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feat(lessons): add lessons from client db

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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<LessonProps> = ({ 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 (
<button
onClick={() => scrollToSection(index)}
className={`flex items-center gap-3 p-3 w-full rounded-lg transition-all ${isActive ? "bg-white shadow-md border border-blue-100" : "hover:bg-slate-100"}`}
>
<div
className={`w-8 h-8 rounded-full flex items-center justify-center shrink-0 ${isActive ? "bg-blue-600 text-white" : isPast ? "bg-blue-400 text-white" : "bg-slate-200 text-slate-500"}`}
>
{isPast ? (
<Check className="w-4 h-4" />
) : (
<Icon className="w-4 h-4" />
)}
</div>
<div className="text-left">
<p
className={`text-sm font-bold ${isActive ? "text-blue-900" : "text-slate-600"}`}
>
{title}
</p>
</div>
</button>
);
};
return (
<div className="flex flex-col lg:flex-row min-h-screen">
<aside className="w-full lg:w-64 lg:fixed lg:top-20 lg:bottom-0 lg:overflow-y-auto p-4 border-r border-slate-200 bg-slate-50 z-0 hidden lg:block">
<nav className="space-y-2">
<SectionMarker index={0} title="Number of Solutions" icon={Grid} />
<SectionMarker index={1} title="Solving Methods" icon={RefreshCw} />
<SectionMarker index={2} title="Practice" icon={BookOpen} />
</nav>
</aside>
<div className="flex-1 lg:ml-64 p-6 md:p-12 max-w-4xl mx-auto">
{/* Section 1: Number of Solutions */}
<section
ref={(el) => {
sectionsRef.current[0] = el;
}}
className="min-h-screen flex flex-col justify-center mb-24 pt-20 lg:pt-0"
>
<h2 className="text-4xl font-extrabold text-slate-900 mb-6">
Systems of Equations: Number of Solutions
</h2>
<div className="prose prose-slate text-lg text-slate-600 mb-8">
<p>
A system of two linear equations represents two lines on a graph.
The <strong>number of solutions</strong> 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.
</p>
</div>
<div className="bg-blue-50 border border-blue-200 rounded-2xl p-6 mb-8 space-y-4">
<h3 className="text-lg font-bold text-blue-900">
The Three Possible Outcomes
</h3>
<div className="overflow-x-auto">
<table className="w-full text-sm border-collapse">
<thead>
<tr className="bg-blue-900 text-white">
<th className="p-3 text-left rounded-tl-lg">Outcome</th>
<th className="p-3 text-left">Geometric Picture</th>
<th className="p-3 text-left">Algebraic Condition</th>
<th className="p-3 text-left rounded-tr-lg">Example</th>
</tr>
</thead>
<tbody className="divide-y divide-blue-100">
<tr className="bg-blue-100">
<td className="p-3 font-bold text-blue-900">
One Solution
</td>
<td className="p-3 text-slate-700">
Lines intersect at exactly one point
</td>
<td className="p-3 text-slate-700">Different slopes</td>
<td className="p-3 font-mono text-xs text-slate-700">
y = 2x + 1<br />y = −x + 4
</td>
</tr>
<tr className="bg-red-50">
<td className="p-3 font-bold text-red-900">No Solution</td>
<td className="p-3 text-slate-700">
Lines are parallel — never meet
</td>
<td className="p-3 text-slate-700">
Same slope, different y-intercept
</td>
<td className="p-3 font-mono text-xs text-slate-700">
y = 2x + 1<br />y = 2x + 5
</td>
</tr>
<tr className="bg-emerald-50">
<td className="p-3 font-bold text-emerald-900">
Infinite Solutions
</td>
<td className="p-3 text-slate-700">
Same line — perfectly overlap
</td>
<td className="p-3 text-slate-700">
Same slope AND same y-intercept (equations are multiples)
</td>
<td className="p-3 font-mono text-xs text-slate-700">
y = 2x + 1<br />
2y = 4x + 2
</td>
</tr>
</tbody>
</table>
</div>
{/* Finding k for specific number of solutions */}
<div className="bg-white rounded-xl p-5 border border-blue-100">
<p className="font-bold text-blue-800 mb-3">
SAT Technique: Finding k for a Specific Number of Solutions
</p>
<div className="space-y-4">
<div className="bg-blue-50 rounded-lg p-4 text-sm">
<p className="font-semibold text-blue-800 mb-2">
Example: For what value of k does 2x + ky = 6 and 4x + 2y =
12 have infinite solutions?
