the multiplication first.

Divide both sides by $180$ :
$\frac{S}{180} = n - 2.$
Then add $2$ to both sides:
$n = \frac{S}{180} + 2.$

The only division was by the number $180$ , which is not zero, so this rearrangement needs no condition on the letters.

Check it with the hexagon, where $S = 720$ :
$\begin{aligned} n &= \frac{720}{180} + 2 \\ &= 4 + 2 \\ &= 6. \end{aligned}$
A hexagon has six sides, so the rearranged formula passes the check.

Distributing first gives the same formula in another form.
From $S = 180n - 360$ , add $360$ to both sides and divide both sides by $180$ to reach $n = \dfrac{S + 360}{180}$ , which splits into $\dfrac{S}{180} + 2$ .

Answer

$n = \dfrac{S}{180} + 2$ (or $n = \dfrac{S + 360}{180}$ ). At $S = 720$ it gives $n = 6$ , the hexagon's number of sides.

Key idea

Undo the operations around the target in the reverse of the order they were applied, then test the result with numbers that fit the original formula.
Problem 2: Mark complete
Problem 3 The distance around a half-disc window

A window has the shape of half a disc: a semicircle of radius $r$ on top of a straight edge of length $2r$ . The distance around the window, the curved edge $\pi r$ plus the straight edge $2r$ , is $P = \pi r + 2r$ . Solve this formula for $r$ .
1. Hint 1
  The target $r$ appears in two terms, $\pi r$ and $2r$ . No single division removes it from both, so first arrange for $r$ to appear only once.
2. Hint 2
  Both terms share the factor $r$ . The distributive law read backward writes $\pi r + 2r$ as $r$ times one quantity, and then one division by that quantity finishes the job.
Answer

$r = \dfrac{P}{\pi + 2}$ (or $r = \dfrac{P}{2 + \pi}$ ).

Full solution

The target $r$ stands in two terms.
Dividing both sides by $\pi$ alone would turn them into $r + \dfrac{2r}{\pi}$ , which still holds $r$ twice, so gather the two copies of $r$ first.

Both terms carry the factor $r$ , so reading the distributive law backward gives
$P = r(\pi + 2).$
Expanding $r(\pi + 2)$ gives back $\pi r + 2r$ , which confirms the factoring.

Now $r$ appears once, multiplied by $\pi + 2$ .
Divide both sides by $\pi + 2$ :
$r = \frac{P}{\pi + 2}.$

The divisor $\pi + 2$ is a fixed number, a little more than $5$ , and it is not zero, so this rearrangement needs no condition on the letters.

Answer

$r = \dfrac{P}{\pi + 2}$ (or $r = \dfrac{P}{2 + \pi}$ ).

Key idea

When the target appears in two terms, factor it out so that it appears once, then divide both sides by the quantity left multiplying it.
Problem 3: Mark complete
Problem 4 A closed box, aimed at its height

A closed box has a rectangular base $l$ by $w$ and height $h$ , all positive lengths, so its six faces have a total area of
$A = 2lw + 2lh + 2wh.$
Solve this formula for $h$ . Then use your formula to find the height of a closed box whose base measures $5$ centimeters by $3$ centimeters and whose surface area is $94$ square centimeters, and check that height in the original formula.
1. Hint 1
  Sort the three terms into those that contain $h$ and the one that does not. Move the term without $h$ away first; the two terms with $h$ are the situation this lesson is about.
2. Hint 2
  After subtracting $2lw$ from both sides, both remaining terms contain $h$ . Factor $h$ out of them so that it appears once, then divide by everything left multiplying it.
3. Hint 3
  For the box, substitute $A = 94$ , $l = 5$ and $w = 3$ , working out the top and the bottom of the fraction separately. Then put all three measurements into the original formula.
Answer

$h = \dfrac{A - 2lw}{2(l + w)}$ (or $\dfrac{A - 2lw}{2l + 2w}$ ). The box is $4$ centimeters tall; the check gives $30 + 40 + 24 = 94$ .

Full solution

Only two of the three terms contain the target, so move the one that does not.
Subtract $2lw$ from both sides:
$A - 2lw = 2lh + 2wh.$

Both terms on the right contain $h$ , and both also carry a $2$ .
Reading the distributive law backward gathers them into one term:
$A - 2lw = 2h(l + w).$

Now $h$ appears once, multiplied by $2(l + w)$ , so divide both sides by that whole quantity, keeping the entire left side over the fraction bar:
$h = \frac{A - 2lw}{2(l + w)}.$

The divisor is built from letters, so the division needs
$2(l + w) \neq 0.$
The base measurements $l$ and $w$ are positive, so $l + w$ is positive and the division is safe.

