[{"_1":2,"_3":-5,"_4":-5},"loaderData",{"_5":6,"_7":8,"_9":-5,"_10":11,"_12":13,"_14":15,"_16":17,"_18":19},"actionData","errors","root",{"_2305":2306,"_2307":2308,"_2309":2310,"_2311":57,"_2312":2313,"_2314":57,"_2315":2316,"_2317":2318,"_2319":55,"_2320":2321,"_2322":-5,"_2323":-5,"_2324":-5,"_2325":-5,"_2326":-5,"_2327":-5,"_2328":2329,"_2330":2331,"_2332":57,"_2333":-5,"_171":57},"routes/__learning-environment",{"_529":1882,"_1883":1884,"_1885":1886,"_171":57},"routes/__learning-environment/course","routes/__learning-environment/course/$courseSlug",{"_489":490,"_491":492,"_493":57,"_494":204,"_171":57},"routes/__learning-environment/course/$courseSlug/books",{"_488":55,"_171":57},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug",{"_163":164,"_165":166,"_167":55,"_168":169,"_170":57,"_171":57},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug",{"_162":-5,"_20":21,"_22":23},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/__chapter-content/faq",{"_20":21,"_22":23,"_24":25,"_26":27},"chapterUuid","9b54da87-dcb9-4deb-8f6a-5bafd9aa5254","chapterTitle","9 Lineaire vergelijkingen","questions",[28,29,30,31,32,33,34,35,36,37,38,39,40,41],"backButtonHref","/course/wiskunde/books/moderne-wiskunde/leerboek-deel-b-2-vwo-13/chapters/3-lineaire-vergelijkingen",{"_42":154,"_44":155,"_46":156,"_48":157,"_50":158,"_52":159,"_54":55,"_56":57,"_58":59,"_60":59,"_61":160,"_63":161},{"_42":147,"_44":148,"_46":149,"_48":150,"_50":149,"_52":151,"_54":55,"_56":57,"_58":59,"_60":59,"_61":152,"_63":153},{"_42":140,"_44":141,"_46":142,"_48":143,"_50":142,"_52":144,"_54":55,"_56":57,"_58":59,"_60":59,"_61":145,"_63":146},{"_42":133,"_44":134,"_46":135,"_48":136,"_50":135,"_52":137,"_54":55,"_56":57,"_58":59,"_60":59,"_61":138,"_63":139},{"_42":125,"_44":126,"_46":127,"_48":128,"_50":129,"_52":130,"_54":55,"_56":57,"_58":59,"_60":59,"_61":131,"_63":132},{"_42":118,"_44":119,"_46":120,"_48":121,"_50":120,"_52":122,"_54":55,"_56":57,"_58":59,"_60":59,"_61":123,"_63":124},{"_42":111,"_44":112,"_46":113,"_48":114,"_50":113,"_52":115,"_54":55,"_56":57,"_58":59,"_60":59,"_61":116,"_63":117},{"_42":103,"_44":104,"_46":105,"_48":106,"_50":107,"_52":108,"_54":55,"_56":57,"_58":59,"_60":59,"_61":109,"_63":110},{"_42":96,"_44":97,"_46":98,"_48":99,"_50":98,"_52":100,"_54":55,"_56":57,"_58":59,"_60":59,"_61":101,"_63":102},{"_42":89,"_44":90,"_46":91,"_48":92,"_50":91,"_52":93,"_54":55,"_56":57,"_58":59,"_60":59,"_61":94,"_63":95},{"_42":81,"_44":82,"_46":83,"_48":84,"_50":85,"_52":86,"_54":55,"_56":57,"_58":59,"_60":59,"_61":87,"_63":88},{"_42":73,"_44":74,"_46":75,"_48":76,"_50":77,"_52":78,"_54":55,"_56":57,"_58":59,"_60":59,"_61":79,"_63":80},{"_42":65,"_44":66,"_46":67,"_48":68,"_50":69,"_52":70,"_54":55,"_56":57,"_58":59,"_60":59,"_61":71,"_63":72},{"_42":43,"_44":45,"_46":47,"_48":49,"_50":51,"_52":53,"_54":55,"_56":57,"_58":59,"_60":59,"_61":62,"_63":64},"uuid","e5743212-6873-4571-b560-ee7e67d52bbc","slug","hoe-los-ik-lineaire-vergelijkingen-op","question","Hoe los ik lineaire vergelijkingen op?","cleanedQuestion","los lineaire vergelijkingen","detailedQuestion","Hoe los je lineaire vergelijkingen op met behulp van de balansmethode, zoals in het voorbeeld $3x + 7 = 22$?","answer","Het oplossen van lineaire vergelijkingen betekent dat je de waarde van de onbekende variabele (vaak $x$, $y$ of $z$) wilt vinden. Dit doe je door de vergelijking in balans te houden, net als een weegschaal. Wat je aan de ene kant van het '=' teken doet, moet je ook aan de andere kant doen.\n\nDe balansmethode werkt in een paar stappen:\n1. **Isoleer de variabele:** Zorg ervoor dat alle termen met de variabele aan één kant van het '=' teken staan en alle losse getallen aan de andere kant.\n2. **Gebruik tegenovergestelde bewerkingen:**\n * Om een getal dat wordt opgeteld (+) weg te halen, trek je het af (–).\n * Om een getal dat wordt afgetrokken (–) weg te halen, tel je het op (+).\n * Om een getal dat wordt vermenigvuldigd (×) weg te halen, deel je erdoor (÷).\n * Om een getal waardoor wordt gedeeld (÷) weg te halen, vermenigvuldig je ermee (×).