_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",{"_1430":1431,"_1432":1433,"_1434":1435,"_1436":52,"_1437":1438,"_1439":1440,"_1441":52,"_1442":1443,"_1444":1445,"_1446":50,"_1447":1448,"_1449":-5,"_1450":-5,"_1451":-5,"_1452":-5,"_1453":-5,"_1454":-5,"_1455":1456,"_1457":1458,"_1459":52,"_1460":-5,"_146":52},"routes/__learning-environment",{"_518":996,"_997":998,"_999":-5,"_1000":1001,"_146":52},"routes/__learning-environment/course","routes/__learning-environment/course/$courseSlug",{"_476":477,"_478":479,"_480":52,"_481":182,"_146":52},"routes/__learning-environment/course/$courseSlug/books",{"_475":50,"_146":52},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug",{"_138":139,"_140":141,"_142":50,"_143":144,"_145":52,"_146":52},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug",{"_130":-5,"_134":135,"_136":137},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug",{"_130":-5,"_131":52,"_132":133,"_22":23},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug/__section-content/faq",{"_22":23,"_24":25,"_26":27},"sectionTitle","1.1 Lineaire vergelijkingen","sectionBaseUrl","/course/wiskunde-b/books/moderne-wiskunde/flex-boek-deel-1-vwo-bovenbouw-121/chapters/1-vergelijkingen/sections/1-lineaire-vergelijkingen","questions",[28,29,30,31,32,33,34,35,36,37],{"_38":122,"_40":123,"_42":124,"_44":125,"_46":126,"_47":127,"_49":50,"_51":52,"_53":54,"_55":54,"_56":128,"_58":129},{"_38":115,"_40":116,"_42":117,"_44":118,"_46":117,"_47":119,"_49":50,"_51":52,"_53":54,"_55":54,"_56":120,"_58":121},{"_38":107,"_40":108,"_42":109,"_44":110,"_46":111,"_47":112,"_49":50,"_51":52,"_53":54,"_55":54,"_56":113,"_58":114},{"_38":100,"_40":101,"_42":102,"_44":103,"_46":102,"_47":104,"_49":50,"_51":52,"_53":54,"_55":54,"_56":105,"_58":106},{"_38":92,"_40":93,"_42":94,"_44":95,"_46":96,"_47":97,"_49":50,"_51":52,"_53":54,"_55":54,"_56":98,"_58":99},{"_38":84,"_40":85,"_42":86,"_44":87,"_46":88,"_47":89,"_49":50,"_51":52,"_53":54,"_55":54,"_56":90,"_58":91},{"_38":76,"_40":77,"_42":78,"_44":79,"_46":80,"_47":81,"_49":50,"_51":52,"_53":54,"_55":54,"_56":82,"_58":83},{"_38":68,"_40":69,"_42":70,"_44":71,"_46":72,"_47":73,"_49":50,"_51":52,"_53":54,"_55":54,"_56":74,"_58":75},{"_38":60,"_40":61,"_42":62,"_44":63,"_46":64,"_47":65,"_49":50,"_51":52,"_53":54,"_55":54,"_56":66,"_58":67},{"_38":39,"_40":41,"_42":43,"_44":45,"_46":43,"_47":48,"_49":50,"_51":52,"_53":54,"_55":54,"_56":57,"_58":59},"uuid","b8345db2-cb9b-4603-b42e-5505d62e7f9d","slug","kan-exact-oplossen-leiden-tot-een-coordinaat","question","Kan exact oplossen leiden tot een coördinaat?","cleanedQuestion","exact oplossen leiden coördinaat","detailedQuestion","answer","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.","isConcept",false,"isSynced",true,"likeCount",0,"dislikeCount","createdAt",["D",1775690329764],"updatedAt",["D",1775690329995],"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],"d92fb354-eaed-4a2e-b587-967e2a809f38","hoe-stel-ik-de-formule-van-een-lijn-op","Hoe stel je de formule van een lineaire functie op?","stel formule lijn","Hoe stel je de formule van een lineaire functie op en hoe bereken je het snijpunt van twee lineaire functies?","De formule van een lineaire functie, ook wel lijn genoemd, heeft de algemene vorm:\n\n$y = ax + b$\n\nHierin is:\n* **$y$**: De afhankelijke variabele. De waarde van $y$ hangt af van de waarde die je voor $x$ kiest.