Literal equations: solving for one variable
A formula gives one quantity once you know the others. What if you know that one, and need one of the others?
Where you are 11 in Linear equations
When you reach for it
The formula d = rt says that distance is rate times time. Give it a rate and a time and it hands back a distance at once. Often the question runs the other way: you know how far a journey is and how fast you will go, and you want the time. You could put the two numbers in and solve the equation that results, but for twenty journeys that is twenty equations, each solved with exactly the same steps.
The better move is to do those steps once, with the letters still in place:
This is the same relationship, rearranged so that t stands alone, and every later question about time is now a single division.
An equation with more than one letter in it is called a literal equation, from the Latin littera, a letter. Formulas are the most familiar kind: P = 2l + 2w, C = 2\pi r, A = P + Prt. Solving a literal equation for one letter means rewriting it as an equivalent equation with that letter alone on one side and nowhere on the other. Reach for it when
- one formula is to be used many times with a different quantity unknown, as in a table or a spreadsheet column;
- you want to see how one quantity depends on the others: \displaystyle w = \frac{P}{2} - l shows at a glance that, for the same perimeter, a longer rectangle is a narrower one;
- an equation in x and y, such as 3x + 2y = 12, is wanted in the form “y equals something in x”, which is the form slope-intercept form and every graphing tool expect.
The procedure
Day 1: every other letter is a number
The whole idea fits in one sentence. To solve for one letter, treat every other letter as a number you happen not to know, and undo what has been done to the letter you want, exactly as in an equation with numbers.
This is allowed because of where the moves come from. The properties of equality say that adding the same number to both sides, subtracting it, multiplying both sides by it, or dividing both sides by it (as long as it is not 0) keeps an equation true. They hold for every number, so they hold for a letter that stands for a number. Subtracting b from both sides is as legitimate as subtracting 5.
Put the two side by side, both solved for x:
| Numbers | Letters | Why the step is allowed |
|---|---|---|
| 3x + 5 = 20 | ax + b = c | the equation to solve |
| 3x = 20 - 5 | ax = c - b | subtract 5, or b, from both sides |
| 3x = 15 | ax = c - b | the numbers combine; the letters cannot |
| \displaystyle x = \frac{15}{3} = 5 | \displaystyle x = \frac{c - b}{a} | divide both sides by 3, or by a |
The steps are the same, in the same order. The order comes from how x was built: in ax + b the x was first multiplied by a and then had b added, so the undoing runs backwards, taking away b first and dividing by a second. Only two things differ.
The arithmetic stops early. 20 - 5 becomes 15, but c - b cannot become anything simpler, because nobody has said what c and b are. So the answer is an expression, not a number, and it is finished once x stands alone.
Every division by a letter carries a condition. Dividing by 3 is always safe. Dividing by a is safe only if a is not 0, and a could be anything. So the full answer is
If a = 0 there is no x left in the equation at all: it says b = c, which is either true for every x or for none. In a formula about lengths, speeds or times the divisor is a positive quantity, so the condition holds automatically, but it is still part of the answer.
A literal answer is checked the way a number answer is, by putting it back in:
Or choose numbers for the other letters and compare: with a = 3, b = 5 and c = 20 the formula gives \displaystyle \frac{20 - 5}{3} = 5, the answer in the left-hand column.
An equation is solved for x when x appears exactly once, alone on one side, and the other side contains no x. Something like \displaystyle x = \frac{c - bx}{a} may be true, but it is not solved: it still needs x in order to find x.
Day 2: when the wanted letter appears twice
Sometimes the letter you want sits in two terms:
Dividing by a does not free x, because bx is still there, and dividing by b fails the same way. With numbers, 3x + 5x = 24, you would add the like terms to get 8x = 24. With letters, a + b cannot be turned into one number, so the adding has to be written in the only form it can take: factor x out.
| Numbers | Letters | Why the step is allowed |
|---|---|---|
| 3x + 5x = 24 | ax + bx = c | the equation to solve |
| (3 + 5)x = 24 | (a + b)x = c | the The distributive property, read right to left |
| x = 3 | \displaystyle x = \frac{c}{a + b} | divide both sides by 8, or by a + b |
This is factoring out a common factor, with the common factor being the letter you want. Once it is out, x appears once, multiplied by the bracket (a + b), and the bracket is just one more number you happen not to know. Dividing by it is the last step, under the condition a + b \neq 0.
