Algebra Foundations for Normal People by Gabriel Beguerie - HTML preview
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Chapter goals
By the end of this chapter, you will be able to:
• substitute values into formulas correctly,
• solve a formula for a chosen variable,
• handle parentheses and fractions carefully while rearranging,
• avoid common denominator mistakes,
• see why formula control matters in science, finance, and statistics.
Prerequisites (if this is hard, go back to Unit 0)
You need equation solving, fractions, exponents, roots, and line formulas from earlier chapters.
Why this matters
Key idea
A formula is an equation that packages a relationship.
In algebra class, formulas are practice. Outside algebra class, formulas are how science, geometry, finance, and statistics talk.
Real life: what it’s for
Formulas tell you area from length and width, distance from rate and time, temperature conversion, interest from principal, rate, and time, and many statistics quantities from data.
If you cannot read and rearrange formulas, later subjects feel harder than they need to.
98
Mode 1: substituting into a formula
99
Mode 1: substituting into a formula
Key idea
Substitution means replacing each variable with its known value and then simplifying carefully.
This sounds simple, but parentheses matter, especially for negative numbers and multi-part expressions.
Recipe (steps that always work)
To use a formula:
1. write the formula clearly,
2. replace each variable with its given value,
3. use parentheses for negative numbers or multi-part values, 4. simplify carefully,
5. report the answer with the right meaning or units.
Worked example (no skipped steps)
Use
𝐴 = 𝑙𝑤
to find the area when 𝑙 = 8 and 𝑤 = 3.
Step 1: substitute the values.
𝐴 = (8)(3)
Step 2: multiply.
𝐴 = 24
Answer: the area is 24 square units.
Worked example (no skipped steps)
Use
𝑑 = 𝑟𝑡
to find the distance when 𝑟 = 60 and 𝑡 = 2 . 5.
Step 1: substitute.
𝑑 = (60)(2 . 5)
Step 2: multiply.
𝑑 = 150
Answer: the distance is 150.
Chapter 12. Formulas and rearranging variables
Worked example (no skipped steps)
Use
𝑦 = 𝑚𝑥 + 𝑏
with 𝑚 = 2, 𝑥 = 5, and 𝑏 = −3.
Step 1: substitute carefully.
𝑦 = (2)(5) + (−3)
Step 2: multiply first.
(2)(5) = 10
So the expression becomes:
𝑦 = 10 + (−3)
Step 3: add the signed numbers.
𝑦 = 7
Answer: 𝑦 = 7.
Mode 2: solving a formula for one variable 101
Worked example (no skipped steps)
Use
𝑃 = 2 𝑙 + 2 𝑤
to find the perimeter when 𝑙 = 7 and 𝑤 = 4.
Step 1: substitute.
𝑃 = 2(7) + 2(4)
Step 2: compute each product.
2(7) = 14 ,
2(4) = 8
So the expression becomes:
𝑃 = 14 + 8
Step 3: add.
𝑃 = 22
Answer: 𝑃 = 22.
Note
If a value is negative, protect it with parentheses.
For example, if 𝑥 = −3, then
𝑥 2 = (−3)2 = 9
but
−32 = −9
because the exponent hits the 3 before the negative sign.
Mode 2: solving a formula for one variable
Key idea
When you solve a formula for one variable, you treat the formula like any other equation.
The job is not mysterious: get the requested variable alone by undoing the operations attached to it.
Recipe (steps that always work)
To solve a formula for a chosen variable:
1. identify which variable you want alone,
2. undo addition and subtraction first,
3. undo multiplication and division next,
4. if powers remain, use the matching inverse operation, 5. check whether the final form makes structural sense.
Chapter 12. Formulas and rearranging variables
Worked example (no skipped steps)
Solve for 𝑏 in
𝑦 = 𝑚𝑥 + 𝑏
Step 1: identify what is attached to 𝑏. The term 𝑚𝑥 is being added to 𝑏.
Step 2: subtract 𝑚𝑥 from both sides.
