import React, { useRef, useState, useEffect } from "react"; import { ArrowDown, Check, BookOpen, Grid, RefreshCw } from "lucide-react"; import SystemVisualizerWidget from "../../../components/lessons/SystemVisualizerWidget"; import Quiz from "../../../components/lessons/Quiz"; import { SYSTEMS_QUIZ_DATA } from "../../../utils/constants"; import { Frac } from "../../../components/Math"; interface LessonProps { onFinish?: () => void; } const SystemsEquationsLesson: React.FC = ({ onFinish }) => { const [activeSection, setActiveSection] = useState(0); const sectionsRef = useRef<(HTMLElement | null)[]>([]); const scrollToSection = (index: number) => { setActiveSection(index); sectionsRef.current[index]?.scrollIntoView({ behavior: "smooth", block: "start", }); }; useEffect(() => { const observer = new IntersectionObserver( (entries) => { entries.forEach((entry) => { if (entry.isIntersecting) { const index = sectionsRef.current.indexOf( entry.target as HTMLElement, ); if (index !== -1) setActiveSection(index); } }); }, { rootMargin: "-20% 0px -60% 0px" }, ); sectionsRef.current.forEach((section) => { if (section) observer.observe(section); }); return () => observer.disconnect(); }, []); const SectionMarker = ({ index, title, icon: Icon, }: { index: number; title: string; icon: any; }) => { const isActive = activeSection === index; const isPast = activeSection > index; return ( ); }; return (
{/* Section 1: Number of Solutions */}
{ sectionsRef.current[0] = el; }} className="min-h-screen flex flex-col justify-center mb-24 pt-20 lg:pt-0" >

Systems of Equations: Number of Solutions

A system of two linear equations represents two lines on a graph. The number of solutions tells you how those lines relate geometrically and algebraically. Every SAT test includes at least one question asking you to identify how many solutions a system has, or to find a constant that produces a specific outcome.

The Three Possible Outcomes

Outcome Geometric Picture Algebraic Condition Example
One Solution Lines intersect at exactly one point Different slopes y = 2x + 1
y = −x + 4
No Solution Lines are parallel — never meet Same slope, different y-intercept y = 2x + 1
y = 2x + 5
Infinite Solutions Same line — perfectly overlap Same slope AND same y-intercept (equations are multiples) y = 2x + 1
2y = 4x + 2
{/* Finding k for specific number of solutions */}

SAT Technique: Finding k for a Specific Number of Solutions

Example: For what value of k does 2x + ky = 6 and 4x + 2y = 12 have infinite solutions?

For infinite solutions, equations must be proportional.

Ratio of x-coefficients: = 2

So all coefficients must scale by 2: k must satisfy{" "} = 2, so k = 2.

Check constants: = 2 ✓

k = 2 → infinite solutions

Example: For what value of k does 3x + 2y = 8 and kx + 4y = 5 have no solution?

No solution → same slope, different intercept.

Same slope means coefficient ratios match for x and y:{" "} = = 2

So k = 6. Check constants: ≠ 2 ✓ (different, confirming no solution)

k = 6 → no solution

Critical Distinction: No Solution vs. Infinite Solutions

No solution: coefficients are proportional but constants are NOT. (Same slope, different lines.)
Infinite solutions: coefficients AND constants are proportional. (Same line, just written differently.) Divide one equation by the other and everything must cancel cleanly.

{/* Section 2: Solving Methods */}
{ sectionsRef.current[1] = el; }} className="min-h-screen flex flex-col justify-center mb-24" >

Solving Methods

The SAT presents systems in many formats. Choose your method based on what form the equations are already in — don't waste time converting unless necessary.

Method 1: Substitution

Best when: one variable is already isolated (e.g., y = 2x − 5) or easy to isolate.

Process: Plug one equation into the other to create a single-variable equation.

Given: y = 2x − 5 and x + y = 7

Substitute y = 2x − 5 into x + y = 7:

x + (2x − 5) = 7

3x = 12 → x = 4

y = 2(4) − 5 = 3

Solution: (4, 3)

Method 2: Elimination (Addition/Subtraction)

Best when: equations are in standard form (Ax + By = C) and coefficients are easy to match.

Process: Multiply one or both equations so one variable's coefficients are equal and opposite, then add the equations.

Given: 2x + y = 10 and 2x − y = 2

Add equations (y terms cancel):

4x = 12 → x = 3

Back-substitute: 2(3) + y = 10 → y = 4

Solution: (3, 4)

Given: 3x + 2y = 16 and x + y = 7

Multiply the second by 2: 2x + 2y = 14

Subtract: (3x + 2y) − (2x + 2y) = 16 − 14

x = 2, then y = 5

Solution: (2, 5)

{/* SAT Speed Tip */}

SAT Speed Strategy: Sum/Difference Shortcut

If the SAT asks for a combination like x + y, or 2x − y, you can often get it directly by adding or subtracting the equations — without finding x and y individually.

Given: 3x + 2y = 14 and x + y = 6. Find 2x + y.

Subtract eq2 from eq1: (3x + 2y) − (x + y) = 14 − 6

2x + y = 8 ← answer directly!

{/* Word Problem Translation */}

Word Problem Setup

Most SAT system word problems follow this template: define two variables, write two equations (one for each constraint), solve.

"A store sells pens for $2 and notebooks for $5. A customer buys 8 items total and spends $28. How many of each did they buy?"

Let p = pens, n = notebooks

p + n = 8 (total items)

2p + 5n = 28 (total cost)

From first: p = 8 − n. Substitute: 2(8 − n) + 5n = 28

16 − 2n + 5n = 28 → 3n = 12 →{" "} n = 4 notebooks, p = 4 pens

{/* Section 3: Quiz */}
{ sectionsRef.current[2] = el; }} className="min-h-screen flex flex-col justify-center" >

Practice Time

{SYSTEMS_QUIZ_DATA.map((quiz, idx) => (
))}

Topic Mastered!

); }; export default SystemsEquationsLesson;