For the box, substitute $A = 94$ , $l = 5$ and $w = 3$ , working out the numerator and the denominator separately:
$\begin{aligned} h &= \frac{94 - 2(5)(3)}{2(5 + 3)} \\ &= \frac{64}{16} \\ &= 4. \end{aligned}$
The height is $4$ centimeters: a length, so centimeters and not square centimeters.

Check in the original formula.
The three pairs of faces have areas $2(5)(3) = 30$ , $2(5)(4) = 40$ and $2(3)(4) = 24$ square centimeters, and $30 + 40 + 24 = 94$ , the surface area given.

Answer

$h = \dfrac{A - 2lw}{2(l + w)}$ (or $\dfrac{A - 2lw}{2l + 2w}$ ). The box is $4$ centimeters tall; the check gives $30 + 40 + 24 = 94$ .

Key idea

Move the terms without the target to the other side, factor the target out of the rest, and divide the whole other side by what is left.
Problem 4: Mark complete
Problem 5 Where a seesaw balances

Two children sit at the two ends of a seesaw that is $L$ meters long. One weighs $w$ kilograms and the other weighs $v$ kilograms, where $L$ , $w$ and $v$ are positive numbers. The seesaw balances when its pivot is $d$ meters from the first child, where
$wd = v(L - d).$
Solve this balance condition for $d$ . Then check your formula in the case where the two children weigh the same.
1. Hint 1
  Expand the right side first, then look at where $d$ appears: once on each side. One division can finish the job only once $d$ stands in a single place.
2. Hint 2
  Add $vd$ to both sides so that every term containing $d$ is on the left. Then factor $d$ out of those terms and divide by what is left multiplying it.
3. Hint 3
  For the check, put $w = v$ into your formula and simplify. Where should the pivot of a seesaw carrying two equal weights be?
Answer

$d = \dfrac{vL}{w + v}$ . When the two children weigh the same it gives $d = \dfrac{L}{2}$ , the middle of the seesaw.

Full solution

Treat $w$ , $v$ and $L$ as fixed numbers.
Expand the right side:
$wd = vL - vd.$

The target now appears on both sides.
Add $vd$ to both sides, so that every term containing $d$ is on the left:
$wd + vd = vL.$

Both terms on the left contain $d$ , so factor it out:
$d(w + v) = vL.$

Divide both sides by $w + v$ .
That division needs $w + v \neq 0$ , and since both weights are positive, $w + v$ is positive, so it is safe:
$d = \frac{vL}{w + v}.$

For the check, put $w = v$ :
$\begin{aligned} d &= \frac{vL}{v + v} \\ &= \frac{vL}{2v} \\ &= \frac{L}{2}. \end{aligned}$
The last step cancels $v$ , which is allowed because $v$ is positive.

So with equal weights the pivot sits at the middle of the seesaw, which is where two equal weights balance.
The formula passes the check.

Answer

$d = \dfrac{vL}{w + v}$ . When the two children weigh the same it gives $d = \dfrac{L}{2}$ , the middle of the seesaw.

Key idea

Collect every term containing the target on one side before factoring it out, and test the result on a case whose answer you already know.
Problem 5: Mark complete
Problem 6 Recovering a list price

A shop takes a fraction $d$ of an item's list price off, where $0 \le d \le 1$ . An item with a list price of $L$ dollars then sells for
$S = L - Ld$
dollars.

Solve this formula for $L$ , and state the value of $d$ it excludes. Use your formula to find the list price of a jacket that sold for $68$ dollars at $d = 0.15$ . Then explain, in terms of the shop's prices, why the sale price at the excluded value of $d$ cannot tell you the list price.
1. Hint 1
  The list price appears in both terms on the right, so it has to be gathered into one place before any division can free it.
2. Hint 2
  Write the lone $L$ as $L \cdot 1$ . Then both terms share the factor $L$ , and factoring it out leaves a $1$ inside the bracket. Divide by that whole bracket.
3. Hint 3
  For the jacket, the number you divide by is $1 - 0.15$ , not $0.15$ . For the last part, work out the sale price at the excluded rate for two items with very different list prices.
Answer

$L = \dfrac{S}{1 - d}$ , excluding $d = 1$ . The jacket's list price was $80$ dollars. At $d = 1$ every item sells for $0$ dollars whatever its list price, so a sale price of $0$ fits every list price.