𝑦 − 𝑚𝑥 = 𝑚𝑥 + 𝑏 − 𝑚𝑥
Step 3: simplify the right side.
𝑦 − 𝑚𝑥 = 𝑏
So:
𝑏 = 𝑦 − 𝑚𝑥
Worked example (no skipped steps)
Solve for 𝑡 in
𝑑 = 𝑟𝑡
Step 1: the variable 𝑡 is being multiplied by 𝑟. Undo that by dividing both sides by 𝑟.
𝑑
𝑟𝑡
=
𝑟
𝑟
Step 2: simplify the right side.
𝑑 = 𝑡
𝑟
So:
𝑑
𝑡 =
( 𝑟 ̸= 0)
𝑟
Worked example (no skipped steps)
Solve for 𝑤 in
𝑃 = 2 𝑙 + 2 𝑤
Step 1: subtract 2 𝑙 from both sides.
𝑃 − 2 𝑙 = 2 𝑙 + 2 𝑤 − 2 𝑙
𝑃 − 2 𝑙 = 2 𝑤
Step 2: divide both sides by 2.
𝑃 − 2 𝑙
2 𝑤
=
2
2
𝑃 − 2 𝑙 = 𝑤
2
So:
𝑃 − 2 𝑙
𝑤 =
2
Mode 2: solving a formula for one variable 103
Worked example (no skipped steps)
Solve for 𝐶 in
9
𝐹 =
𝐶 + 32
5
Step 1: subtract 32 from both sides.
9
𝐹 − 32 =
𝐶
5
Step 2: ask what undoes multiplication by 9 . The inverse is multiplication by its reciprocal 5
5 .
9
Step 3: multiply both sides by 5 .
9
5
5
9
( 𝐹 − 32) =
· 𝐶
9
9
5
Step 4: simplify the right side. Because
5
9
·
= 1 ,
9
5
we get:
5 ( 𝐹 − 32) = 𝐶
9
So:
5
𝐶 =
( 𝐹 − 32)
9
Note
When a formula contains 𝜋, treat 𝜋 like a known nonzero number. It is approximately 3 . 14, but in algebra it is usually cleaner to leave it as 𝜋.
Also keep track of restrictions created by division. If you divide by 𝑟, ℎ, 𝑎, or any other variable, the step assumes that variable is not zero.
Chapter 12. Formulas and rearranging variables
Worked example (no skipped steps)
Solve for 𝑟 in
𝐴 = 𝜋𝑟 2
Step 1: divide both sides by 𝜋.
𝐴 = 𝑟 2
𝜋
This division is safe because 𝜋 ̸= 0.
Step 2: undo the square by taking square roots.
√︃
𝐴
𝑟 =
𝜋
In geometry, we usually keep the positive root because a radius is a length.
Fraction-heavy formulas
Key idea
Fraction formulas cause mistakes when students try to “move pieces around” instead of doing legal operations to both sides.
Do not break the denominator. Treat the whole fraction as one object.
Worked example (no skipped steps)
Solve for 𝑥 in
𝑥 − 𝑎 = 𝑐
𝑏
This formula assumes 𝑏 ̸= 0 because 𝑏 is a denominator.
Step 1: identify the outermost operation. The numerator 𝑥 − 𝑎 is being divided by 𝑏.
Step 2: undo that division by multiplying both sides by 𝑏.
𝑥 − 𝑎
𝑏 ·
= 𝑏𝑐
𝑏
Step 3: simplify the left side.
𝑥 − 𝑎 = 𝑏𝑐
Step 4: add 𝑎 to both sides.
𝑥 = 𝑏𝑐 + 𝑎
105
Worked example (no skipped steps)
Solve for 𝑏 in
1
𝐴 =
𝑏ℎ
2
Step 1: multiply both sides by 2.
2 𝐴 = 𝑏ℎ
Step 2: divide both sides by ℎ.