\n* **$x$**: De onafhankelijke variabele. Je kiest zelf een waarde voor $x$.\n* **$a$**: De **richtingscoëfficiënt** (RC). Dit getal vertelt je hoe steil de lijn is en of de lijn stijgt ($a$ positief) of daalt ($a$ negatief). Het geeft aan hoeveel $y$ verandert als $x$ met 1 toeneemt. Je berekent $a$ met de formule:\n $a = \\frac{\\text{verandering in y}}{\\text{verandering in x}} = \\frac{y_2 - y_1}{x_2 - x_1}$\n* **$b$**: Het **startgetal** of de constante term. Dit getal geeft aan waar de lijn de verticale as (de $y$-as) snijdt. Het snijpunt met de $y$-as heeft altijd de coördinaten $(0, b)$.\n\nHieronder lees je hoe je de formule van een lineaire functie kunt opstellen in verschillende situaties en hoe je het snijpunt van twee van deze functies berekent.\n\n### 1. De richtingscoëfficiënt ($a$) berekenen uit twee punten\n\nAls je twee punten van een lijn weet, bijvoorbeeld $P_1(x_1, y_1)$ en $P_2(x_2, y_2)$, kun je de richtingscoëfficiënt ($a$) berekenen met de formule:\n\n$a = \\frac{y_2 - y_1}{x_2 - x_1}$\n\n**Voorbeeld:** Een lijn gaat door de punten $(2, 5)$ en $(5, 14)$.\n* $x_1 = 2, y_1 = 5$\n* $x_2 = 5, y_2 = 14$\n* $a = \\frac{14 - 5}{5 - 2} = \\frac{9}{3} = 3$.\nDe richtingscoëfficiënt is 3.\n\n### 2. De formule van een lijn opstellen als je twee punten weet\n\n1. **Bereken de richtingscoëfficiënt ($a$)** met de formule: $a = \\frac{y_2 - y_1}{x_2 - x_1}$.\n2. **Vul $a$ en één van de punten in de algemene formule in** ($y = ax + b$) om $b$ te vinden.\n * Kies één van de twee punten (bijvoorbeeld $(x_1, y_1)$) en vul de $x_1$, $y_1$ en de berekende $a$ in de formule $y = ax + b$.\n * Los de vergelijking op voor $b$.\n3. **Schrijf de volledige formule op** met de gevonden $a$ en $b$.\n\n**Voorbeeld:** Een lijn gaat door de punten $(2, 5)$ en $(5, 14)$.\n1. We berekenden al $a = 3$.\n2. Gebruik punt $(2, 5)$ en $a=3$ in $y = ax + b$:\n $5 = 3 \\times 2 + b$\n $5 = 6 + b$\n $b = 5 - 6 = -1$.\n3. De formule is $y = 3x - 1$.\n\n### 3. De formule van een lijn opstellen als je de richtingscoëfficiënt ($a$) en één punt weet\n\n1. **Vul de gegeven $a$ en het punt in de algemene formule in** ($y = ax + b$) om $b$ te vinden.\n * **Voorbeeld:** Stel je weet $a=4$ en de lijn gaat door punt $(3, 10)$.\n * Vul in $y = ax + b$: $10 = 4 \\times 3 + b$.\n * $10 = 12 + b$.\n * $b = 10 - 12 = -2$.\n2. **Schrijf de volledige formule op** met de gegeven $a$ en de gevonden $b$.\n * De formule is $y = 4x - 2$.\n\n### 4. Speciale gevallen van lineaire formules\n\n#### a. Horizontale lijnen\n* Een horizontale lijn heeft altijd een richtingscoëfficiënt van 0, omdat de lijn niet stijgt of daalt.\n* De formule is dan van de vorm $y = b$.\n* **Voorbeeld:** De lijn $y = 5$ is een horizontale lijn die door alle punten gaat waar de $y$-coördinaat 5 is.