If the wanted letter is on both sides, first gather it on one side by subtracting the same term from both sides, and move every term without it to the other side. Then factor, then divide. Example 5 does exactly this.
A line’s equation such as 3x + 2y = 12 is a literal equation too, and solving it for y is Example 6. Example 7 shows a rearranged formula paying for itself over several uses.
Worked examples
Examples 1 to 4 are the first day’s work, and Examples 5 to 7 the second’s.
Time from distance and rate
Solve d = rt for t. Then use the result: Tavita drives 333 km at an average speed of 74 km per hour. How long does the drive take?
SolutionIn d = rt the letter t has been multiplied by r. Undo that by dividing both sides by r, which is allowed because a speed of travel is not 0:
Swapping the two sides is allowed because an equation says the same thing read either way (the symmetric property of equality), so
For Tavita’s drive, d = 333 and r = 74:
The drive takes 4.5 hours, which is 4 hours 30 minutes. Check in the original formula: 74 \times 4.5 = 333. The units agree too: kilometers divided by kilometers per hour leaves hours, as Speed, distance and time explains.
The width of a rectangle from its perimeter
Solve P = 2l + 2w for w. Then find the width of a rectangular pen that uses 52 m of fencing and is 17 m long.
The Perimeter is two lengths and two widths. Here w was multiplied by 2 and then had 2l added, so undo the addition first. Subtract 2l from both sides:
The left side does not simplify, because P and 2l are not like terms. Now divide both sides by 2:
Dividing each term of the numerator by 2 gives a second form of the same answer:
The second form has a meaning you can see in the figure. Half the perimeter, \displaystyle \frac{P}{2}, is one length and one width, the walk from a corner to the opposite corner along the fence. Take away the length and the width is left.
For the pen, P = 52 and l = 17, so w = 26 - 17 = 9. The pen is 9 m wide. Check: 2(17) + 2(9) = 34 + 18 = 52.
The rearranged formula also shows a limit the original kept quiet about: w is positive only when \displaystyle l < \frac{P}{2}. With 52 m of fencing, the length must be under 26 m.
The radius from the circumference
Solve C = 2\pi r for r. Then find the radius of a circular ice rink whose edge barrier is 38 m long, to the nearest centimeter.
Solution\pi is written as a letter, but it is not a variable: it is one fixed number, about 3.14159. So 2\pi r is r multiplied by the number 2\pi, and one division undoes it. Divide both sides by 2\pi, which is not 0:
For the rink, C = 38:
The radius is about 6.05 m. Keeping \displaystyle \frac{19}{\pi} exact until the last step means rounding happens only once. Check: \displaystyle 2\pi \cdot \frac{19}{\pi} = 38. The formula being turned round is the one Circumference derives.
Fahrenheit to Celsius
A thermometer reads 59^\circ\text{F}. Solve \displaystyle F = \frac{9}{5}C + 32 for C, then find the Celsius reading.
SolutionC was multiplied by \displaystyle \frac{9}{5} and then had 32 added, so undo in reverse order. Subtract 32 from both sides:
Then multiply both sides by \displaystyle \frac{5}{9}, which undoes the multiplication by \displaystyle \frac{9}{5} because \displaystyle \frac{5}{9} \cdot \frac{9}{5} = 1:
The bracket matters: the \displaystyle \frac{5}{9} multiplies the whole difference. At 59^\circ\text{F}, \displaystyle C = \frac{5}{9}(27) = 15, so the reading is 15^\circ\text{C}. Check: \displaystyle \frac{9}{5}(15) + 32 = 27 + 32 = 59. Any other reading now takes one line with no new solving: 5^\circ\text{F} gives \displaystyle \frac{5}{9}(-27) = -15, that is -15^\circ\text{C}. Converting between Celsius and Fahrenheit puts both forms to work.
The wanted letter in two terms
(a) Solve A = P + Prt for P. Then find the amount that must be invested at simple interest of 5\% a year to grow to 1430 dollars in 2 years.
(b) Solve kx - 3 = 2x + m for x.
