[{"_1":2,"_3":-5,"_4":-5},"loaderData",{"_5":6,"_7":8,"_9":-5,"_10":11,"_12":13,"_14":15,"_16":17,"_18":19,"_20":21},"actionData","errors","root",{"_2223":2224,"_2225":2226,"_2227":2228,"_2229":53,"_2230":2231,"_2232":53,"_2233":2234,"_2235":2236,"_2237":51,"_2238":2239,"_2240":-5,"_2241":-5,"_2242":-5,"_2243":-5,"_2244":-5,"_2245":-5,"_2246":2247,"_2248":2249,"_2250":53,"_2251":-5,"_143":53},"routes/__learning-environment",{"_444":1800,"_1801":1802,"_1803":1804,"_143":53},"routes/__learning-environment/course","routes/__learning-environment/course/$courseSlug",{"_403":404,"_405":406,"_407":53,"_408":409,"_143":53},"routes/__learning-environment/course/$courseSlug/books",{"_402":51,"_143":53},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug",{"_135":136,"_137":138,"_139":51,"_140":141,"_142":53,"_143":53},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug",{"_128":-5,"_131":132,"_133":134},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug",{"_128":-5,"_129":130,"_22":23},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug/__section-content/faq",{"_22":23,"_24":25,"_26":27},"sectionTitle","HAVO Formules, vergelijkingen en ongelijkheden","sectionBaseUrl","/course/wiskunde/books/kern/wiskunde-2a-vmbo-theoretischhavo-2/chapters/3-lineaire-verbanden/sections/6-formules-vergelijkingen-en-ongelijkheden","questions",[28,29,30,31,32,33,34,35,36,37],{"_38":121,"_40":122,"_42":123,"_44":124,"_46":123,"_48":125,"_50":51,"_52":53,"_54":55,"_56":55,"_57":126,"_59":127},{"_38":114,"_40":115,"_42":116,"_44":117,"_46":116,"_48":118,"_50":51,"_52":53,"_54":55,"_56":55,"_57":119,"_59":120},{"_38":107,"_40":108,"_42":109,"_44":110,"_46":109,"_48":111,"_50":51,"_52":53,"_54":55,"_56":55,"_57":112,"_59":113},{"_38":100,"_40":101,"_42":102,"_44":103,"_46":102,"_48":104,"_50":51,"_52":53,"_54":55,"_56":55,"_57":105,"_59":106},{"_38":92,"_40":93,"_42":94,"_44":95,"_46":96,"_48":97,"_50":51,"_52":53,"_54":55,"_56":55,"_57":98,"_59":99},{"_38":84,"_40":85,"_42":86,"_44":87,"_46":88,"_48":89,"_50":51,"_52":53,"_54":55,"_56":55,"_57":90,"_59":91},{"_38":76,"_40":77,"_42":78,"_44":79,"_46":80,"_48":81,"_50":51,"_52":53,"_54":55,"_56":55,"_57":82,"_59":83},{"_38":69,"_40":70,"_42":71,"_44":72,"_46":71,"_48":73,"_50":51,"_52":53,"_54":55,"_56":55,"_57":74,"_59":75},{"_38":61,"_40":62,"_42":63,"_44":64,"_46":65,"_48":66,"_50":51,"_52":53,"_54":55,"_56":55,"_57":67,"_59":68},{"_38":39,"_40":41,"_42":43,"_44":45,"_46":47,"_48":49,"_50":51,"_52":53,"_54":55,"_56":55,"_57":58,"_59":60},"uuid","55124698-8c80-4bca-9c38-034c07815e66","slug","welke-methoden-vergelijkingen-oplossen-wanneer-gebruiken","question","Welke methoden zijn er om vergelijkingen op te lossen en wanneer gebruik je welke?","cleanedQuestion","Welke methoden vergelijkingen lossen wanneer gebruik welke","detailedQuestion","Welke methoden zijn er om vergelijkingen op te lossen en wanneer gebruik je welke methode?","answer","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.","isConcept",false,"isSynced",true,"likeCount",0,"dislikeCount","createdAt",["D",1769645719409],"updatedAt",["D",1769645719841],"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],"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],"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],"e5743212-6873-4571-b560-ee7e67d52bbc","hoe-los-ik-lineaire-vergelijkingen-op","Hoe los ik lineaire vergelijkingen op?","los lineaire vergelijkingen","Hoe los je lineaire vergelijkingen op met behulp van de balansmethode, zoals in het voorbeeld $3x + 7 = 22$?","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.",["D",1772497042168],["D",1772497042424],"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],"7e0a2801-0ef5-42cf-9070-21739c7e0fac","rekenen-met-gemengde-breuken","Hoe reken ik met gemengde breuken?","reken gemengde breuken","Om te rekenen met gemengde breuken, zoals $2 \\frac{1}{5}$, moet je deze eerst omzetten naar een onechte breuk. Een onechte breuk is een breuk waarbij de teller (het getal boven de streep) groter is dan de noemer (het getal onder de streep).