Unit 2 · Lesson 3 of 6
Formulas: solving for a letter
You'll be able to: rearrange a formula such as \(P = 2l + 2w\) so that a chosen letter is alone on one side.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first; it is fine to be slow.
Solve \(\dfrac{x}{3} - \dfrac{1}{2} = \dfrac{x}{6}\). Type the value of \(x\).
The LCD is 6. Multiply every term by 6: \(2x - 3 = x\).
Subtract \(x\): \(x - 3 = 0\), so \(x = 3\).
Evaluate \(b^2 - 4ac\) when \(a = 2\), \(b = -3\), and \(c = -1\).
Use parentheses: \((-3)^2 - 4(2)(-1)\).
\((-3)^2 = 9\) and \(4(2)(-1) = -8\).
So \(9 - (-8) = 9 + 8 = 17\).
Step 2 of 5The idea
A formula like \(P = 2l + 2w\) is a literal equation: an equation with several letters. "Solve for \(w\)" means get \(w\) alone on one side.
You already know how. Use the same moves as in a normal equation. The only change: the other letters stay in the answer, because you cannot combine them with numbers.
To solve for one letter:
- Circle the letter you want. Every other letter acts like a number.
- Add or subtract to move the terms without your letter to the other side.
- Divide both sides by whatever is multiplying your letter (or multiply to clear a fraction).
Compare: \(3x + 5 = 11\) gives \(3x = 11 - 5\), then \(x = \dfrac{11 - 5}{3}\). In the same way, \(ax + b = c\) gives \(ax = c - b\), then \(x = \dfrac{c - b}{a}\).
Why this shows up on the placement test: the test asks you to solve formulas for a letter, and you will do this every time you rewrite a line as \(y = mx + b\).
Step 3 of 5Worked examples
Example 1: solve \(P = 2l + 2w\) for \(w\)
- The letter we want is \(w\). Treat \(P\) and \(l\) like numbers.
- Subtract \(2l\) from both sides: \(P - 2l = 2w\). The \(2l\) term has no \(w\), so it moves away.
- Divide both sides by 2: \(\dfrac{P - 2l}{2} = w\).
- Answer: \(w = \dfrac{P - 2l}{2}\). Same thing written another way: \(w = \dfrac{P}{2} - l\). Both are correct.
Example 2: solve \(y = mx + b\) for \(m\)
- The letter we want is \(m\). It is multiplied by \(x\).
- Subtract \(b\) from both sides: \(y - b = mx\).
- Divide both sides by \(x\): \(\dfrac{y - b}{x} = m\).
- Answer: \(m = \dfrac{y - b}{x}\). Quick check with numbers: if \(m = 2, x = 3, b = 1\), then \(y = 7\), and \(\dfrac{7 - 1}{3} = 2\).
Step 4 of 5Practice
Type only the right side of your answer (for example, for \(w = \dfrac{P - 2l}{2}\) type (P - 2l)/2). Capital and small letters count as different letters, so copy them exactly.
The area of a rectangle is \(A = lw\). Solve for \(w\).
\(w\) is multiplied by \(l\), so divide both sides by \(l\): \(\dfrac{A}{l} = w\).
Answer: \(w = \dfrac{A}{l}\).
Distance is \(d = rt\) (rate times time). Solve for \(t\).
\(t\) is multiplied by \(r\), so divide both sides by \(r\).
Answer: \(t = \dfrac{d}{r}\).
The area of a triangle is \(A = \dfrac{1}{2}bh\). Solve for \(h\).
Clear the fraction: multiply both sides by 2 to get \(2A = bh\).
Divide both sides by \(b\): \(\dfrac{2A}{b} = h\).
Answer: \(h = \dfrac{2A}{b}\).
A taxi ride costs \(C = 3 + 2m\) dollars, where \(m\) is the number of miles. Solve for \(m\).
Subtract 3 from both sides: \(C - 3 = 2m\).
Divide the whole side by 2: \(\dfrac{C - 3}{2} = m\).
Answer: \(m = \dfrac{C - 3}{2}\).
Solve \(3x + 2y = 6\) for \(y\).
Subtract \(3x\) from both sides: \(2y = 6 - 3x\).
Divide both sides by 2: \(y = \dfrac{6 - 3x}{2}\).
Answer: \(y = \dfrac{6 - 3x}{2}\), which is the same as \(y = -\dfrac{3}{2}x + 3\).
Temperature: \(F = \dfrac{9}{5}C + 32\). Solve for \(C\).
Subtract 32 from both sides: \(F - 32 = \dfrac{9}{5}C\).
Multiply both sides by \(\dfrac{5}{9}\), the reciprocal of \(\dfrac{9}{5}\): \(\dfrac{5}{9}(F - 32) = C\).
Answer: \(C = \dfrac{5}{9}(F - 32)\).
Solve \(2x - 5y = 10\) for \(y\).
Subtract \(2x\) from both sides: \(-5y = 10 - 2x\).
Divide both sides by \(-5\): \(y = \dfrac{10 - 2x}{-5}\). Dividing each term by \(-5\): \(\dfrac{10}{-5} = -2\) and \(\dfrac{-2x}{-5} = \dfrac{2}{5}x\).
Answer: \(y = \dfrac{2}{5}x - 2\), which is the same as \(y = \dfrac{2x - 10}{5}\).
Which is \(P = 2l + 2w\) solved for \(l\)?
- \(l = \dfrac{P}{2} - 2w\)
- \(l = \dfrac{P + 2w}{2}\)
- \(l = \dfrac{P - 2w}{2}\)
- \(l = P - \dfrac{2w}{2}\)
Subtract \(2w\) from both sides: \(P - 2w = 2l\). Divide the whole side by 2: \(l = \dfrac{P - 2w}{2}\). Answer (c).
(a) divides only \(P\) by 2, (b) adds \(2w\) instead of subtracting it, and (d) divides only the \(2w\) by 2.
Step 5 of 5Wrap-up
You can now solve a formula for any letter: move the terms without that letter away, then divide the whole side by what multiplies it.
Read or watch next: OpenStax Intermediate Algebra 2e, 2.3 Solve a Formula for a Specific Variable (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Algebra 2: Manipulating formulas (short videos). Quick rules: Equations and inequalities cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why can't I cancel the 2s in (P - 2l)/2?" It knows this course and can explain things another way.