\n\n#### b. Verticale lijnen\n* Een verticale lijn heeft **geen richtingscoëfficiënt** (deze is ongedefinieerd, omdat de verandering in $x$ nul is).\n* De formule is van de vorm $x = c$ (waarbij $c$ een constante is).\n* **Voorbeeld:** De lijn $x = 3$ is een verticale lijn die door alle punten gaat waar de $x$-coördinaat 3 is.\n\n#### c. Proportionele verbanden\n* Een proportioneel verband is een speciaal type lineair verband waarbij het startgetal $b = 0$.\n* De formule is dan van de vorm $y = ax$.\n* **Uitleg:** De grafiek van een proportioneel verband gaat altijd door de oorsprong $(0, 0)$.\n\n### 5. Het snijpunt van twee lineaire functies berekenen\n\nOm het snijpunt van twee lineaire functies te berekenen, stel je de formules van de twee lijnen aan elkaar gelijk. Op het snijpunt zijn de $y$-waarden (en dus ook de $x$-waarden) van beide functies gelijk.\n\n**Stappenplan:**\n1. **Stel de formules aan elkaar gelijk.**\n * Als je twee formules hebt, bijvoorbeeld $y = a_1x + b_1$ en $y = a_2x + b_2$, dan stel je ze aan elkaar gelijk: $a_1x + b_1 = a_2x + b_2$.\n2. **Breng alle $x$-termen naar één kant van het gelijkteken.**\n * Gebruik de balansmethode: wat je aan de ene kant van de vergelijking doet, moet je ook aan de andere kant doen. Als je een term verplaatst, verandert het teken.\n3. **Breng alle losse getallen naar de andere kant van het gelijkteken.**\n * Ook hier gebruik je de balansmethode.\n4. **Los de vergelijking op voor $x$.**\n * Deel beide kanten door het getal dat voor de $x$ staat.\n5. **Vul de gevonden $x$-waarde in één van de oorspronkelijke formules in om de $y$-coördinaat te vinden.**\n * Het maakt niet uit welke formule je kiest, de $y$-waarde moet hetzelfde zijn.\n6. **Noteer het snijpunt als een coördinaat $(x, y)$.**\n\n**Voorbeeld:** Bereken het snijpunt van Lijn 1: $y = 2x + 3$ en Lijn 2: $y = x + 7$.\n\n1. **Formules gelijkstellen:**\n $2x + 3 = x + 7$\n2. **$x$-termen naar één kant:** Trek $x$ af aan beide kanten.\n $2x - x + 3 = x - x + 7$\n $x + 3 = 7$\n3. **Losse getallen naar de andere kant:** Trek $3$ af aan beide kanten.\n $x + 3 - 3 = 7 - 3$\n $x = 4$\n4. **$x$ berekenen:** De $x$-coördinaat is $4$.\n5. **$y$ berekenen:** Vul $x=4$ in Lijn 2 ($y = x + 7$).\n $y = 4 + 7$\n $y = 11$\n6. **Snijpunt noteren:** Het snijpunt is $(4, 11)$.\n\nDoor deze stappen en concepten te volgen, kun je de formule van de meeste lineaire functies opstellen en snijpunten berekenen.",["D",1765065787283],["D",1789082177484],"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],"35e8e19e-f970-4ab2-aaf2-a37d8011ae0c","wanneer-draait-teken-om-bij-ongelijkheden","Hoe bepaal je de richting van het ongelijkheidsteken bij het oplossen van ongelijkheden?","Wanneer draait teken oplossen ongelijkheden","Hoe bepaal je de richting van het ongelijkheidsteken bij het oplossen van ongelijkheden, en wanneer draait het teken om?","Bij het oplossen van ongelijkheden is het cruciaal om te weten hoe je de juiste richting van het ongelijkheidsteken bepaalt. Dit betekent dat je moet vaststellen of de oplossing 'groter dan' ($>$), 'kleiner dan' ($<$), 'groter of gelijk aan' ($\\ge$) of 'kleiner of gelijk aan' ($\\le$) moet zijn.