Solution(a) P appears twice, once alone and once in Prt, and neither term can be divided away on its own. The lone P is P \cdot 1, so factoring P out of both terms gives
Now P appears once, multiplied by the bracket. Divide both sides by it:
The condition 1 + rt \neq 0 always holds here: a rate and a time are not negative, so 1 + rt is at least 1. For the investment, 1 + rt = 1 + 0.05 \times 2 = 1.1, and
The investment is 1300 dollars. Check: 1300 \times 0.05 \times 2 = 130, and 1300 + 130 = 1430. Simple interest works with this factored form.
(b) Here x is on both sides. Gather the x terms on the left by subtracting 2x from both sides, and send the -3 to the right by adding 3 to both sides:
Factor x out of the left side, then divide by the bracket:
This time the condition is a real one: nothing stops k from being 2. If it is, the x terms cancel and the equation says -3 = m. That is true for every x when m = -3 and for no x otherwise, the two cases of No solution, one solution, or every number. Check with numbers: k = 5 and m = 9 give \displaystyle x = \frac{12}{3} = 4, and indeed 5(4) - 3 = 17 and 2(4) + 9 = 17.
A line’s equation, solved for y
Solve 3x + 2y = 12 for y. Use the result to make a table of points on the line and to find where it crosses each axis.
Subtract 3x from both sides, then divide both sides by 2. Dividing a difference by 2 divides each of its terms:
With the x term written first, as is usual,
Now every choice of x gives its y by one substitution, with no equation to solve:
| x | -2 | 0 | 2 | 4 | 6 |
|---|---|---|---|---|---|
| y | 9 | 6 | 3 | 0 | -3 |
The line crosses the y-axis where x = 0, at (0, 6), and the x-axis where y = 0, at (4, 0). Each time x goes up by 2, y goes down by 3: that steady change, \displaystyle -\frac{3}{2} for each unit of x, is the number in front of x, and the 6 is where the line crosses the y-axis. Reading those two numbers straight off the equation is what slope-intercept form is for. Check one row in the original: 3(2) + 2(3) = 6 + 6 = 12.
The original, in standard form, is no less correct; it is simply solved for neither letter. Solved for x instead it gives \displaystyle x = 4 - \frac{2}{3}y, the better form when the y values are the ones you are given.
One rearrangement, many trapezoids
The area of a trapezoid with parallel sides a and b and height h is \displaystyle A = \frac{1}{2}(a + b)h. Ingrid-Sol is designing four trapezoid tiles, each with an area of 36 square centimeters. Their parallel sides are 4 and 8, 3 and 6, 7 and 9, and 10 and 14 centimeters. Find the height each tile needs.
SolutionRather than solve four equations, solve the formula for h once. Multiply both sides by 2 to undo the half, then divide both sides by the bracket a + b, which is positive because the sides are lengths:
Here 2A = 72 every time, so each height is 72 divided by the sum of the parallel sides:
| Parallel sides (cm) | a + b | \displaystyle h = \frac{72}{a + b} (cm) |
|---|---|---|
| 4 and 8 | 12 | 6 |
| 3 and 6 | 9 | 8 |
| 7 and 9 | 16 | 4.5 |
| 10 and 14 | 24 | 3 |
The rearranged formula shows what the four separate answers would hide. The height depends on a + b alone, so a tile with sides 5 and 7 needs the same height as one with 4 and 8; and doubling the sum from 12 to 24 halves the height from 6 to 3. Check the third tile: \displaystyle \frac{1}{2}(7 + 9)(4.5) = 8 \times 4.5 = 36.
Boundaries
Dividing by a letter needs that letter, or bracket, to be non-zero. The answer \displaystyle x = \frac{c - b}{a} holds only for a \neq 0, and \displaystyle x = \frac{c}{a + b} only for a + b \neq 0. When the divisor is 0, the wanted letter has dropped out of the equation, and what is left is either true for every value or for none.
Solved means alone, and only once. The wanted letter must stand by itself on one side and be absent from the other. If it is still on both sides, the collecting and factoring are not finished.
The other letters must not secretly stand for the wanted one. Treating l as a number while solving P = 2l + 2w for w works because l does not depend on w. If the length is known to be twice the width, then l is 2w in disguise and must be replaced first.