\n3. **Bereken de waarde van de variabele:** Als de variabele geïsoleerd is, kun je de waarde ervan berekenen.\n\n**Voorbeeld:**\nLaten we de vergelijking $3x + 7 = 22$ oplossen.\n\n1. **Haal de losse getallen naar één kant:** We willen de $+7$ aan de linkerkant weghalen. De tegenovergestelde bewerking van optellen is aftrekken. Dus, we trekken aan beide kanten 7 af:\n $3x + 7 - 7 = 22 - 7$\n $3x = 15$\n\n2. **Isoleer de variabele $x$:** Nu staat er '3 keer $x$ is 15'. Om $x$ alleen te krijgen, moeten we de vermenigvuldiging met 3 ongedaan maken. De tegenovergestelde bewerking van vermenigvuldigen is delen. Dus, we delen beide kanten door 3:\n $\\frac{3x}{3} = \\frac{15}{3}$\n $x = 5$\n\nDe waarde van $x$ in de vergelijking $3x + 7 = 22$ is dus 5. Door deze stappen te volgen, kun je de meeste lineaire vergelijkingen oplossen.","isConcept",false,"isSynced",true,"likeCount",0,"dislikeCount","createdAt",["D",1772497042168],"updatedAt",["D",1772497042424],"a616ca56-23a8-47ef-8978-7bd38f0242ef","wat-betekent-exact-oplossen","Wat betekent exact oplossen?","betekent exact oplossen","Wat betekent exact oplossen en kan dit leiden tot een coördinaat?","\"Exact oplossen\" betekent dat je de precieze waarde(n) van de onbekende in een vergelijking vindt, zonder afrondingen. Je zoekt dus naar de exacte oplossing die de vergelijking kloppend maakt. Dit doe je met methoden zoals de balansmethode, ontbinden in factoren of de ABC-formule. Bij sommige vergelijkingen, zoals wortelvergelijkingen, is het extra belangrijk om je gevonden oplossingen te controleren, omdat niet elke uitkomst een werkelijke oplossing hoeft te zijn.\n\nJazeker, exact oplossen kan zeker leiden tot een coördinaat! Wanneer je bijvoorbeeld twee lineaire vergelijkingen exact oplost, ben je vaak op zoek naar het snijpunt van de twee lijnen die bij die vergelijkingen horen. Het snijpunt is een specifieke plek in een assenstelsel, en die plek geef je aan met een coördinaat.\n\nJe vindt dan eerst de x-waarde door de vergelijkingen aan elkaar gelijk te stellen. Zodra je de x-waarde hebt, vul je die in één van de oorspronkelijke vergelijkingen in om de bijbehorende y-waarde te vinden. Samen vormen die x- en y-waarden dan de coördinaat van het snijpunt!",["D",1775690331349],["D",1775690331545],"ef8f8e2a-756c-4cff-adab-6f65a63333c9","lineaire-vergelijkingen-met-breuken-oplossen","Hoe los je lineaire vergelijkingen met breuken op met de balansmethode?","los lineaire vergelijkingen breuken","Hoe los je lineaire vergelijkingen met breuken op met behulp van de balansmethode?","Het oplossen van lineaire vergelijkingen met breuken doe je door de balansmethode toe te passen. De balansmethode houdt in dat je aan beide kanten van het '=' teken dezelfde bewerking uitvoert om de vergelijking in evenwicht te houden, net als bij een weegschaal. Het belangrijkste is om de breuken weg te werken, want dat maakt het rekenen een stuk makkelijker.\n\nDit doe je door het kleinste gemeenschappelijke veelvoud (KGV) van alle noemers in de vergelijking te vinden. De noemers zijn de getallen onder de streep van de breuken.\n\nZodra je het KGV hebt gevonden, vermenigvuldig je *alle* termen in de vergelijking met dit KGV. Hierdoor verdwijnen de breuken en houd je een vergelijking over zonder breuken, die je vervolgens op de gebruikelijke manier met de balansmethode kunt oplossen.\n\n**Voorbeeld:**\nLaten we de vergelijking $\\frac{1}{6}x+\\frac12=4-\\frac23x+\\frac43$ oplossen met de balansmethode.\n\n1. **Identificeer de noemers:** De noemers in deze vergelijking zijn 6, 2 en 3.\n2. **Vind het KGV:** Het kleinste gemeenschappelijke veelvoud van 6, 2 en 3 is 6.