Full solution

The target $L$ stands in both terms on the right.
The lone term is $L \cdot 1$ , so factoring out $L$ leaves a $1$ behind in the bracket:
$S = L(1 - d).$
Expanding $L(1 - d)$ gives back $L - Ld$ , which confirms the factoring.

Now $L$ appears once, multiplied by $1 - d$ , so divide both sides by $1 - d$ :
$L = \frac{S}{1 - d}.$
The divisor $1 - d$ is zero exactly when $d = 1$ , so the formula excludes $d = 1$ .

For the jacket, $S = 68$ and $d = 0.15$ , so the divisor is $1 - 0.15 = 0.85$ :
$\begin{aligned} L &= \frac{68}{0.85} \\ &= 80. \end{aligned}$
The list price was $80$ dollars.
Running it forward confirms it: the discount is $80(0.15) = 12$ dollars, and $80 - 12 = 68$ dollars, the sale price.

At $d = 1$ the shop takes the whole list price off, so every item sells for $L - L(1) = 0$ dollars.
A coat listed at $200$ dollars and a scarf listed at $12$ dollars would both sell for $0$ dollars.

The single sale price $0$ fits the coat and the scarf alike, so it cannot say which list price it came from.
The excluded value marks the one rate at which the sale price no longer carries the list price.

Answer

$L = \dfrac{S}{1 - d}$ , excluding $d = 1$ . The jacket's list price was $80$ dollars. At $d = 1$ every item sells for $0$ dollars whatever its list price, so a sale price of $0$ fits every list price.

Key idea

The value a rearranged formula excludes can be a real limit of the situation: at $d = 1$ the sale price no longer tells you the list price.
Problem 6: Mark complete
Problem 7 Three answers and one test

A phone's battery starts at $F$ percent and drops by $r$ percentage points for each hour of use, where $r > 0$ , so after $h$ hours it shows $B = F - rh$ percent. Three students solved this formula for $h$ . Priya wrote $h = \dfrac{F - B}{r}$ , Tom wrote $h = F - \dfrac{B}{r}$ , and Lena wrote $h = \dfrac{B - F}{r}$ .

Test all three forms with a battery that starts at $100$ percent and drops $10$ percentage points an hour for $4$ hours, to find which of them the test rules out. Then rearrange the formula yourself to decide whether any form that passes is correct.
1. Hint 1
  A fair test uses numbers that fit the original formula. Work out what the battery shows from the original, then see which forms give back the number of hours you started with.
2. Hint 2
  With $F = 100$ , $r = 10$ and $h = 4$ , the original formula gives the reading $B$ . Put that reading, with $F = 100$ and $r = 10$ , into each student's form.
3. Hint 3
  To rearrange, add $rh$ to both sides so that the $h$ term is positive, move $B$ to the other side, and then divide the whole side by $r$ .
Answer

The test gives $4$ hours for Priya's form, $94$ hours for Tom's and $-4$ hours for Lena's, so it rules out Tom's and Lena's. Rearranging gives $h = \dfrac{F - B}{r}$ (with $r > 0$ , as given), so Priya's form is correct.

Full solution

A fair test starts from numbers that fit the original formula.
With $F = 100$ , $r = 10$ and $h = 4$ , the battery loses $10(4) = 40$ percentage points in the $4$ hours, so it shows $B = 60$ percent.

Put $F = 100$ , $B = 60$ and $r = 10$ into each form.
Priya's gives $\dfrac{100 - 60}{10} = 4$ , the $4$ hours that went in.
Tom's gives
$100 - \dfrac{60}{10} = 94.$
Lena's gives
$\dfrac{60 - 100}{10} = -4.$

So the test rules out Tom's and Lena's forms, since neither returns the $4$ hours.
Tom divided only $B$ by $r$ and left $F$ undivided.
Lena lost a sign, and her form even gives a negative number of hours.

Priya's form passes, but one passing test does not prove a formula right, since a wrong formula can agree with the right one at particular numbers.
Rearranging settles it.
Add $rh$ to both sides, then subtract $B$ from both sides:
$\begin{aligned} B + rh &= F \\ rh &= F - B. \end{aligned}$

Divide both sides by $r$ , which is not zero because $r > 0$ :
$h = \frac{F - B}{r}.$
That is Priya's form, so hers is the correct rearrangement.

Answer

The test gives $4$ hours for Priya's form, $94$ hours for Tom's and $-4$ hours for Lena's, so it rules out Tom's and Lena's. Rearranging gives $h = \dfrac{F - B}{r}$ (with $r > 0$ , as given), so Priya's form is correct.