A rearranged formula keeps the original’s limits. \displaystyle w = \frac{P}{2} - l gives a negative “width” when \displaystyle l > \frac{P}{2}, and \displaystyle t = \frac{d}{r} means nothing for r = 0. The algebra does not know what the letters measure; the situation does.
Different-looking answers can be equal. \displaystyle \frac{P - 2l}{2} and \displaystyle \frac{P}{2} - l are the same expression, as are \displaystyle \frac{12 - 3x}{2} and \displaystyle -\frac{3}{2}x + 6. Two answers agree when one turns into the other by allowed rewriting, or, as a quick test, when they give the same value for the same numbers.
This lesson keeps the wanted letter to the first power. If it is squared, as in A = \pi r^2 solved for r, the last step is a square root, \displaystyle r = \sqrt{\frac{A}{\pi}}, and the positive root is the right one only because a radius is positive. When the letter could be negative, both roots count; that belongs to solving with square roots.
Where this goes next
Solving an equation in x and y for y puts it in the shape y = mx + b, and slope-intercept form reads the slope and the starting value straight off that shape. Going from standard form to slope-intercept form is this lesson’s move with y as the wanted letter.
The same move builds inverse functions: solving y = 3x + 5 for x, to get \displaystyle x = \frac{y - 5}{3}, gives the rule that turns the output back into the input. In eliminating a parameter, one equation is solved for the shared letter t and the result substituted into the other.
In trigonometry, \displaystyle \sin A = \frac{a}{c} is solved for whichever side is missing, which is how a missing side is found from a ratio. And when the wanted letter sits in a denominator, clearing the fractions first turns the problem into one of this lesson’s, which is part of solving rational equations.
Exercises
Day 1: one appearance of the wanted letter
Solve each equation for the letter named, and state any condition the answer needs.
- A = lw, for l
- I = Prt, for r
- y = mx + b, for b
- P = a + b + c, for b
- y = mx + b, for x
- \displaystyle A = \frac{1}{2}bh, for h
- \displaystyle \frac{x}{a} + 4 = b, for x
- \displaystyle m = \frac{a + b}{2}, for b
Day 1: side by side
- Solve 7x - 4 = 24 for x, and next to it solve px - q = r for x, showing that the steps match.
Day 1: rearrange, then use
- Solve d = rt for r. Ruairi cycles 54 km in 2.25 hours. What is his average speed?
- Solve P = 2l + 2w for l. A rectangular noticeboard has perimeter 2.6 m and width 0.45 m. How long is it?
- Solve C = 2\pi r for r. A round tablecloth has a hem 470 cm long. Find its radius, exactly and to two decimal places.
Day 2: the wanted letter in two terms
- ax - 5x = 12, for x
- 3y + ky = k, for y
- S = 2lw + 2lh + 2wh, the surface area of a box, for h
- a(x - 2) = 3x, for x
- A = P + Prt, for r. (Compare with Example 5: which letter you want decides whether factoring is needed.)
Day 2: lines
Solve each equation for y.
- 4x + 5y = 20
- 2x - 3y = 9
- ax + by = c. What kind of line is it when b = 0?
Day 2: several values at once
- Nnamdi has 35 euros saved and adds 15 euros each week, so after t weeks he has A = 35 + 15t euros. Solve for t, then find how many weeks it takes to reach 95, 155 and 260 euros.
- A box has volume V = lwh. Solve for h. Leopold’s boxes all have a base 8 cm by 5 cm; find the heights that give volumes of 240, 320 and 440 cubic centimeters.
Reasoning
- Ruairi solves P = 2l + 2w for w and gets \displaystyle w = \frac{P - 2l}{2}. Tavita gets \displaystyle w = \frac{P}{2} - l. Who is right? Explain.
- Marisol-Kai solves ax + bx = c for x and writes \displaystyle x = \frac{c - bx}{a}. Every step she took was correct. Explain why the equation is still not solved for x, and finish the job.
- Leopold says that \displaystyle x = \frac{c}{a + b} solves ax + bx = c for every choice of a, b and c. Explain when it fails, and what the equation’s solutions are then.
Answers
- \displaystyle l = \frac{A}{w}, with w \neq 0. Divide both sides by w.
