Algebra Foundations for Normal People by Gabriel Beguerie - HTML preview
Download the book in PDF, ePub, Kindle for a complete version.
Chapter goals
By the end of this chapter, you will be able to:
• solve equations with parentheses,
• solve equations with variables on both sides,
• clear fractions safely,
• recognize no-solution and infinitely-many-solutions cases,
• stay organized when several algebra moves must happen in sequence.
Prerequisites (if this is hard, go back to Unit 0)
You need distribution from Chapter 5 and one-step/two-step equations from Chapter 6.
Why this matters
Key idea
Multi-step equations are not a new kind of math. They are several familiar moves in a row.
The panic usually comes from trying to do everything at once.
Real life: what it’s for
When a pricing formula has a setup fee, a rate, a discount, and a tax, the equation is naturally multi-step. When a science formula has fractions and several variables, the equation is naturally multi-step.
So this chapter is not “extra algebra.” It is the point where algebra starts to look like real formulas.
60
61
Recipe (steps that always work)
A reliable order is:
1. simplify each side first,
2. distribute if needed,
3. combine like terms on each side,
4. move variable terms to one side,
5. move constants to the other side,
6. divide or multiply to isolate the variable,
7. check.
Case 1: parentheses first
Key idea
If parentheses are present, the first danger is pretending they are not there. Usually you must distribute before the equation becomes easy to read.
Worked example (no skipped steps)
Solve:
3( 𝑥 + 2) = 18
Step 1: distribute.
3 𝑥 + 6 = 18
Step 2: subtract 6 from both sides.
3 𝑥 = 12
Step 3: divide by 3.
𝑥 = 4
Check:
3(4 + 2) = 3(6) = 18
Correct.
Chapter 7. Multi-step equations
Worked example (no skipped steps)
Solve:
5 − 2( 𝑥 + 1) = −9
Step 1: distribute the −2 carefully.
5 − 2 𝑥 − 2 = −9
Step 2: combine like terms on the left.
3 − 2 𝑥 = −9
Step 3: subtract 3 from both sides.
−2 𝑥 = −12
Step 4: divide by −2.
𝑥 = 6
Check:
5 − 2(6 + 1) = 5 − 14 = −9
Correct.
Note
The line
−2( 𝑥 + 1) = −2 𝑥 − 2
is where many students lose the problem. The negative sign must hit every term inside the parentheses.
Case 2: variables on both sides
Key idea
When variables appear on both sides, the goal is still the same: gather all variable terms on one side and all constant terms on the other.
Worked example (no skipped steps)
Solve:
5 𝑥 − 7 = 2 𝑥 + 8
Step 1: subtract 2 𝑥 from both sides.
5 𝑥 − 2 𝑥 − 7 = 2 𝑥 − 2 𝑥 + 8
3 𝑥 − 7 = 8
Step 2: add 7 to both sides.
3 𝑥 = 15
Case 3: fractions in equations
63
Step 3: divide by 3.
𝑥 = 5
Check:
5(5) − 7 = 25 − 7 = 18 ,
2(5) + 8 = 10 + 8 = 18
Both sides match.
Worked example (no skipped steps)
Solve:
2( 𝑥 − 3) + 4 = 𝑥 + 9
Step 1: distribute.
2 𝑥 − 6 + 4 = 𝑥 + 9
Step 2: combine like terms on the left.
2 𝑥 − 2 = 𝑥 + 9
Step 3: subtract 𝑥 from both sides.
𝑥 − 2 = 9
Step 4: add 2.
𝑥 = 11
Check:
2(11 − 3) + 4 = 2(8) + 4 = 20 ,
11 + 9 = 20
Correct.
Case 3: fractions in equations
Key idea
Fractions are not forbidden. They are just inconvenient.
If the fractions are making the equation hard to read, multiply every term by the least common denominator.
Worked example (no skipped steps)
Solve:
𝑥
𝑥
+
= 5
2
3
Step 1: decide on a least common denominator. The denominators are 2 and 3. The smallest positive number both of them divide evenly is 6, so the least common denominator is 6.
Step 2: multiply every term by 6.
𝑥
𝑥
6 ·
+ 6 ·
= 6 · 5
2
3
Step 3: simplify each term. Because 6 ÷ 2 = 3 and 6 ÷ 3 = 2, we get: 3 𝑥 + 2 𝑥 = 30
Chapter 7. Multi-step equations
Step 4: combine like terms.
5 𝑥 = 30
Step 5: divide by 5.