Key idea

A test with numbers that fit the original can rule a rearrangement out; to show that a form that passes is right, rearrange the formula yourself.
Problem 7: Mark complete
Problem 8 Dana's price for a school group

A museum charges $p$ dollars for each of the $n$ students in a school group, and $3$ dollars more than that for each of the $2$ teachers with them, so the group pays $T = np + 2(p + 3)$ dollars in all.

Asked to solve this formula for $p$ , Dana writes
$p = \frac{T - 2p - 6}{n}.$
She checks it with a group of $10$ students at a student price of $15$ dollars, which pays $186$ dollars in all. Her right side gives $\dfrac{186 - 30 - 6}{10} = 15$ , so her check passes.

Decide whether Dana has solved the formula for $p$ , explain why her check passed, and then solve the formula for $p$ yourself.
1. Hint 1
  Nothing Dana did breaks a rule of algebra. The question to ask is a different one: what does solving for $p$ require of the finished equation, and does hers meet it?
2. Hint 2
  Her equation came from the formula by legal moves, so any numbers that fit the formula also fit her equation. Did her test numbers fit the formula?
3. Hint 3
  To solve properly, expand the bracket and move the constant term away first, then gather the two terms that contain $p$ into one term by factoring, before you divide by anything.
Answer

Not solved: $p$ is still on her right side. Her equation follows from the formula by legal moves, so any numbers that fit the formula, hers included, fit it too. Solved for $p$ : $p = \dfrac{T - 6}{n + 2}$ .

Full solution

Dana's equation is true.
Expanding the bracket gives
$T = np + 2p + 6.$
Subtracting $2p$ and $6$ from both sides gives $T - 2p - 6 = np$ , and dividing both sides by $n$ , which is not zero for a group with students in it, gives exactly what she wrote.

Every step she took was a legal move, so any numbers that fit the original formula also fit her equation.
Her test numbers fit the original: the students pay $10(15) = 150$ dollars, the teachers pay $2(15 + 3) = 36$ dollars, and
$150 + 36 = 186.$
So her check could not have failed, for this group or for any other whose total fits the museum's prices.

This is not a check that passed on a wrong value: nothing in her work is wrong.
The trouble is that it is unfinished.
To solve for $p$ is to reach an equation with $p$ alone on one side and nowhere on the other, so that $p$ can be computed from $T$ and $n$ .

Her right side still contains $p$ .
For the group of $10$ students that paid $186$ dollars it reads $p = \dfrac{186 - 2p - 6}{10}$ , which is an equation still to be solved, not a price.
So Dana has not solved the formula for $p$ .

To solve it properly, start from the expanded formula $T = np + 2p + 6$ and subtract $6$ from both sides, so that only the terms containing $p$ are left on the right:
$T - 6 = np + 2p.$
Both of those terms carry the factor $p$ , so
$T - 6 = p(n + 2).$
Divide both sides by $n + 2$ , which is at least $2$ for any group, so the division is safe:
$p = \frac{T - 6}{n + 2}.$

For the same group this gives the price directly from $T$ and $n$ , with no $p$ needed on the right:
$\begin{aligned} p &= \frac{186 - 6}{10 + 2} \\ &= \frac{180}{12} \\ &= 15. \end{aligned}$
That is the $15$ dollars a student that the group paid.

Answer

Not solved: $p$ is still on her right side. Her equation follows from the formula by legal moves, so any numbers that fit the formula, hers included, fit it too. Solved for $p$ : $p = \dfrac{T - 6}{n + 2}$ .

Key idea

A rearranged equation can be true without being solved: the target has to stand alone on one side and appear nowhere on the other.
Problem 8: Mark complete
Problem 9 A letter inside the coefficient

Here $k$ stands for a fixed number. Solve
$k(x - 3) = 2x + 1$
for $x$ , and state the value of $k$ your answer excludes. Then find every number $x$ that satisfies the original equation when $k$ has that excluded value.
1. Hint 1
  Treat $k$ like any other constant. Expand the bracket, then gather every term containing $x$ on one side, just as you would with numbers.
2. Hint 2
  After collecting, the $x$ terms are $kx$ and $-2x$ . Factor $x$ out of them and divide by what is left; that divisor is where the excluded value comes from.
3. Hint 3
  At the excluded value, expand the original equation and subtract the $x$ term from both sides. Ask whether any number $x$ can make the statement that remains true.
Answer

$x = \dfrac{3k + 1}{k - 2}$ , excluding $k = 2$ . At $k = 2$ the equation has no solution.