\n3. **Vermenigvuldig alle termen met het KGV (balansmethode stap 1):** Vermenigvuldig elke term in de vergelijking met 6 om de breuken weg te werken:\n $6 \\cdot \\frac{1}{6}x + 6 \\cdot \\frac12 = 6 \\cdot 4 - 6 \\cdot \\frac23x + 6 \\cdot \\frac43$\n Dit vereenvoudigt tot:\n $1x + 3 = 24 - 4x + 8$\n4. **Vereenvoudig beide zijden:**\n $x + 3 = 32 - 4x$\n5. **Breng alle termen met $x$ naar één kant (balansmethode stap 2):** Tel $4x$ op aan beide zijden van de vergelijking:\n $x + 4x + 3 = 32 - 4x + 4x$\n $5x + 3 = 32$\n6. **Breng alle constante termen naar de andere kant (balansmethode stap 3):** Trek 3 af van beide zijden van de vergelijking:\n $5x + 3 - 3 = 32 - 3$\n $5x = 29$\n7. **Isoleer $x$ (balansmethode stap 4):** Deel beide zijden door 5:\n $\\frac{5x}{5} = \\frac{29}{5}$\n $x = \\frac{29}{5}$ of $x = 5\\frac{4}{5}$\n\nDoor deze stappen te volgen, kun je lineaire vergelijkingen met breuken systematisch oplossen met behulp van de balansmethode.",["D",1760659390438],["D",1776813458462],"55124698-8c80-4bca-9c38-034c07815e66","welke-methoden-vergelijkingen-oplossen-wanneer-gebruiken","Welke methoden zijn er om vergelijkingen op te lossen en wanneer gebruik je welke?","Welke methoden vergelijkingen lossen wanneer gebruik welke","Welke methoden zijn er om vergelijkingen op te lossen en wanneer gebruik je welke methode?","Er zijn verschillende methoden om vergelijkingen op te lossen, afhankelijk van het type vergelijking. Twee veelgebruikte methoden zijn de balansmethode en de inklemmethode.\n\n### De Balansmethode\nDe balansmethode is een handige manier om lineaire vergelijkingen op te lossen. Het werkt net als een weegschaal die in evenwicht moet blijven: wat je aan de ene kant van het '=' teken doet, moet je ook aan de andere kant doen. Zo los je een vergelijking stap voor stap op.\n\n**Voorbeeld van de balansmethode:**\nStel je hebt de vergelijking: $4x + 3 = 2x + 7$\n\n1. **Stel de vergelijking op:** Je hebt de vergelijking $4x + 3 = 2x + 7$. Hierbij staat 'x' voor een onbekende waarde en de getallen voor bekende waarden.\n2. **Haal de getallen aan één kant weg:** Je wilt alle losse getallen aan één kant van de vergelijking hebben. Trek bijvoorbeeld 3 van beide kanten af:\n $4x + 3 - 3 = 2x + 7 - 3$\n $4x = 2x + 4$\n3. **Haal de letters aan de andere kant weg:** Nu wil je alle 'x'en aan de andere kant hebben. Trek 2x van beide kanten af:\n $4x - 2x = 2x + 4 - 2x$\n $2x = 4$\n4. **Deel door het getal voor de letter:** Je wilt weten wat één 'x' is. Deel daarom beide kanten door het getal dat voor de 'x' staat, in dit geval 2:\n $2x / 2 = 4 / 2$\n $x = 2$\nDe oplossing is dus $x = 2$.\n\n### Wanneer gebruik je welke methode?\n* **Voor lineaire vergelijkingen** (zoals $4x + 3 = 2x + 7$): De balansmethode is vaak het handigst en sneller, omdat je hiermee de precieze oplossing vindt.\n* **Voor moeilijkere vergelijkingen** (met bijvoorbeeld kwadraten, wortels of machten): Dan is de **inklem- of inklemmethode** vaak de beste methode, omdat de balansmethode dan te ingewikkeld kan worden. Bij inklemmen probeer je verschillende waarden voor 'x' totdat je dicht genoeg bij de oplossing bent.\n\nVoor de vergelijkingen die je op de middelbare school leert, is de balansmethode vaak de meest directe en efficiënte manier om een exacte oplossing te vinden.",["D",1769645719409],["D",1769645719841],"775c1658-194e-4f8f-b27a-7278a975d408","hoe-werkt-de-balansmethode-voor-het-oplossen-van-vergelijkingen","Hoe werkt de balansmethode voor het oplossen van vergelijkingen?","werkt balansmethode oplossen vergelijkingen","De balansmethode is een handige manier om wiskundige vergelijkingen op te lossen. Het idee is dat je een vergelijking ziet als een weegschaal die in balans is. Wat je aan de ene kant doet, moet je ook aan de andere kant doen om de balans te houden. Zo kom je erachter wat de onbekende letter (vaak 'x') waard is.\n\nStel je voor dat je een weegschaal hebt. Aan de linkerkant liggen bijvoorbeeld 4 rode blokjes en 3 gele blokjes. Aan de rechterkant liggen 2 rode blokjes en 7 gele blokjes. De weegschaal is in evenwicht.\n\n* Een **geel blokje** weegt 1 kilo.\n* Een **rood blokje** is onbekend, daarvoor gebruiken we de letter 'x'.