Systems of Equations — Algebra Review

🎯

The Big Idea

A solution (x, y) must make BOTH equations true at the same time

A system of 2 equations in 2 unknowns looks like this:

{ 2x + y = 7
x − y = 2

💡

The solution is an ordered pair (x, y)

You're looking for ONE specific point that lies on both lines. Plug your answer back into both original equations to verify — if both are true, you've got it!

✅

One Solution

Lines intersect at exactly one point. Most common case.
e.g. (3, 1)

∞

Infinite Solutions

Lines are identical — same equation in disguise. Variables cancel entirely and you get 0 = 0.

🚫

No Solution

Lines are parallel — never meet. Variables cancel and you get a false statement like 0 = 5.

🔄

Substitution Method

Solve one equation for one variable, then plug into the other

When to use substitution

Best when one equation already has an isolated variable (like y = 2x + 1) or when it's easy to isolate one.

1

Isolate one variable in either equation. Pick whichever is easiest — look for a coefficient of 1.

2

Substitute that expression into the OTHER equation. Now you have one equation with one unknown.

3

Solve for the remaining variable.

4

Back-substitute to find the other variable. Then check in both original equations.

📌 Example 1 — Substitution: { y = 2x − 1 and 3x + y = 9 }

1

Already isolated: y = 2x − 1. Use this as the substitution expression.

2

Substitute into 3x + y = 9: 3x + (2x − 1) = 9

3

Simplify: 5x − 1 = 9 → 5x = 10 → x = 2

4

Back-substitute: y = 2(2) − 1 = 4 − 1 = y = 3

✓

Check eq 1: 3 = 2(2)−1 = 3 ✅ Check eq 2: 3(2)+3 = 6+3 = 9 ✅

Solution: (x, y) = (2, 3)

📌 Example 2 — Substitution: { x + 2y = 8 and 3x − y = 3 }

1

Isolate x in equation 1 (coefficient is 1): x = 8 − 2y

2

Substitute into equation 2: 3(8 − 2y) − y = 3

3

Distribute: 24 − 6y − y = 3 → 24 − 7y = 3 → −7y = −21 → y = 3

4

Back-substitute: x = 8 − 2(3) = 8 − 6 = x = 2

✓

Check eq 1: 2+2(3)=8 ✅ Check eq 2: 3(2)−3=3 ✅

Solution: (x, y) = (2, 3)

⚡

Elimination Method

Add or subtract equations to cancel out one variable

When to use elimination

Best when both equations are in standard form (ax + by = c) and coefficients line up nicely — especially when you spot matching or opposite coefficients.

1

Line up the equations with like terms in columns (x under x, y under y).

2

Multiply one or both equations (if needed) to make one variable's coefficients equal or opposite.

3

Add or subtract the equations to eliminate one variable.

4

Solve for the remaining variable, then substitute back to find the other.

➕ Add the Equations

Use when one variable has opposite coefficients (e.g. +3y and −3y). Adding eliminates that variable.


          2x + 3y = 11

        + x − 3y =  1

          ——————————

          3x         = 12

          x = 4

➖ Subtract the Equations

Use when one variable has equal coefficients (e.g. both have 2y). Subtracting eliminates that variable.


          5x + 2y = 14

        − 3x + 2y =  6

          ——————————

          2x         =  8

          x = 4

✖️ Multiply, Then Add

When coefficients don't match, multiply one equation by a constant first to create opposites, then add.


          x + 2y = 7  × 3

          3x + y = 11

          ——————————

          3x + 6y = 21

        + 3x +  y = 11  ← negate

✖️✖️ Multiply Both, Then Add

When neither matches, multiply both equations by different constants to get opposite coefficients, then add.


          2x + 3y = 12  × 2

          3x + 2y = 13  × 3

          ——————————

          4x + 6y = 24

        + 9x − 6y = 39

📌 Example 3 — Elimination by Addition: { 2x + 3y = 11 and x − 3y = 1 }

1

Spot opposite coefficients: +3y and −3y — adding will cancel y.

2

Add the equations: (2x + 3y) + (x − 3y) = 11 + 1 → 3x = 12 → x = 4

3

Substitute x = 4 into equation 2: 4 − 3y = 1 → −3y = −3 → y = 1

✓

Check eq 1: 2(4)+3(1)=11 ✅ Check eq 2: 4−3(1)=1 ✅

Solution: (x, y) = (4, 1)

📌 Example 4 — Elimination by Subtraction: { 5x + 2y = 14 and 3x + 2y = 6 }

1

Spot equal coefficients: both have 2y — subtracting will cancel y.

