Solving Equations: Linear, Systems & the Quadratic Formula

Q: How do I solve a linear equation?

Apply inverse operations to both sides, undoing addition before multiplication: 3x + 5 = 20 → subtract 5: 3x = 15 → divide by 3: x = 5. Variables on both sides: 5x − 3 = 2x + 9 → subtract 2x: 3x − 3 = 9 → add 3: 3x = 12 → x = 4. Golden rule: whatever you do to one side, do to the other — the equation is a balance.

Q: How do I solve a system of two equations?

Elimination: multiply equations to match a coefficient, then add/subtract to eliminate one variable. x + y = 10 and x − y = 2: add → 2x = 12 → x = 6, so y = 4. Substitution: isolate one variable (y = 10 − x), plug into the other. Elimination wins with matching coefficients; substitution wins when a variable has coefficient 1. Both give the same answer.

Q: How do I use the quadratic formula?

For ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) ÷ 2a. Example 2x² − 4x − 6 = 0 (a=2, b=−4, c=−6): discriminant = 16 + 48 = 64, √64 = 8 → x = (4 ± 8) ÷ 4 → x = 3 or x = −1. Identify a, b, c carefully WITH their signs — most errors are sign slips on b or c, not the formula itself.

Q: What does the discriminant tell me?

b² − 4ac predicts the roots before solving: positive → two distinct real roots (curve crosses x-axis twice); zero → one repeated root (tangent to the axis); negative → no real roots (two complex conjugates, curve never crosses). x² + 2x + 5: discriminant 4 − 20 = −16 → no real solution. A 10-second discriminant check saves solving dead ends on exams.

Q: How do I turn a word problem into an equation?

Four steps: (1) name the unknown — let x = what the question asks; (2) translate phrases — "5 more than" = +5, "3 times as much" = 3×, "per/each" signals multiplication or division, "is/equals" = the equals sign; (3) write the equation from the relationship; (4) solve and CHECK against the story. "A number tripled plus 7 gives 22": 3x + 7 = 22 → x = 5.

Q: What are the classic equation-solving errors?

(1) Dividing only part of a side: 3x + 6 = 9 → x + 6 = 3 (must divide ALL terms: x + 2 = 3); (2) sign errors moving terms: x − 5 = 3 → x = 3 − 5 is wrong (add 5: x = 8); (3) forgetting ± on square roots: x² = 9 → x = ±3; (4) dividing by a variable that might be zero — x² = 3x → x = 3 OR x = 0; (5) not substituting back to verify.

Key Takeaways

Linear: same operation on both sides until x stands alone — 3x+5 = 20 → 3x = 15 → x = 5

Quadratic: x = (−b ± √(b²−4ac)) / 2a — discriminant b²−4ac decides 2, 1 or 0 real roots

Systems: eliminate one variable (add equations) or substitute — 2 equations solve 2 unknowns

Always check: substitute the answer back — catches sign errors instantly

Core answer: Solving an equation means isolating the unknown with inverse operations — whatever you do to one side, do to the other. Linear: 3x + 5 = 20 → x = 5. Quadratic: ax² + bx + c = 0 → x = (−b ± √(b²−4ac)) / 2a. The discriminant b²−4ac tells you before solving: positive = two real roots, zero = one, negative = none real.

Linear equations: one unknown, one answer

3x + 5 = 20 → subtract 5: 3x = 15 → divide by 3: x = 5. Check: 3×5 + 5 = 20 ✓. Always substitute back — 10 seconds, catches 90% of errors.

Everyday example — break-even. A café sells coffee at ¥25, cost ¥10, rent ¥9,000/month. Solve 25n − 10n = 9,000 → 15n = 9,000 → n = 600 cups to break even.

Quadratic equations: the formula that always works

For ax² + bx + c = 0:

Discriminant Δ = b² − 4ac	Roots
Δ > 0	two distinct real roots
Δ = 0	one repeated root
Δ < 0	no real roots (complex pair)

Example. x² − 5x + 6 = 0: Δ = 25 − 24 = 1 → x = (5 ± 1)/2 → x = 3 or x = 2. Factoring check: (x−2)(x−3) ✓.