\n\nDe vergelijking die hierbij hoort, is:\n$4x + 3 = 2x + 7$\n\nOm deze vergelijking met de balansmethode op te lossen, volg je de volgende stappen:\n\n1. **Breng alle termen met 'x' naar één kant:**\n Trek $2x$ af van beide zijden van de vergelijking om de 'x'-termen aan één kant te krijgen:\n $4x - 2x + 3 = 2x - 2x + 7$\n $2x + 3 = 7$\n\n2. **Breng alle constante termen naar de andere kant:**\n Trek 3 af van beide zijden van de vergelijking om de constante termen aan de andere kant te krijgen:\n $2x + 3 - 3 = 7 - 3$\n $2x = 4$\n\n3. **Isoleer 'x':**\n Deel beide zijden door 2 om de waarde van 'x' te vinden:\n $\\frac{2x}{2} = \\frac{4}{2}$\n $x = 2$\n\nDe waarde van het rode blokje (x) is dus 2 kilo. Door deze stappen te volgen, houd je de 'weegschaal' in balans en vind je de oplossing van de vergelijking.",["D",1786231061219],["D",1786231061453],"80e54b46-8112-4fee-9427-2cc5a49d22db","wat-betekent-een-ongelijkheid","Wat betekent een ongelijkheid?","betekent ongelijkheid","Een ongelijkheid is een wiskundige uitspraak die aangeeft wanneer de ene uitdrukking groter of kleiner is dan de andere, in plaats van precies gelijk. Waar je bij een gewone vergelijking (zoals $f(x) = g(x)$) zoekt wanneer twee uitdrukkingen precies gelijk aan elkaar zijn (denk aan de snijpunten van grafieken), kijk je bij een ongelijkheid naar wanneer de ene uitdrukking groter of kleiner is dan de andere.\n\nDe tekens die hiervoor worden gebruikt zijn:\n* $>$ (groter dan)\n* $<$ (kleiner dan)\n* $\\ge$ (groter dan of gelijk aan)\n* $\\le$ (kleiner dan of gelijk aan)\n\nAls je bijvoorbeeld $f(x) > g(x)$ ziet staan, dan zoek je naar alle $x$-waarden waarvoor de grafiek van $f$ boven de grafiek van $g$ ligt. Als er $f(x) < g(x)$ staat, zoek je naar de $x$-waarden waar de grafiek van $f$ onder die van $g$ ligt.\n\nHet gaat er dus om een bereik van waarden te vinden, in plaats van één of enkele specifieke punten.",["D",1765756863743],["D",1765756863994],"5fc3bbe3-9493-4170-aeb4-3cac51731281","hoe-werkt-balansmethode","Hoe werkt de balansmethode met variabelen links en rechts?","werkt balansmethode","Hoe werkt de balansmethode om vergelijkingen op te lossen, en hoe ga ik om met variabelen aan beide kanten?","De balansmethode is een manier om vergelijkingen stap voor stap op te lossen, net zoals je een weegschaal in evenwicht houdt. Wat je aan de ene kant van het gelijkteken ($=$) doet, moet je ook aan de andere kant doen om de vergelijking in balans te houden. Dit is vooral handig wanneer er variabelen (zoals $x$) aan beide kanten van de vergelijking staan.\n\nLaten we dit uitleggen met het voorbeeld: $4x + 3 = 2x + 7$. Hierbij staat '$x$' voor een onbekende waarde en de getallen voor bekende waarden. Het doel is om uiteindelijk te weten wat $x$ is.\n\nHier is hoe je de balansmethode gebruikt:\n\n1. **Stel de vergelijking op:** Je hebt de vergelijking: $4x + 3 = 2x + 7$.\n\n2. **Haal de losse getallen aan één kant weg:** Je wilt alle losse getallen (zonder $x$) aan één kant van het gelijkteken hebben.\n * Kijk naar de linkerkant: $4x + 3$. Om de '+3' weg te krijgen, zodat er alleen $x$-en overblijven aan die kant, moet je het tegenovergestelde doen. Het tegenovergestelde van optellen is aftrekken. Dus, we trekken 3 af van de linkerkant.\n * Omdat het een balans is, moet je precies hetzelfde doen aan de andere kant (de rechterkant) om de weegschaal in evenwicht te houden. Dus, we trekken ook 3 af van de rechterkant.\n * Dit geeft:\n $4x + 3 - 3 = 2x + 7 - 3$\n $4x = 2x + 4$\n\n3. **Haal de variabelen ($x$-termen) aan de andere kant weg:** Nu wil je alle $x$-termen aan één kant hebben. De $x$-termen staan nu aan beide kanten ($4x$ links en $2x$ rechts).\n * We willen de $2x$ aan de rechterkant weghalen. Dat doen we door er $2x$ van af te trekken.\n * Om de balans te houden, moeten we ook $2x$ aftrekken van de linkerkant.\n * Dit geeft:\n $4x - 2x = 2x + 4 - 2x$\n $2x = 4$\n\n4. **Deel door het getal voor de variabele:** Je wilt weten wat één $x$ is. Deel daarom beide kanten door het getal dat voor de $x$ staat, in dit geval 2.\n * $2x / 2 = 4 / 2$\n * $x = 2$\n\nZo zie je dat $x = 2$ de oplossing is voor de vergelijking $4x + 3 = 2x + 7$.\n\nDoor deze stappen te volgen, houd je de vergelijking in balans en vind je de waarde van de variabele.",["D",1769645721160],["D",1780355743523],"86861efa-0ed2-40c6-91ff-e48873fbd29b","hoe-werkt-de-kettingregel","Hoe werkt de kettingregel in wiskunde?","werkt kettingregel wiskunde","De kettingregel is een belangrijke regel in de differentiaalrekening die je gebruikt om de afgeleide te bepalen van een samengestelde functie. Een samengestelde functie is een functie die bestaat uit twee of meer functies die in elkaar geschoven zijn, bijvoorbeeld $f(x) = (g(x))^n$ of $f(x) = \\sin(g(x))$.