2

Subtract eq 2 from eq 1: (5x + 2y) − (3x + 2y) = 14 − 6 → 2x = 8 → x = 4

3

Substitute x = 4 into equation 2: 3(4) + 2y = 6 → 12 + 2y = 6 → 2y = −6 → y = −3

✓

Check eq 1: 5(4)+2(−3)=20−6=14 ✅ Check eq 2: 3(4)+2(−3)=12−6=6 ✅

Solution: (x, y) = (4, −3)

📌 Example 5 — Multiply One Equation: { x + 2y = 7 and 3x − y = 7 }

1

Goal: eliminate y. Eq 1 has +2y, eq 2 has −y. Multiply eq 2 by 2: 2(3x − y = 7) → 6x − 2y = 14

2

Now add: (x + 2y) + (6x − 2y) = 7 + 14 → 7x = 21 → x = 3

3

Substitute x = 3 into eq 1: 3 + 2y = 7 → 2y = 4 → y = 2

✓

Check eq 1: 3+2(2)=7 ✅ Check eq 2: 3(3)−2=7 ✅

Solution: (x, y) = (3, 2)

📌 Example 6 — Multiply Both Equations: { 2x + 3y = 12 and 3x + 2y = 13 }

1

Goal: eliminate y. LCM of 3 and 2 is 6. Multiply eq 1 by 2: 4x + 6y = 24. Multiply eq 2 by 3: 9x + 6y = 39.

2

Subtract (both have +6y): (9x + 6y) − (4x + 6y) = 39 − 24 → 5x = 15 → x = 3

3

Substitute x = 3 into eq 1: 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2

✓

Check eq 1: 2(3)+3(2)=6+6=12 ✅ Check eq 2: 3(3)+2(2)=9+4=13 ✅

Solution: (x, y) = (3, 2)

⚠️

No Solution & Infinite Solutions

When variables cancel completely — watch what's left

🚫 No Solution — Parallel Lines

?

System: { 2x + y = 5 and 4x + 2y = 3 }

1

Multiply eq 1 by 2: 4x + 2y = 10

2

Subtract eq 2: (4x + 2y) − (4x + 2y) = 10 − 3

!

0 = 7 ← FALSE — no solution exists. Lines are parallel.

∞ Infinite Solutions — Same Line

?

System: { x + 2y = 4 and 2x + 4y = 8 }

1

Multiply eq 1 by 2: 2x + 4y = 8

2

Subtract eq 2: (2x + 4y) − (2x + 4y) = 8 − 8

!

0 = 0 ← TRUE — infinite solutions. Same line.

Situation	Best Method	Operation
One equation has y = … or x = …	Substitution	Plug in directly
Coefficients are opposite (+3y / −3y)	Elimination	➕ Add equations
Coefficients are equal (2y / 2y)	Elimination	➖ Subtract equations
Coefficients don't match — one is a multiple	Elimination	✖️ Multiply one, then add/subtract
No easy shortcut	Elimination	✖️✖️ Multiply both, then add/subtract
Variables cancel → false (0 = 5)	🚫 No solution	Parallel lines
Variables cancel → true (0 = 0)	∞ Solutions	Same line (identical equations)

✅

Always check your answer!

Plug your (x, y) back into both original equations. If both sides balance, you're correct. This takes 20 seconds and catches all arithmetic errors.

🌹

Solving Systems like r = 2sinθ and r = 1 + cosθ

Set equal · use trig identities · find all θ in [0, 2π] · check for pole

💡

The core idea: both curves share the same (r, θ)

A solution is a point where both polar curves pass through the same location. Since both expressions equal r, you can set them equal to each other and solve for θ. Then plug θ back in to find r.

Step-by-Step Process

Works for any two polar/trig equations where both r and θ are unknown

1

Identify the type:
• Type A (r sinθ = a, r cosθ = b): divide to get tanθ = a/b, then solve for θ, then back-sub for r.
• Type B (r = f(θ), r = g(θ)): set equal to get one equation in θ, solve for θ, then find r.