𝑥 = 6
Check:
6
6
+
= 3 + 2 = 5
2
3
Correct.
Worked example (no skipped steps)
Solve:
2 𝑥 − 1 = 5
3
Step 1: multiply both sides by 3 to clear the denominator.
2 𝑥 − 1
3 ·
= 3 · 5
3
Step 2: simplify.
2 𝑥 − 1 = 15
Step 3: add 1.
2 𝑥 = 16
Step 4: divide by 2.
𝑥 = 8
Check:
2(8) − 1
16 − 1
15
=
=
= 5
3
3
3
Correct.
Note
When clearing fractions, multiply every term by the least common denominator, not just the first term you notice.
Case 4: special outcomes
Key idea
Sometimes the algebra collapses into a false statement or an always-true statement.
That is not a mistake. It is the answer.
65
Worked example (no skipped steps)
Solve:
2 𝑥 + 5 = 2 𝑥 − 1
Step 1: subtract 2 𝑥 from both sides.
5 = −1
This is false.
Answer: the equation has no solution.
Worked example (no skipped steps)
Solve:
4( 𝑥 + 1) = 4 𝑥 + 4
Step 1: distribute on the left.
4 𝑥 + 4 = 4 𝑥 + 4
Step 2: subtract 4 𝑥 from both sides.
4 = 4
This is always true.
Answer: the equation has infinitely many solutions.
Bridge (why this matters later)
This chapter protects the rest of the series more than it may seem.
Functions uses multi-step equations when solving for intercepts and when comparing expressions. Trigonometry hides algebra inside identities and equations. Calculus constantly asks you to simplify before the real idea can happen. Statistics formulas often require exactly this kind of calm symbolic control.
Common mistakes (and how to avoid them)
• skipping distribution,
• moving terms across the equal sign without doing a legal operation to both sides,
• multiplying only some terms when clearing fractions,
• forgetting to check,
• forcing a single numerical answer when the equation actually has no solution or infinitely many solutions.
Recap: what you’re taking away
When the equation looks crowded, do not speed up. Slow down.
Your job is not to be clever. Your job is to keep the sequence visible:
• simplify,
Chapter 7. Multi-step equations
• isolate,
• solve,
• check.
Problems
7.1. Solve: 3( 𝑥 + 2) = 18.
7.2. Solve: 4(2 𝑦 − 1) = 20.
7.3. Solve: 5 − 2( 𝑥 + 1) = −9.
7.4. Solve: 2( 𝑥 − 3) + 4 = 10.
7.5. Solve: 5 𝑥 − 7 = 2 𝑥 + 8.
7.6. Solve: 6 𝑎 + 3 = 2 𝑎 + 15.
7.7. Solve: 3 𝑥 + 4 = 2 𝑥 + 11.
7.8. Solve: 4( 𝑥 − 1) + 3 = 2 𝑥 + 9.
7.9. Solve: 2( 𝑥 + 3) − 4 = 𝑥 + 9.
7.10. Solve: 3 − (2 𝑥 − 5) = 12.
7.11. Solve: 𝑥 + 𝑥 = 5.
2
3
7.12. Solve: 𝑥 + 5 = 9.
3
7.13. Solve: 2 𝑥−1 = 5.
3
7.14. Solve: 𝑥+4 = 7.
2
7.15. Solve: 2 𝑥 + 5 = 2 𝑥 − 1.
7.16. Solve: 4( 𝑥 + 1) = 4 𝑥 + 4.
7.17. Solve: 3( 𝑥 − 2) = 3 𝑥 + 5.
7.18. Solve: 7 − 2( 𝑦 + 3) = 1.
7.19. Solve: 5( 𝑥 − 2) = 3 𝑥 + 6.
7.20. Solve: 3 𝑥+6 = 8.
3
7.21. Solve: 2(3 𝑚 − 1) + 5 = 15.
7.22. Solve: 4 − 3(2 𝑥 − 1) = −11.
7.23. Decide whether 4 𝑎 + 7 = 4 𝑎 − 2 has one solution, no solution, or infinitely many solutions.
7.24. Decide whether 3( 𝑏 − 2) = 3 𝑏 − 6 has one solution, no solution, or infinitely many solutions.
67
Quick checklist (if you miss it, reread and try again)
• I can distribute before solving when parentheses appear.
• I can gather variables on one side and constants on the other.
• I can clear fractions by multiplying every term by the least common denominator.
• I can recognize no-solution and infinitely-many-solutions cases.
Inequalities without sign mistakes