2 𝐴 = 𝑏
ℎ
So:
2 𝐴
𝑏 =
( ℎ ̸= 0)
ℎ
Worked example (no skipped steps)
Solve for 𝑥 in
𝑥 − 𝑦
𝑧 =
3
Step 1: multiply both sides by 3.
3 𝑧 = 𝑥 − 𝑦
Step 2: add 𝑦 to both sides.
3 𝑧 + 𝑦 = 𝑥
So:
𝑥 = 3 𝑧 + 𝑦
Note
A very common wrong move is this:
𝑥 − 𝑎
𝑐
= 𝑐
=⇒
𝑥 − 𝑎 =
𝑏
𝑏
That is wrong.
To clear a denominator, multiply by it. Do not divide by it again.
Bridge (why this matters later)
Formula control matters in every later book.
• Functions uses formulas to define relationships and transform expressions.
• Calculus uses formulas constantly when simplifying, substituting, and solving for one quantity in terms of another.
• Statistics uses formulas all the time, and students often struggle not because statistics is mysterious but because the algebra inside the formula is weak.
Chapter 12. Formulas and rearranging variables
Common mistakes (and how to avoid them)
• substituting without parentheses,
• solving for the wrong variable,
• dropping the square-root step,
• breaking a denominator incorrectly,
• forgetting that context can matter (for example, a radius should not be negative).
Recap: what you’re taking away
Formulas are not a different species of algebra. They are equations with names and jobs.
If you can substitute carefully and isolate the requested variable step by step, you already have the main skill formulas demand.
Problems
12.1. Use 𝐴 = 𝑙𝑤 with 𝑙 = 8 and 𝑤 = 3.
12.2. Use 𝑑 = 𝑟𝑡 with 𝑟 = 60 and 𝑡 = 2 . 5.
12.3. Use 𝑦 = 𝑚𝑥 + 𝑏 with 𝑚 = 2, 𝑥 = 5, 𝑏 = −3.
12.4. Solve for 𝑏 in 𝑦 = 𝑚𝑥 + 𝑏.
12.5. Solve for 𝑡 in 𝑑 = 𝑟𝑡.
12.6. Solve for 𝑤 in 𝑃 = 2 𝑙 + 2 𝑤.
12.7. Solve for 𝐶 in 𝐹 = 9 𝐶 + 32.
5
12.8. Solve for 𝑟 in 𝐴 = 𝜋𝑟 2.
12.9. Solve for 𝑥 in 𝑥− 𝑎 = 𝑐.
𝑏
12.10. Solve for 𝑏 in 𝐴 = 1 𝑏ℎ.
2
12.11. Solve for 𝑥 in 𝑧 = 𝑥− 𝑦 .
3
12.12. Use 𝑃 = 2 𝑙 + 2 𝑤 with 𝑙 = 7, 𝑤 = 4.
12.13. Use 𝐼 = 𝑃 𝑟𝑡 with 𝑃 = 500, 𝑟 = 0 . 08, 𝑡 = 2.
12.14. Solve for 𝑃 in 𝐼 = 𝑃 𝑟𝑡.
12.15. Solve for ℎ in 𝐴 = 1 𝑏ℎ.
2
12.16. Solve for 𝑥 in 𝑦 = 𝑎𝑥 + 𝑏.
12.17. Solve for 𝑟 in 𝐶 = 2 𝜋𝑟.
12.18. Use 𝐶 = 2 𝜋𝑟 with 𝑟 = 3.
107
12.19. Use 𝑠 = 𝑢𝑡 + 1 𝑎𝑡 2 with 𝑢 = 4, 𝑎 = 2, 𝑡 = 3.
2
12.20. Solve for 𝑎 in 𝑠 = 𝑢𝑡 + 1 𝑎𝑡 2.
2
Quick checklist (if you miss it, reread and try again)
• I can substitute values without losing signs or parentheses.
• I can isolate a chosen variable in a formula.
• I can treat fractions as whole objects instead of breaking them apart illegally.
Word problems without guessing