2

For Type A — divide the equations: (r sinθ)/(r cosθ) = a/b → tanθ = a/b. The r cancels, leaving just θ.

3

Determine the correct quadrant for θ by checking the signs of r sinθ (= y) and r cosθ (= x). Positive x and y → Q1. Use this to pick the right solution.

4

Find r: substitute θ into either original equation. Or use the shortcut: r = √((r sinθ)² + (r cosθ)²) = √(a² + b²).

5

For Type B — check the pole (origin): if either curve passes through r = 0, the pole might be an intersection even if θ values differ. Test r = 0 in each equation separately.

6

Write solutions as (r, θ) pairs and verify in both original equations.

📌 Example 1 — r = 2sinθ and r = 2cosθ (Find all intersections)

1

Set equal: 2sinθ = 2cosθ

2

Divide both sides by 2cosθ: sinθ/cosθ = 1 → tanθ = 1

3

Solve for θ in [0, 2π]: tanθ = 1 when θ = π/4 and θ = 5π/4

4

Find r at θ = π/4: r = 2sin(π/4) = 2·(√2/2) = √2

5

Find r at θ = 5π/4: r = 2sin(5π/4) = 2·(−√2/2) = −√2 → negative r means same physical point as (√2, π/4). Not a new intersection.

6

Check the pole: r = 2sinθ = 0 when θ = 0 or π. r = 2cosθ = 0 when θ = π/2 or 3π/2. Different θ values, but both reach r = 0 → the pole is also an intersection.

Solutions: (√2, π/4) and the pole (0, 0)

📌 Example 2 — Given: r sinθ = 2√2 and r cosθ = 2√2 (Solve for both r and θ)

1

Recognize the form: these look like rectangular conversion formulas. Recall: y = r sinθ and x = r cosθ. So we have x = 2√2 and y = 2√2.

2

Divide equation 1 by equation 2: (r sinθ)/(r cosθ) = (2√2)/(2√2) → tanθ = 1

3

Solve for θ in [0, 2π]: tanθ = 1 at θ = π/4 and θ = 5π/4. Since both r sinθ and r cosθ are positive, the point is in Q1 → θ = π/4

4

Substitute θ = π/4 into equation 1: r sin(π/4) = 2√2 → r·(√2/2) = 2√2 → r = 2√2 · (2/√2) = 2√2 · √2 = r = 4

✓

Check eq 2: 4·cos(π/4) = 4·(√2/2) = 2√2 ✅

Solution: r = 4, θ = π/4 → point (4, π/4) in polar = (2√2, 2√2) in rectangular

📌 Example 3 — Given: r sinθ = 3 and r cosθ = 3√3 (Solve for both r and θ)

1

Divide equation 1 by equation 2: (r sinθ)/(r cosθ) = 3/(3√3) → tanθ = 1/√3

2

Solve for θ: tanθ = 1/√3 at θ = π/6 and θ = 7π/6. Since both equations give positive values, point is in Q1 → θ = π/6

3

Substitute θ = π/6 into equation 2: r cos(π/6) = 3√3 → r·(√3/2) = 3√3 → r = 3√3·(2/√3) = 3·2 = r = 6

4

Find r using Pythagorean identity as a check: r² = (r sinθ)² + (r cosθ)² = 3² + (3√3)² = 9 + 27 = 36 → r = 6 ✅

✓

Check eq 1: 6·sin(π/6) = 6·(1/2) = 3 ✅ Check eq 2: 6·cos(π/6) = 6·(√3/2) = 3√3 ✅

Solution: r = 6, θ = π/6 → point (6, π/6) in polar = (3√3, 3) in rectangular

🔑

Shortcut — use r² = x² + y²

When you have r sinθ = a and r cosθ = b, you can always find r directly without needing θ first:
r² = (r sinθ)² + (r cosθ)² = a² + b² → r = √(a² + b²)
Then find θ from tanθ = a/b (y over x). This is the fastest path when the numbers are messy.

📐 Systems for Geometric Sequences & Series

📐

Finding a and r from Two Given Terms

Set up two equations using aₙ = a·rⁿ⁻¹ · then divide to eliminate a

📋

The key formula: aₙ = a · rⁿ⁻¹

Every term in a geometric sequence is: aₙ = a · rⁿ⁻¹ where a = first term and r = common ratio. If you're given two terms, you get two equations. Dividing them eliminates a and lets you solve for r first.