\n\nDe kettingregel stelt dat als je een functie hebt die geschreven kan worden als $y = f(u)$ en $u = g(x)$, dan is de afgeleide van $y$ naar $x$ gelijk aan het product van de afgeleide van $y$ naar $u$ en de afgeleide van $u$ naar $x$. In formulevorm ziet dat er zo uit:\n\n$\\frac{dy}{dx} = \\frac{dy}{du} \\cdot \\frac{du}{dx}$\n\nEen andere manier om dit te onthouden is: \"buitenste functie afleiden, binnenste functie laten staan, dan vermenigvuldigen met de afgeleide van de binnenste functie\".\n\n**Voorbeeld:**\nStel, je hebt de functie $f(x) = (2x + 3)^4$.\nHierin is de \"buitenste\" functie iets tot de macht 4, en de \"binnenste\" functie is $2x + 3$.\n\n1. **Identificeer de binnenste en buitenste functie:**\n Laat $u = 2x + 3$ (de binnenste functie).\n Dan wordt de functie $y = u^4$ (de buitenste functie).\n\n2. **Bereken de afgeleide van de buitenste functie naar $u$:**\n $\\frac{dy}{du} = \\frac{d}{du}(u^4) = 4u^3$\n\n3. **Bereken de afgeleide van de binnenste functie naar $x$:**\n $\\frac{du}{dx} = \\frac{d}{dx}(2x + 3) = 2$\n\n4. **Vermenigvuldig de twee afgeleiden en vervang $u$ terug:**\n $\\frac{dy}{dx} = \\frac{dy}{du} \\cdot \\frac{du}{dx} = (4u^3) \\cdot 2$\n Nu vervangen we $u$ weer door $2x + 3$:\n $\\frac{dy}{dx} = 4(2x + 3)^3 \\cdot 2$\n $\\frac{dy}{dx} = 8(2x + 3)^3$\n\nDit is de afgeleide van de functie $f(x) = (2x + 3)^4$ met behulp van de kettingregel. De kettingregel is essentieel voor het differentiëren van complexere functies.",["D",1786230918119],["D",1786230918450],"b8345db2-cb9b-4603-b42e-5505d62e7f9d","kan-exact-oplossen-leiden-tot-een-coordinaat","Kan exact oplossen leiden tot een coördinaat?","exact oplossen leiden coördinaat","Wanneer je exact oplost, kun je zeker op een coördinaat uitkomen. Dit gebeurt bijvoorbeeld wanneer je het snijpunt van twee lijnen zoekt. Als je twee lineaire vergelijkingen exact oplost, ben je op zoek naar de specifieke plek in een assenstelsel waar de twee lijnen elkaar kruisen. Deze plek wordt aangegeven met een coördinaat, bestaande uit een x-waarde en een y-waarde.\n\nOm dit te doen, stel je de twee vergelijkingen aan elkaar gelijk om de x-waarde van het snijpunt te vinden. Zodra je de x-waarde hebt berekend, vul je deze in één van de oorspronkelijke vergelijkingen in om de bijbehorende y-waarde te bepalen. De gevonden x- en y-waarden vormen samen de coördinaat van het snijpunt.",["D",1775690329764],["D",1775690329995],"0553f851-5ba2-4308-a8b4-a8278aaf4b53","hoe-bepaal-je-wanneer-lineaire-groei-groter-is-dan-ander","Hoe bepaal je in welk jaar lineaire groei groter is dan een andere?","bepaal welk jaar lineaire groei groter","Hoe reken ik bij lineaire groei uit in welk jaar het een groter is dan het ander?","Om bij lineaire groei uit te rekenen in welk jaar het ene groter is dan het andere, maak je gebruik van lineaire ongelijkheden. Dit betekent dat je twee lineaire functies met elkaar vergelijkt om te bepalen wanneer de waarde van de ene functie die van de andere overtreft.\n\nJe doorloopt hiervoor de volgende stappen:\n\n1. **Stel de ongelijkheid op:** Begin met het opstellen van een ongelijkheid die de situatie beschrijft. Dit houdt in dat je de twee lineaire groeiformules (bijvoorbeeld $y = ax + b$) tegenover elkaar zet met een ongelijkheidsteken ertussen. De veelgebruikte ongelijkheidstekens zijn:\n * $<$ (kleiner dan)\n * $>$ (groter dan)\n * $\\le$ (kleiner dan of gelijk aan)\n * $\\ge$ (groter dan of gelijk aan)\n\n *Voorbeeld:* Stel, je hebt twee situaties:\n * Situatie A: $y_A = 10x + 50$ (waarbij $x$ het aantal jaren is en $y_A$ de waarde van situatie A)\n * Situatie B: $y_B = 15x + 20$ (waarbij $x$ het aantal jaren is en $y_B$ de waarde van situatie B)\n Als je wilt weten wanneer situatie A groter is dan situatie B, stel je de ongelijkheid op: $10x + 50 > 15x + 20$.