</p>
<div className="font-mono space-y-1 text-slate-700">
<p>
For infinite solutions, equations must be proportional.
</p>
<p>
Ratio of x-coefficients: <Frac n="4" d="2" /> = 2
</p>
<p>
So all coefficients must scale by 2: k must satisfy{" "}
<Frac n="2k" d="2" /> = 2, so k = 2.
</p>
<p>
Check constants: <Frac n="12" d="6" /> = 2 ✓
</p>
<p className="text-blue-700 font-bold">
k = 2 → infinite solutions
</p>
</div>
</div>
<div className="bg-red-50 rounded-lg p-4 text-sm">
<p className="font-semibold text-red-800 mb-2">
Example: For what value of k does 3x + 2y = 8 and kx + 4y =
5 have no solution?
</p>
<div className="font-mono space-y-1 text-slate-700">
<p>No solution → same slope, different intercept.</p>
<p>
Same slope means coefficient ratios match for x and y:{" "}
<Frac n="k" d="3" /> = <Frac n="4" d="2" /> = 2
</p>
<p>
So k = 6. Check constants: <Frac n="5" d="8" /> ≠ 2 ✓
(different, confirming no solution)
</p>
<p className="text-red-700 font-bold">
k = 6 → no solution
</p>
</div>
</div>
</div>
</div>
<div className="bg-red-50 border border-red-200 rounded-xl p-4 text-sm">
<p className="font-bold text-red-800 mb-1">
Critical Distinction: No Solution vs. Infinite Solutions
</p>
<p className="text-slate-700">
<strong>No solution</strong>: coefficients are proportional but
constants are NOT. (Same slope, different lines.)
<br />
<strong>Infinite solutions</strong>: coefficients AND constants
are proportional. (Same line, just written differently.) Divide
one equation by the other and everything must cancel cleanly.
</p>
</div>
</div>
<SystemVisualizerWidget />
<button
onClick={() => scrollToSection(1)}
className="mt-12 group flex items-center text-blue-600 font-bold hover:text-blue-800 transition-colors"
>
Next: Solving Methods{" "}
<ArrowDown className="ml-2 w-5 h-5 group-hover:translate-y-1 transition-transform" />
</button>
</section>
{/* Section 2: Solving Methods */}
<section
ref={(el) => {
sectionsRef.current[1] = el;
}}
className="min-h-screen flex flex-col justify-center mb-24"
>
<h2 className="text-4xl font-extrabold text-slate-900 mb-6">
Solving Methods
</h2>
<div className="prose prose-slate text-lg text-slate-600 mb-6">
<p>
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.
</p>
</div>
<div className="bg-blue-50 border border-blue-200 rounded-2xl p-6 mb-8 space-y-5">
<h3 className="text-lg font-bold text-blue-900">
Method 1: Substitution
</h3>
<div className="bg-white rounded-xl p-5 border border-blue-100">
<p className="text-slate-600 text-sm mb-3">
<strong>Best when:</strong> one variable is already isolated
(e.g., y = 2x − 5) or easy to isolate.
</p>
<p className="text-slate-600 text-sm mb-3">
<strong>Process:</strong> Plug one equation into the other to
create a single-variable equation.
</p>
<div className="bg-blue-50 rounded-lg p-4 font-mono text-sm space-y-1 text-slate-700">
<p>Given: y = 2x − 5 and x + y = 7</p>
<p>Substitute y = 2x − 5 into x + y = 7:</p>
<p>x + (2x − 5) = 7</p>
<p>3x = 12 → x = 4</p>
<p>y = 2(4) − 5 = 3</p>
<p className="text-blue-700 font-bold">Solution: (4, 3)</p>
</div>
</div>
<h3 className="text-lg font-bold text-blue-900">
Method 2: Elimination (Addition/Subtraction)
</h3>
<div className="bg-white rounded-xl p-5 border border-blue-100">
<p className="text-slate-600 text-sm mb-3">
<strong>Best when:</strong> equations are in standard form (Ax +
By = C) and coefficients are easy to match.