Step-by-Step Process

Use when you know two terms of a geometric sequence and need to find a (first term) and r (common ratio)

1

Write two equations using aₙ = a · rⁿ⁻¹ for each given term.

2

Divide equation 2 by equation 1 (larger index ÷ smaller index). This cancels a completely.

3

Solve for r by taking the appropriate root.

4

Substitute r back into either original equation to solve for a.

5

Check: verify that your a and r produce both given terms correctly.

📌 Example 1 — a₂ = 6 and a₅ = 48 (Find a and r)

1

Write two equations:
a₂ = a · r¹ = 6 → ar = 6 …(eq 1)
a₅ = a · r⁴ = 48 → ar⁴ = 48 …(eq 2)

2

Divide eq 2 by eq 1: ar⁴ / ar = 48/6 → r³ = 8

3

Solve for r: r = ∛8 = r = 2

4

Substitute r = 2 into eq 1: a(2) = 6 → a = 3

✓

Check: a₂ = 3·2¹ = 6 ✅ a₅ = 3·2⁴ = 3·16 = 48 ✅

First term a = 3 · Common ratio r = 2 · Sequence: 3, 6, 12, 24, 48…

📌 Example 2 — a₁ = 80 and a₄ = 10 (Decreasing sequence)

1

Write two equations:
a₁ = a · r⁰ = 80 → a = 80 …(eq 1)
a₄ = a · r³ = 10 → ar³ = 10 …(eq 2)

2

Since a = 80 from eq 1, substitute directly into eq 2: 80r³ = 10

3

Solve for r: r³ = 10/80 = 1/8 → r = ∛(1/8) = r = 1/2

4

a is already known: a = 80

✓

Check: a₁ = 80 ✅ a₄ = 80·(1/2)³ = 80·(1/8) = 10 ✅

First term a = 80 · Common ratio r = 1/2 · Sequence: 80, 40, 20, 10…

📌 Example 3 — a₂ = −6 and a₄ = −54 (Negative terms)

1

Write two equations:
a₂ = ar¹ = −6 …(eq 1)
a₄ = ar³ = −54 …(eq 2)

2

Divide eq 2 by eq 1: ar³/ar = −54/−6 → r² = 9

3

Solve for r: r = ±3. Both are valid — check which makes the sequence consistent. Since a₂ and a₄ are both negative and terms multiply by r twice, r = 3 keeps them negative if a is negative. We'll find a next.

4

Use r = 3 in eq 1: a(3) = −6 → a = −2
Use r = −3 in eq 1: a(−3) = −6 → a = 2

✓

Both work! Two valid sequences: −2, −6, −18, −54… (r=3) or 2, −6, 18, −54… (r=−3)

Two solutions: (a=−2, r=3) or (a=2, r=−3). Both produce a₂=−6 and a₄=−54.

📌 Example 4 — Bonus: Use sum formula with system S∞ = 12, a₁ = 4. Find r.

1

Recall: Infinite geometric series sum: S∞ = a/(1−r) when |r| < 1

2

Substitute known values: 12 = 4/(1−r)

3

Multiply both sides by (1−r): 12(1−r) = 4 → 12 − 12r = 4 → −12r = −8

4

Solve: r = 8/12 = r = 2/3

✓

Check: S∞ = 4/(1 − 2/3) = 4/(1/3) = 4 × 3 = 12 ✅ |r| = 2/3 < 1 ✅

Common ratio r = 2/3 · Sequence: 4, 8/3, 16/9, … · Sum converges to 12

Given Info	Equations to Write	Key Step
Two terms aₘ and aₙ	arᵐ⁻¹ = value₁ arⁿ⁻¹ = value₂	Divide bigger by smaller → solves for r
First term a₁ given	a = known arⁿ⁻¹ = value	Substitute a directly, solve for r
Sum S∞ and a given	S∞ = a/(1−r)	One equation, one unknown — solve for r
r² after division	r = ±√(value)	Check both ± — may yield 2 valid sequences

Systems ofEquations

The solution is an ordered pair (x, y)

Always check your answer!

The core idea: both curves share the same (r, θ)

Shortcut — use r² = x² + y²

The key formula: aₙ = a · rⁿ⁻¹

Systems of
Equations