- \displaystyle r = \frac{I}{Pt}, with P \neq 0 and t \neq 0. r is multiplied by Pt, so divide both sides by Pt.
- b = y - mx. Subtract mx from both sides. No division, so no condition.
- b = P - a - c. Subtract a and c from both sides.
- \displaystyle x = \frac{y - b}{m}, with m \neq 0. Subtract b from both sides, then divide by m.
- \displaystyle h = \frac{2A}{b}, with b \neq 0. Multiply both sides by 2, giving 2A = bh, then divide by b.
- x = a(b - 4). Subtract 4: \displaystyle \frac{x}{a} = b - 4. Multiply both sides by a. The equation already needs a \neq 0, since it divides by a.
- b = 2m - a. Multiply both sides by 2: 2m = a + b. Subtract a.
- x = 4 and \displaystyle x = \frac{r + q}{p}, with p \neq 0. Add 4, or q, to both sides: 7x = 28 and px = r + q. Divide by 7, or by p.
- \displaystyle r = \frac{d}{t}, and Ruairi averages 24 km per hour. \displaystyle \frac{54}{2.25} = 24.
- \displaystyle l = \frac{P}{2} - w, and the noticeboard is 0.85 m long. Subtract 2w and divide by 2: \displaystyle l = \frac{P - 2w}{2}. Then 1.3 - 0.45 = 0.85.
- \displaystyle r = \frac{C}{2\pi}, so the radius is \displaystyle \frac{235}{\pi} \approx 74.80 cm. \displaystyle \frac{470}{2\pi} simplifies to \displaystyle \frac{235}{\pi}.
- \displaystyle x = \frac{12}{a - 5}, with a \neq 5. Factor: x(a - 5) = 12. Divide by a - 5.
- \displaystyle y = \frac{k}{3 + k}, with k \neq -3. Factor: y(3 + k) = k. Divide by 3 + k.
- \displaystyle h = \frac{S - 2lw}{2l + 2w}. Subtract 2lw: S - 2lw = 2lh + 2wh = h(2l + 2w). Divide by 2l + 2w, which is positive for a real box.
- \displaystyle x = \frac{2a}{a - 3}, with a \neq 3. Expand: ax - 2a = 3x. Gather: ax - 3x = 2a. Factor: x(a - 3) = 2a.
- \displaystyle r = \frac{A - P}{Pt}, with P \neq 0 and t \neq 0. r appears only once, so no factoring: subtract P, giving A - P = Prt, then divide by Pt.
- \displaystyle y = -\frac{4}{5}x + 4. 5y = 20 - 4x, then divide by 5.
- \displaystyle y = \frac{2}{3}x - 3. -3y = 9 - 2x. Dividing by -3 changes every sign: \displaystyle y = -3 + \frac{2}{3}x.
- \displaystyle y = \frac{c - ax}{b}, with b \neq 0, which is \displaystyle y = -\frac{a}{b}x + \frac{c}{b}. When b = 0 the equation is ax = c, so (for a \neq 0) \displaystyle x = \frac{c}{a}: a vertical line, which cannot be solved for y at all.
- \displaystyle t = \frac{A - 35}{15}. It takes 4 weeks to reach 95 euros, 8 weeks to reach 155, and 15 weeks to reach 260. For example, \displaystyle \frac{260 - 35}{15} = \frac{225}{15} = 15.
- \displaystyle h = \frac{V}{lw}. Here lw = 40, so the heights are 6 cm, 8 cm and 11 cm: \displaystyle \frac{240}{40}, \displaystyle \frac{320}{40} and \displaystyle \frac{440}{40}.
- Both are right. Dividing a difference by 2 divides each term (the distributive property), so \displaystyle \frac{P - 2l}{2} = \frac{P}{2} - \frac{2l}{2} = \frac{P}{2} - l.
- The right side still contains x, so the formula needs x to find x. Solved means x appears once, alone, and not on the other side. Gather both x terms on one side and factor: x(a + b) = c, so \displaystyle x = \frac{c}{a + b}, for a + b \neq 0.
- It fails when a + b = 0, because the last step divides by a + b. Then ax + bx = (a + b)x = 0 for every x, so the equation says 0 = c. If c = 0 that is true whatever x is, and every number is a solution; if c \neq 0 it is never true, and there is no solution.