\n\nEr zijn twee belangrijke manieren om de richting van het ongelijkheidsteken te bepalen:\n\n1. **Grafisch kijken:**\n * Stel je voor dat je de grafieken van beide kanten van de ongelijkheid tekent.\n * Het snijpunt van deze grafieken is de $x$-waarde waarbij de twee kanten van de ongelijkheid precies gelijk zijn.\n * Kijk dan naar het oorspronkelijke ongelijkheidsteken.\n * Als er bijvoorbeeld staat $A < B$, dan zoek je naar de $x$-waarden waar de grafiek van $A$ **onder** de grafiek van $B$ ligt.\n * Als er staat $A > B$, dan zoek je waar de grafiek van $A$ **boven** de grafiek van $B$ ligt.\n * De 'of gelijk aan' tekens ($\\ge, \\le$) betekenen dat het snijpunt zelf ook meetelt in de oplossing.\n\n2. **Algebraïsch (let op de regels!):**\n * Tijdens het algebraïsch oplossen van de ongelijkheid, behandel je het vaak net als een vergelijking om het snijpunt te vinden.\n * **Heel belangrijk:** 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.\n* Als 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* Maar als je beide kanten vermenigvuldigt met een negatief getal, bijvoorbeeld -1:\n $2 \\times (-1)$ en $5 \\times (-1)$ worden $-2$ en $-5$.\n Nu 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.\n\n**Samenvattend:**\n* Als je deelt of vermenigvuldigt met een **positief getal**, blijft het ongelijkheidsteken hetzelfde.\n* Als je deelt of vermenigvuldigt met een **negatief getal**, klapt het ongelijkheidsteken om.\n\nHet snijpunt zelf blijft altijd de grens, of het nu positief of negatief is. Het is de richting van het ongelijkheidsteken die verandert als je met een negatief getal werkt.",["D",1777072735394],["D",1790205623088],"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],"da0746a3-a968-4444-81d9-1ba047399478","wat-is-het-verschil-tussen-een-ongelijkheid-en-een-vergelijking","Wat is het verschil tussen een ongelijkheid en een vergelijking?","verschil tussen ongelijkheid vergelijking","Een ongelijkheid en een vergelijking zijn beide wiskundige uitdrukkingen, maar ze hebben een fundamenteel verschil in wat ze aangeven en wat voor soort oplossing je zoekt.\n\n### Wat is een vergelijking?\nEen **vergelijking** is een wiskundige uitdrukking waarbij je twee zijden aan elkaar gelijkstelt met een is-teken ($=$). Bij een vergelijking zoek je naar de specifieke waarde(n) van de variabele(n) waarvoor de linker- en rechterkant precies gelijk zijn. De oplossing van een vergelijking is meestal één of enkele specifieke getallen.\n* **Voorbeeld:** $2x + 1 = 7$\n Hier zoek je de exacte waarde van $x$ waarvoor de uitdrukking klopt.\n $2x = 6$\n $x = 3$\n De oplossing is één specifiek punt: $x=3$.\n\n### Wat is een ongelijkheid?\nEen **ongelijkheid** is een wiskundige uitdrukking waarbij je twee functies of expressies met elkaar vergelijkt met behulp van een ongelijkheidsteken. Je zoekt hier niet naar een exact gelijk punt, maar naar een bereik van waarden waarvoor de ene uitdrukking groter of kleiner is dan de andere.