\n\n2. **Los de bijbehorende vergelijking op:** Om het 'omschakelpunt' te vinden, los je eerst de vergelijking op die hoort bij de ongelijkheid. Je vervangt het ongelijkheidsteken dus tijdelijk door een is-gelijk-teken (=). Dit snijpunt is het moment waarop de twee precies aan elkaar gelijk zijn.\n\n *Voorbeeld:* Voor de ongelijkheid $10x + 50 > 15x + 20$ los je eerst de vergelijking op:\n $10x + 50 = 15x + 20$\n $50 - 20 = 15x - 10x$\n $30 = 5x$\n $x = 6$\n Dit betekent dat na 6 jaar de waarden van situatie A en situatie B precies gelijk zijn.\n\n3. **Bepaal de oplossing van de ongelijkheid:** Nadat je het snijpunt hebt gevonden, kijk je wanneer de ene functie daadwerkelijk groter is dan de andere. Dit kun je doen door een waarde te testen die voor of na het snijpunt ligt, of door naar de grafieken van de functies te kijken.\n\n *Voorbeeld:* We weten dat bij $x=6$ de waarden gelijk zijn. We willen weten wanneer $10x + 50 > 15x + 20$.\n * Test een waarde kleiner dan 6, bijvoorbeeld $x=5$:\n * $y_A = 10(5) + 50 = 50 + 50 = 100$\n * $y_B = 15(5) + 20 = 75 + 20 = 95$\n Hier is $100 > 95$, dus $y_A > y_B$ als $x=5$.\n * Test een waarde groter dan 6, bijvoorbeeld $x=7$:\n * $y_A = 10(7) + 50 = 70 + 50 = 120$\n * $y_B = 15(7) + 20 = 105 + 20 = 125$\n Hier is $120 < 125$, dus $y_A < y_B$ als $x=7$.\n\n Uit de testwaarden blijkt dat situatie A groter is dan situatie B wanneer $x < 6$. Omdat $x$ vaak staat voor jaren en niet negatief kan zijn in dit soort contexten, is de oplossing $0 \\le x < 6$. Als de vraag specifiek is naar 'in welk jaar', dan zijn dit de jaren 0, 1, 2, 3, 4 en 5.\n\nDoor deze stappen te volgen, kun je bepalen in welke periode of welk jaar de ene lineaire groei groter is dan de andere.",["D",1779405373672],["D",1779405373904],"154d6306-a210-492c-afe2-bb3e80f51fbf","basisprincipes-rekenen-met-letters","Wat zijn de basisprincipes van rekenen met letters?","basisprincipes rekenen letters","Rekenen met letters, ook wel algebra genoemd, is een belangrijk onderdeel van de wiskunde waarbij je letters gebruikt om onbekende getallen of variabele waarden voor te stellen. Dit maakt het mogelijk om algemene regels en formules op te stellen die voor veel verschillende situaties gelden.\n\nDe basisprincipes van rekenen met letters omvatten:\n\n1. **Variabelen:** Letters zoals $x$, $y$, $a$, $b$ worden gebruikt als variabelen. Een variabele is een symbool dat een waarde kan aannemen die kan veranderen.\n2. **Termen:** Een term is een deel van een wiskundige uitdrukking dat bestaat uit getallen, letters of een combinatie daarvan, vermenigvuldigd met elkaar. Voorbeelden zijn $3x$, $5y^2$, $7$.\n3. **Coëfficiënten:** Het getal dat voor een letter staat, wordt de coëfficiënt genoemd. Bijvoorbeeld, in $3x$ is $3$ de coëfficiënt. Als er geen getal voor de letter staat, is de coëfficiënt $1$ (bijvoorbeeld $x$ is hetzelfde als $1x$).\n4. **Gelijksoortige termen:** Dit zijn termen die precies dezelfde letters (met dezelfde machten) bevatten. Je kunt alleen gelijksoortige termen bij elkaar optellen of van elkaar aftrekken.\n * **Voorbeeld:** $3x + 5x = 8x$. Hier zijn $3x$ en $5x$ gelijksoortige termen.\n * **Voorbeeld:** $4y - 2y = 2y$.\n * **Voorbeeld:** $2x + 3y$ kan niet verder worden vereenvoudigd, omdat $2x$ en $3y$ geen gelijksoortige termen zijn.\n5. **Vermenigvuldigen met letters:**\n * Bij het vermenigvuldigen van letters tel je de machten bij elkaar op als de letters hetzelfde zijn.\n * **Voorbeeld:** $x \\cdot x = x^2$.\n * **Voorbeeld:** $x^2 \\cdot x^3 = x^{(2+3)} = x^5$.\n * Vermenigvuldig getallen met getallen en letters met letters.\n * **Voorbeeld:** $3x \\cdot 2y = (3 \\cdot 2) \\cdot (x \\cdot y) = 6xy$.\n * **Voorbeeld:** $4a \\cdot 5a = (4 \\cdot 5) \\cdot (a \\cdot a) = 20a^2$.\n6. **Delen met letters:**\n * Bij het delen van letters trek je de machten van elkaar af als de letters hetzelfde zijn.\n * **Voorbeeld:** $x^5 / x^2 = x^{(5-2)} = x^3$.\n * **Voorbeeld:** $6x / 2 = 3x$.\n * **Voorbeeld:** $10xy / 5x = 2y$.\n7. **Haakjes wegwerken:** Gebruik de distributieve eigenschap om haakjes weg te werken. Alles binnen de haakjes wordt vermenigvuldigd met wat ervoor staat.\n * **Voorbeeld:** $2(x + 3) = 2 \\cdot x + 2 \\cdot 3 = 2x + 6$.