\n\n**Stap 1: Gemengde breuk omzetten naar een onechte breuk**\nOm een gemengde breuk om te zetten, vermenigvuldig je het hele getal met de noemer en tel je de teller erbij op. De noemer blijft hetzelfde.\n* Voorbeeld: Zet $2 \\frac{1}{5}$ om.\n * Vermenigvuldig het hele getal (2) met de noemer (5): $2 \\times 5 = 10$.\n * Tel de teller (1) erbij op: $10 + 1 = 11$.\n * De noemer blijft 5.\n * Dus, $2 \\frac{1}{5}$ wordt $\\frac{11}{5}$.\n\n**Stap 2: Reken met de onechte breuken**\nNadat je alle gemengde breuken hebt omgezet naar onechte breuken, kun je ermee rekenen zoals je met gewone breuken zou doen. Dit betekent dat je, afhankelijk van de bewerking (optellen, aftrekken, vermenigvuldigen, delen), de volgende stappen volgt:\n\n* **Gelijknamig maken (voor optellen en aftrekken):** Zoek de kleinste gemeenschappelijke noemer (KGN) van alle breuken. Maak elke breuk gelijknamig door de teller en noemer met hetzelfde getal te vermenigvuldigen.\n * Voorbeeld: Voor de som $\\frac{11}{5} - \\frac{3}{4} + \\frac{7}{10}$ is de KGN van 5, 4 en 10 gelijk aan 20.\n * $\\frac{11}{5} = \\frac{11 \\times 4}{5 \\times 4} = \\frac{44}{20}$\n * $\\frac{3}{4} = \\frac{3 \\times 5}{4 \\times 5} = \\frac{15}{20}$\n * $\\frac{7}{10} = \\frac{7 \\times 2}{10 \\times 2} = \\frac{14}{20}$\n\n* **Optellen en aftrekken:** Tel de tellers bij elkaar op of trek ze van elkaar af, terwijl de noemer hetzelfde blijft.\n * Voorbeeld: $\\frac{44}{20} - \\frac{15}{20} + \\frac{14}{20}$\n * Eerst aftrekken: $\\frac{44 - 15}{20} = \\frac{29}{20}$\n * Dan optellen: $\\frac{29 + 14}{20} = \\frac{43}{20}$\n\n**Stap 3: Resultaat eventueel terug omzetten naar een gemengde breuk**\nAls je een onechte breuk als antwoord hebt, kun je deze vaak terug omzetten naar een gemengde breuk voor de duidelijkheid.\n* Voorbeeld: Zet $\\frac{43}{20}$ om.\n * Deel de teller (43) door de noemer (20): $43 \\div 20 = 2$ met een rest van $3$.\n * Het hele getal is 2, de nieuwe teller is 3 en de noemer blijft 20.\n * Dus, $\\frac{43}{20}$ wordt $2 \\frac{3}{20}$.",["D",1767225664163],["D",1767225664657],"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],"52853cac-de3b-4024-b21e-d16102d65ca4","kleinste-gemeenschappelijke-noemer-vinden","Hoe vind ik de kleinste gemeenschappelijke noemer?","vind kleinste gemeenschappelijke noemer","De kleinste gemeenschappelijke noemer (KGN) is het kleinste getal dat een veelvoud is van alle noemers van de breuken die je wilt optellen of aftrekken. Je hebt de KGN nodig om breuken gelijknamig te maken, zodat je ze kunt optellen of aftrekken.\n\n**Hoe vind je de kleinste gemeenschappelijke noemer?**\n\nDe meest gebruikelijke methode, vooral bij kleinere getallen, is het opschrijven van de veelvouden van elke noemer totdat je het eerste getal vindt dat in alle rijtjes voorkomt.\n\n**Stappenplan:**\n1. Schrijf de veelvouden op van de eerste noemer.\n2. Schrijf de veelvouden op van de tweede noemer.\n3. Ga door met alle noemers die je hebt.\n4. Zoek het kleinste getal dat in *alle* rijtjes van veelvouden voorkomt. Dit is de KGN.\n\n**Voorbeeld:** Vind de kleinste gemeenschappelijke noemer voor de breuken met noemers 5, 3 en 6.\n1. **Veelvouden van 5:** 5, 10, 15, 20, 25, **30**, 35, ...\n2. **Veelvouden van 3:** 3, 6, 9, 12, 15, 18, 21, 24, 27, **30**, 33, ...\n3. **Veelvouden van 6:** 6, 12, 18, 24, **30**, 36, ...\n\nHet kleinste getal dat in alle drie de rijtjes voorkomt, is 30. Dus de kleinste gemeenschappelijke noemer voor 5, 3 en 6 is 30.\n\nZodra je de KGN hebt gevonden, kun je elke breuk gelijknamig maken door de teller en de noemer met hetzelfde getal te vermenigvuldigen, zodat de noemer de KGN wordt. Dit is essentieel om breuken correct op te tellen of af te trekken.",["D",1767225662807],["D",1767225663452],"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 $$x > -3$$. 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