\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)\nDe oplossing van een ongelijkheid is een *bereik* van waarden, bijvoorbeeld $x < 2$ of $x \\ge -5$.\n\n### Hoe los je een ongelijkheid op en wat is de rol van het '=' teken?\nWanneer je een ongelijkheid oplost, is het vaak handig om eerst te bepalen waar de twee zijden van de ongelijkheid precies aan elkaar **gelijk** zijn. Dit punt, het snijpunt van de grafieken van de twee uitdrukkingen, is een cruciaal moment. Op dit punt verandert de relatie tussen de twee zijden: de ene kant gaat van 'groter dan' naar 'kleiner dan' of andersom.\n\nJe gebruikt het '=' teken dus om dat specifieke snijpunt te vinden, maar de oorspronkelijke vraag blijft een ongelijkheid. Je behandelt het even 'als een soort vergelijking' om dat ene punt te vinden, maar de uiteindelijke oplossing van de ongelijkheid is een *bereik* van waarden, en niet één specifiek getal.\n\n**Voorbeeld van het oplossen van een ongelijkheid:**\nStel je hebt de ongelijkheid $2x + 1 < -3x + 11$.\n\n1. **Vind het snijpunt (behandel als vergelijking):**\n $2x + 1 = -3x + 11$\n $5x = 10$\n $x = 2$\n Dit is het punt waar de twee uitdrukkingen gelijk zijn.\n\n2. **Bepaal de richting van de ongelijkheid:**\n * **Grafisch kijken:** Als je de grafieken van $y=2x+1$ en $y=-3x+11$ zou tekenen, zie je dat de grafiek van $y=2x+1$ onder de grafiek van $y=-3x+11$ ligt als $x < 2$.\n * **Algebraïsch (belangrijke regel):** Tijdens het algebraïsch oplossen van de ongelijkheid, behandel je het vaak net als een vergelijking, maar er is één cruciale regel:\n * Als je deelt of vermenigvuldigt met een **negatief getal**, dan moet het ongelijkheidsteken **omklappen**!\n\n Laten we het voorbeeld afmaken:\n $2x + 1 < -3x + 11$\n $5x < 10$ (hier heb je $3x$ opgeteld, teken blijft hetzelfde)\n $x < 2$ (hier heb je gedeeld door $5$, een positief getal, dus teken blijft hetzelfde)\n De oplossing is $x < 2$.\n\n **Voorbeeld van omklappen van het teken:**\n Stel je hebt de ongelijkheid $-3x < 9$.\n Om $x$ te vinden, deel je beide kanten door $-3$. Omdat je deelt door een **negatief getal**, klapt het ongelijkheidsteken om!\n $x > \\frac{9}{-3}$\n $x > -3$\n\nSamenvattend: een vergelijking zoekt naar specifieke punten waar twee uitdrukkingen gelijk zijn, terwijl een ongelijkheid zoekt naar een bereik van waarden waar de ene uitdrukking groter of kleiner is dan de andere. Het '=' teken wordt gebruikt als een hulpmiddel om het kritieke punt (snijpunt) te vinden bij het oplossen van een ongelijkheid.",["D",1790205623503],["D",1790205623856],"8c81dcdb-37c6-4c74-8272-491169f27045","hoe-bereken-je-snijpunt-twee-lijnen","Hoe bereken je het snijpunt van twee lineaire functies?","bereken snijpunt twee lijnen","Hoe bereken je het snijpunt van twee lineaire functies of lijnen, inclusief een gedetailleerd stappenplan voor het oplossen van de vergelijking?","Om het snijpunt van twee lineaire functies (ook wel lijnen genoemd) te berekenen, volg je een aantal stappen. Het snijpunt is het punt waar de twee lijnen elkaar kruisen en waar de $x$- en $y$-waarden voor beide formules hetzelfde zijn.