\n * **Voorbeeld:** $x(y - 4) = x \\cdot y - x \\cdot 4 = xy - 4x$.\n\nDeze basisprincipes zijn essentieel voor het oplossen van vergelijkingen, het werken met formules en het begrijpen van complexere algebraïsche concepten.",["D",1786230920118],["D",1786230920416],"3c30e11a-a22a-462c-ba8b-d81e770a2a8a","ongelijkheden-getallenlijn-balansmethode","Hoe los ik ongelijkheden op met een getallenlijn en balansmethode?","los ongelijkheden getallenlijn balansmethode","Het oplossen van ongelijkheden met een getallenlijn en de balansmethode is een handige manier om de oplossing visueel te maken. De balansmethode wordt gebruikt om de 'scheidingspunten' te vinden, dit zijn de x-waarden waar de twee kanten van de ongelijkheid aan elkaar gelijk zijn. De getallenlijn helpt je vervolgens om te bepalen welke intervallen de oplossing vormen.\n\n### De balansmethode bij ongelijkheden\nDe balansmethode gebruik je om de kritieke punten te vinden. Dit zijn de x-waarden waar $f(x)$ en $g(x)$ aan elkaar gelijk zijn, oftewel waar de grafieken elkaar snijden.\n\nDe stappen zijn grotendeels hetzelfde als bij vergelijkingen:\n1. **Neem de ongelijkheid over.** Bijvoorbeeld: $f(x) > g(x)$.\n2. **Maak er eerst een vergelijking van:** $f(x) = g(x)$.\n3. **Los deze vergelijking op met de balansmethode** om de x-waarden te vinden waar $f(x)$ en $g(x)$ gelijk zijn. Dit zijn je 'kritieke' punten.\n\n**Belangrijk verschil bij ongelijkheden:**\nAls je tijdens het oplossen van een ongelijkheid beide kanten vermenigvuldigt of deelt door een **negatief getal**, moet je het ongelijkheidsteken omdraaien! Dit is cruciaal om de juiste oplossing te krijgen.\n\n### Met een getallenlijn\nZodra je de 'kritieke' x-waarden hebt gevonden (met de balansmethode), kun je deze gebruiken op een getallenlijn:\n1. **Teken een getallenlijn** en markeer hierop de gevonden 'kritieke' x-waarden. Deze punten verdelen de getallenlijn in verschillende intervallen.\n2. **Kies testpunten:** Kies in elk interval een willekeurige x-waarde (een 'testpunt') en vul deze in de oorspronkelijke ongelijkheid in.\n3. **Controleer de ongelijkheid:** Kijk of de ongelijkheid waar is voor dat testpunt. Als de ongelijkheid waar is, dan is het hele interval waar dat testpunt in ligt, onderdeel van de oplossing.\n4. **Markeer de oplossing:**\n * Als de ongelijkheid een strikt teken heeft ($<$ of $>$), gebruik je een **open bolletje** bij de kritieke punten op de getallenlijn, omdat deze punten zelf niet meedoen.\n * Als de ongelijkheid een 'gelijk aan' teken heeft ($\\le$ of $\\ge$), gebruik je een **dicht bolletje**, omdat de kritieke punten dan wel meedoen.\n * Kleur vervolgens de intervallen in die voldoen aan de ongelijkheid.",["D",1773447601292],["D",1773447601575],"35e8e19e-f970-4ab2-aaf2-a37d8011ae0c","wanneer-draait-teken-om-bij-ongelijkheden","Wanneer draait het teken om bij het oplossen van ongelijkheden?","Wanneer draait teken oplossen ongelijkheden","Het teken van een ongelijkheid draait om wanneer je beide kanten van de ongelijkheid vermenigvuldigt of deelt door een **negatief getal**. Dit is een cruciale regel om te onthouden, want als je dit niet doet, klopt je oplossing niet.\n\n**Waarom draait het teken om?**\nDenk aan een eenvoudige ongelijkheid: $2 < 5$. Dit is waar.\nAls je beide kanten vermenigvuldigt met een positief getal, bijvoorbeeld 3:\n$2 \\times 3 < 5 \\times 3 \\Rightarrow 6 < 15$. De ongelijkheid blijft waar en het teken blijft hetzelfde.\n\nMaar als je beide kanten vermenigvuldigt met een negatief getal, bijvoorbeeld -1:\n$2 \\times (-1)$ en $5 \\times (-1)$ worden $-2$ en $-5$.\nNu is $-2$ *groter* dan $-5$. Dus, om de ongelijkheid waar te houden, moet het teken omdraaien: $-2 > -5$.\n\n**Voorbeeld:**\nStel je hebt de ongelijkheid $$-3x < 9$$.\nOm $x$ te isoleren, moet je beide kanten delen door $$-3$$. Omdat $$-3$$ een negatief getal is, moet je het ongelijkheidsteken omklappen.