</p>
<p className="text-slate-600 text-sm mb-3">
<strong>Process:</strong> Multiply one or both equations so one
variable's coefficients are equal and opposite, then add the
equations.
</p>
<div className="bg-blue-50 rounded-lg p-4 font-mono text-sm space-y-1 text-slate-700">
<p>Given: 2x + y = 10 and 2x − y = 2</p>
<p>Add equations (y terms cancel):</p>
<p>4x = 12 → x = 3</p>
<p>Back-substitute: 2(3) + y = 10 → y = 4</p>
<p className="text-blue-700 font-bold">Solution: (3, 4)</p>
</div>
<div className="mt-3 bg-blue-50 rounded-lg p-4 font-mono text-sm space-y-1 text-slate-700">
<p>Given: 3x + 2y = 16 and x + y = 7</p>
<p>Multiply the second by 2: 2x + 2y = 14</p>
<p>Subtract: (3x + 2y) − (2x + 2y) = 16 − 14</p>
<p>x = 2, then y = 5</p>
<p className="text-blue-700 font-bold">Solution: (2, 5)</p>
</div>
</div>
{/* SAT Speed Tip */}
<div className="bg-sky-50 border border-sky-200 rounded-xl p-5">
<p className="font-bold text-sky-900 mb-2">
SAT Speed Strategy: Sum/Difference Shortcut
</p>
<p className="text-slate-700 text-sm mb-3">
If the SAT asks for a <em>combination</em> like x + y, or 2x −
y, you can often get it directly by adding or subtracting the
equations — without finding x and y individually.
</p>
<div className="bg-white rounded-lg p-3 font-mono text-sm space-y-1 text-slate-700">
<p>Given: 3x + 2y = 14 and x + y = 6. Find 2x + y.</p>
<p>Subtract eq2 from eq1: (3x + 2y) − (x + y) = 14 − 6</p>
<p className="text-sky-700 font-bold">
2x + y = 8 ← answer directly!
</p>
</div>
</div>
{/* Word Problem Translation */}
<div className="bg-white rounded-xl p-5 border border-blue-100">
<p className="font-bold text-blue-800 mb-3">Word Problem Setup</p>
<p className="text-slate-600 text-sm mb-3">
Most SAT system word problems follow this template: define two
variables, write two equations (one for each constraint), solve.
</p>
<div className="bg-blue-50 rounded-lg p-4 text-sm">
<p className="text-slate-700 mb-2 italic">
"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?"
</p>
<div className="font-mono space-y-1 text-slate-700">
<p>Let p = pens, n = notebooks</p>
<p>p + n = 8 (total items)</p>
<p>2p + 5n = 28 (total cost)</p>
<p>From first: p = 8 − n. Substitute: 2(8 − n) + 5n = 28</p>
<p>
16 − 2n + 5n = 28 → 3n = 12 →{" "}
<strong className="text-blue-700">
n = 4 notebooks, p = 4 pens
</strong>
</p>
</div>
</div>
</div>
</div>
<button
onClick={() => scrollToSection(2)}
className="mt-12 group flex items-center text-blue-600 font-bold hover:text-blue-800 transition-colors"
>
Next: Practice Quiz{" "}
<ArrowDown className="ml-2 w-5 h-5 group-hover:translate-y-1 transition-transform" />
</button>
</section>
{/* Section 3: Quiz */}
<section
ref={(el) => {
sectionsRef.current[2] = el;
}}
className="min-h-screen flex flex-col justify-center"
>
<h2 className="text-4xl font-extrabold text-slate-900 mb-8">
Practice Time
</h2>
{SYSTEMS_QUIZ_DATA.map((quiz, idx) => (
<div key={quiz.id} className="mb-12">
<Quiz data={quiz} />
</div>
))}
<div className="p-8 bg-blue-900 rounded-2xl text-white text-center mt-12">
<h3 className="text-2xl font-bold mb-4">Topic Mastered!</h3>
<button
onClick={onFinish}
className="px-6 py-3 bg-white text-blue-900 font-bold rounded-full hover:bg-blue-50 transition-colors"
>
Finish Lesson ✓
</button>
</div>
</section>
</div>
</div>
);
};
export default SystemsEquationsLesson;