Full solution

Treat $k$ as a fixed number and expand the left side:
$kx - 3k = 2x + 1.$

Subtract $2x$ from both sides and add $3k$ to both sides, so that the $x$ terms are on the left and everything else is on the right:
$kx - 2x = 3k + 1.$

Both terms on the left contain $x$ , so factor it out:
$x(k - 2) = 3k + 1.$

Divide both sides by $k - 2$ , which requires $k \neq 2$ :
$x = \frac{3k + 1}{k - 2}.$
A check at $k = 3$ : the formula gives $x = 10$ , and the original equation has $3(10 - 3) = 21$ on the left and $2(10) + 1 = 21$ on the right.

Now put the excluded value $k = 2$ into the original equation and expand the left side:
$2x - 6 = 2x + 1.$
Subtracting $2x$ from both sides leaves $-6 = 1$ , a false statement with no $x$ in it.
No number satisfies the equation, so it has no solution.

That is why $k = 2$ had to be excluded.
A formula for $x$ hands back one number, but at $k = 2$ there is no solution for it to hand back.
The collected equation says the same, since at $k = 2$ it reads
$x \cdot 0 = 7.$

Answer

$x = \dfrac{3k + 1}{k - 2}$ , excluding $k = 2$ . At $k = 2$ the equation has no solution.

Key idea

A value that a rearrangement excludes needs its own look in the original equation, which here has no solution at that value.
Problem 9: Mark complete
Problem 10 Does rearranging only rewrite?

Jess claims that solving a formula for a different letter only rewrites it, so any numbers that fit the original formula also fit the rearranged one.

Test her claim on the formula $M = ab - 4a$ and its rearrangement $a = \dfrac{M}{b - 4}$ . Find numbers for $M$ , $a$ and $b$ that fit the first but not the second, and say what each form does with them. Then correct her claim about this formula so that it is true.
1. Hint 1
  Her claim is about every set of numbers, so one set that breaks it is enough. The rearranged form can fail to name a number in one place only: where its denominator is zero.
2. Hint 2
  The denominator $b - 4$ is zero at $b = 4$ . Choose any value for $a$ , work out $M$ from the first form at $b = 4$ , and then put $M$ and $b$ into the second form.
3. Hint 3
  For the correction, find the step of the rearranging that divided by an expression containing a letter, and say what the claim needs to assume about that expression.
Answer

For example, $b = 4$ , $M = 0$ and any $a$ , such as $a = 5$ , fit $M = ab - 4a$ , while the second form's right side becomes $\dfrac{0}{0}$ , which names no number. Corrected: the two forms fit exactly the same numbers when $b \neq 4$ .

Full solution

The claim is about every set of numbers at once, so one set that breaks it is enough.
The second form fails to name a number only where its denominator $b - 4$ is zero, that is, at $b = 4$ .

Take $b = 4$ and any value of $a$ , say $a = 5$ .
The first form gives
$\begin{aligned} M &= 5(4) - 4(5) \\ &= 0. \end{aligned}$
So $M = 0$ , $a = 5$ and $b = 4$ fit
$M = ab - 4a.$

Put the same numbers into the second form: its right side becomes $\dfrac{0}{4 - 4}$ , that is, $\dfrac{0}{0}$ .
Division by zero names no number, so the second form gives no value of $a$ at all here, let alone $a = 5$ .
These numbers fit the first form and not the second, so the claim is false.

The rearranging took two steps.
Factoring $ab - 4a$ as $a(b - 4)$ rewrites a side as an equal expression, which is true for every value of the letters.
Dividing both sides by $b - 4$ is allowed only when $b - 4 \neq 0$ , and when it is allowed it can be undone by multiplying back by $b - 4$ .

So the claim becomes true once the condition is attached: the two forms fit exactly the same numbers when $b \neq 4$ .
The rearranged formula has to carry that condition, because at $b = 4$ and $M = 0$ the original holds for every value of $a$ , while the rearranged form names none.

Answer

For example, $b = 4$ , $M = 0$ and any $a$ , such as $a = 5$ , fit $M = ab - 4a$ , while the second form's right side becomes $\dfrac{0}{0}$ , which names no number. Corrected: the two forms fit exactly the same numbers when $b \neq 4$ .

Key idea

A rearrangement that divides by an expression containing a letter says nothing where that expression is zero, so its condition has to be written with it.
Problem 10: Mark complete

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