\n\nStap 1: Begin met de ongelijkheid:\n$$-3x < 9$$\n\nStap 2: Deel beide zijden door $$-3$$. Omdat je deelt door een negatief getal, draai je het ongelijkheidsteken om (van $<$ naar $>$):\n$$\\frac{-3x}{-3} > \\frac{9}{-3}$$\n\nStap 3: Vereenvoudig de uitdrukking:\n$$x > -3$$\n\nDe oplossing voor de ongelijkheid $$-3x < 9$$ is dus $$x > -3$$. Dit betekent dat elke waarde van $x$ die groter is dan $-3$ de oorspronkelijke ongelijkheid waar maakt.",["D",1777072735394],["D",1777072735657],"0e579489-cf76-4cab-b037-6e43579b406f","wat-is-de-balansmethode","Wat is de balansmethode?","balansmethode","Ik vroeg me af wat de balansmethode precies inhoudt en hoe ik deze kan gebruiken om wiskundige vergelijkingen op te lossen, waarbij ik beide kanten van de vergelijking in balans moet houden, net als een weegschaal.","De balansmethode is een techniek die wordt gebruikt om wiskundige vergelijkingen op te lossen. Het basisprincipe is dat je beide kanten van de vergelijking altijd in evenwicht houdt, vergelijkbaar met een weegschaal. Dit betekent dat elke bewerking die je aan de ene kant van de vergelijking uitvoert (bijvoorbeeld iets aftrekken), je ook aan de andere kant moet uitvoeren om het evenwicht te bewaren.",["D",1758758541024],["D",1758758541162],"compositeLecture","content",[271,272,273,274,275,276],"book",{"_42":172,"_44":173,"_63":174,"_61":-5,"_54":55,"_175":176,"_177":178,"_179":55,"_180":181,"_182":183,"_184":185,"_186":187,"_188":55,"_189":190,"_191":192,"_193":194,"_195":196,"_197":198,"_199":200},"userHasSelectedBook","alerts",[],"myBooksLinkToAddBook","success","ddebdd78-5862-4040-99f4-8d7e5ff9e2b1","leerboek-deel-b-2-vwo-13",["D",1787042505574],"chapterPrefix","","configurationStep","FINISHED","digitalBookIsIndexed","methodSlug","moderne-wiskunde","type","EXTERNAL","methodTitle","Moderne 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Kijk op kans ","hasJoJoLists","sections",[283,284,285,286],{"_42":309,"_44":310,"_218":311,"_290":312,"_292":313,"_294":55,"_295":55,"_296":55,"_297":59,"_298":-5,"_280":55},{"_42":304,"_44":305,"_218":306,"_290":307,"_292":308,"_294":55,"_295":55,"_296":55,"_297":59,"_298":-5,"_280":55},{"_42":299,"_44":300,"_218":301,"_290":302,"_292":303,"_294":55,"_295":55,"_296":55,"_297":59,"_298":-5,"_280":55},{"_42":287,"_44":288,"_218":289,"_290":291,"_292":293,"_294":55,"_295":55,"_296":55,"_297":59,"_298":-5,"_280":55},"1856d686-3e12-4a6c-a8ab-ae9cdb681266","4-wisselende-kansen","Wisselende kansen","positionName","12.4","fullTitle","12.4 Wisselende kansen","checked","hasLecture","hasSummary","numberOfExercises","numberOfTasks","f065cde6-22fc-4cf3-adf3-ed8ae021a335","3-tellen-en-machten","Tellen en machten","12.3","12.3 Tellen en machten","84e30a1b-2c96-4dfa-b1ae-b0ee0abab535","2-boomdiagrammen","Boomdiagrammen","12.2","12.2 Boomdiagrammen","b2247162-0750-49af-b7d1-9cd8868f4a71","1-rekenen-met-kansen","Rekenen met kansen","12.1","12.1 Rekenen met kansen","8c8ff5ad-5d9f-479c-92b0-f12af84e8b0e","5-ontbinden-in-factoren","11 Ontbinden in factoren",[318,319,320,321,322],{"_42":345,"_44":346,"_218":347,"_290":348,"_292":349,"_294":55,"_295":57,"_296":57,"_297":334,"_298":-5,"_280":55},{"_42":340,"_44":341,"_218":342,"_290":343,"_292":344,"_294":55,"_295":57,"_296":57,"_297":334,"_298":-5,"_280":55},{"_42":335,"_44":336,"_218":337,"_290":338,"_292":339,"_294":55,"_295":57,"_296":57,"_297":334,"_298":-5,"_280":55},{"_42":329,"_44":330,"_218":331,"_290":332,"_292":333,"_294":55,"_295":57,"_296":57,"_297":334,"_298":-5,"_280":55},{"_42":323,"_44":324,"_218":325,"_290":326,"_292":327,"_294":55,"_295":57,"_296":57,"_297":328,"_298":-5,"_280":55},"f0b9a0fb-b837-4ae9-8e2d-0ad18d5bb9e5","5-kwadratische-vergelijkingen","Kwadratische vergelijkingen","11.5","11.5 Kwadratische vergelijkingen",10,"e805483f-0946-4ceb-a884-1d990a7b8c89","4-a-x-b-0","A x B = 0","11.4","11.4 A x B = 0",3,"309c82e4-c533-4487-9de7-05ef285e4ef2","3-ontbinden-van-drietermen","Ontbinden van drietermen","11.3","11.3 Ontbinden van drietermen","7ace63b0-6eb9-4f8d-84bf-eeffd9bef742","2-ontbinden-van-tweetermen","Ontbinden van tweetermen","11.2","11.2 Ontbinden van tweetermen","a07d0ee8-c192-4b4c-9f5c-2dad4c2342d4","1-ontbinden-in-factoren","Ontbinden in factoren","11.1","11.1 Ontbinden in factoren","178fd721-c7ec-4034-baef-b0c81af28b3b","4-ruimtemeetkunde","10 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