Ballistics example. A ball thrown up at 20 m/s: height h = 20t − 5t². When does it land? 20t − 5t² = 0 → 5t(4 − t) = 0 → t = 4 s. Max height at t = 2 s: h = 20 m.

Systems: two equations, two unknowns

Substitution or elimination. 2x + y = 10, x − y = 2: add the equations → 3x = 12 → x = 4, y = 2.

Shopping example. 3 apples + 2 oranges = ¥19; 2 apples + 4 oranges = ¥22. Doubling the first: 6a + 4o = 38; subtract second: 4a = 16 → apple ¥4, orange ¥3.5.

Common mistakes and myths

Dividing by a variable that could be 0 — x² = 3x divided by x loses the root x = 0; factor instead: x(x − 3) = 0.
Sign flips when moving terms — moving +5 across "=" makes −5; the most common student error by far.
Forgetting ± on square roots — x² = 9 has two solutions (3 and −3); √9 = 3, but the equation has both.
Applying the quadratic formula with wrong a, b, c — standardize first: 5x = 3 − x² becomes x² + 5x − 3 = 0 before reading coefficients.
Not checking extraneous roots — squaring both sides (radical equations) and clearing denominators (rational equations) can introduce fake solutions; always substitute back.

Frequently Asked Questions

How do I solve a linear equation?

Apply inverse operations to both sides, undoing addition before multiplication: 3x + 5 = 20 → subtract 5: 3x = 15 → divide by 3: x = 5. Variables on both sides: 5x − 3 = 2x + 9 → subtract 2x: 3x − 3 = 9 → add 3: 3x = 12 → x = 4. Golden rule: whatever you do to one side, do to the other — the equation is a balance.

How do I solve a system of two equations?

Elimination: multiply equations to match a coefficient, then add/subtract to eliminate one variable. x + y = 10 and x − y = 2: add → 2x = 12 → x = 6, so y = 4. Substitution: isolate one variable (y = 10 − x), plug into the other. Elimination wins with matching coefficients; substitution wins when a variable has coefficient 1. Both give the same answer.

How do I use the quadratic formula?

For ax² + bx + c = 0: x = (−b ± √(b² − 4ac)) ÷ 2a. Example 2x² − 4x − 6 = 0 (a=2, b=−4, c=−6): discriminant = 16 + 48 = 64, √64 = 8 → x = (4 ± 8) ÷ 4 → x = 3 or x = −1. Identify a, b, c carefully WITH their signs — most errors are sign slips on b or c, not the formula itself.

What does the discriminant tell me?

b² − 4ac predicts the roots before solving: positive → two distinct real roots (curve crosses x-axis twice); zero → one repeated root (tangent to the axis); negative → no real roots (two complex conjugates, curve never crosses). x² + 2x + 5: discriminant 4 − 20 = −16 → no real solution. A 10-second discriminant check saves solving dead ends on exams.

How do I turn a word problem into an equation?

Four steps: (1) name the unknown — let x = what the question asks; (2) translate phrases — "5 more than" = +5, "3 times as much" = 3×, "per/each" signals multiplication or division, "is/equals" = the equals sign; (3) write the equation from the relationship; (4) solve and CHECK against the story. "A number tripled plus 7 gives 22": 3x + 7 = 22 → x = 5.

What are the classic equation-solving errors?

(1) Dividing only part of a side: 3x + 6 = 9 → x + 6 = 3 (must divide ALL terms: x + 2 = 3); (2) sign errors moving terms: x − 5 = 3 → x = 3 − 5 is wrong (add 5: x = 8); (3) forgetting ± on square roots: x² = 9 → x = ±3; (4) dividing by a variable that might be zero — x² = 3x → x = 3 OR x = 0; (5) not substituting back to verify.