\n\nHier is het stappenplan:\n\n**Stap 1: Zorg dat je de formules van beide lijnen hebt**\nElke lineaire lijn heeft een formule in de vorm $y = ax + b$, waarbij $a$ de richtingscoëfficiënt is en $b$ het snijpunt met de $y$-as. Als je de formules nog niet hebt, moet je deze eerst opstellen.\n\n**Stap 2: Stel de formules aan elkaar gelijk**\nOmdat het snijpunt een punt is waar de $y$-waarden van beide lijnen gelijk zijn, kun je de twee formules aan elkaar gelijkstellen. Dit geeft je een vergelijking met alleen $x$.\n\n**Stap 3: Los de vergelijking op voor $x$ met de balansmethode**\nNu je de twee formules aan elkaar gelijk hebt gesteld, heb je een vergelijking met alleen $x$. Om deze op te lossen, gebruik je de balansmethode:\n* **Breng alle termen met $x$ naar één kant van het isgelijkteken.** Dit doe je door aan beide kanten van de vergelijking dezelfde term op te tellen of af te trekken. Als je bijvoorbeeld een term als $+3x$ van de rechterkant wilt verwijderen, trek je aan beide kanten $3x$ af. Als je $-2x$ wilt verwijderen, tel je aan beide kanten $2x$ op.\n* **Breng alle losse getallen (termen zonder $x$) naar de andere kant van het isgelijkteken.** Ook dit doe je met de balansmethode. Als je bijvoorbeeld een term als $+5$ van de linkerkant wilt verwijderen, trek je aan beide kanten $5$ af. Als je $-7$ wilt verwijderen, tel je aan beide kanten $7$ op.\n* **Isoleer $x$.** Als je eenmaal een vergelijking hebt in de vorm $Cx = D$ (waarbij $C$ en $D$ getallen zijn), deel je beide kanten van de vergelijking door het getal dat voor de $x$ staat (de coëfficiënt $C$). Dit geeft je de waarde van $x$. Dit is de $x$-coördinaat van het snijpunt.\n\n**Stap 4: Vul de gevonden $x$-waarde in één van de oorspronkelijke formules in**\nNu je de $x$-coördinaat hebt, kun je deze invullen in één van de twee oorspronkelijke lijnformules. Het maakt niet uit welke je kiest, want in het snijpunt zijn de $y$-waarden hetzelfde. Reken de $y$-waarde uit. Dit is de $y$-coördinaat van het snijpunt.\n\n**Stap 5: Noteer het snijpunt als een coördinaatpaar $(x, y)$**\nSchrijf de gevonden $x$- en $y$-waarden op als een coördinaatpaar $(x, y)$. Dit is het snijpunt van de twee lijnen.\n\n**Voorbeeld:**\nStel je hebt lijn $k$ met de formule $y = x + 24$ en lijn $l$ met de formule $y = -1 \\frac{2}{3}x - 3$.\n\n1. **Formules zijn gegeven:**\n * Lijn $k: y = x + 24$\n * Lijn $l: y = -1 \\frac{2}{3}x - 3$\n\n2. **Stel de formules aan elkaar gelijk:**\n $x + 24 = -1 \\frac{2}{3}x - 3$\n\n3. **Los de vergelijking op voor $x$ met de balansmethode:**\n * **Breng alle termen met $x$ naar één kant.** Tel $1 \\frac{2}{3}x$ op aan beide kanten:\n $x + 24 + 1 \\frac{2}{3}x = -1 \\frac{2}{3}x - 3 + 1 \\frac{2}{3}x$\n $2 \\frac{2}{3}x + 24 = -3$\n * **Breng alle losse getallen naar de andere kant.** Trek $24$ af aan beide kanten:\n $2 \\frac{2}{3}x + 24 - 24 = -3 - 24$\n $2 \\frac{2}{3}x = -27$\n * **Isoleer $x$.** Deel beide kanten door $2 \\frac{2}{3}$.\n $2 \\frac{2}{3}$ kun je schrijven als een breuk: $\\frac{8}{3}$.\n $x = -27 \\div \\frac{8}{3}$\n $x = -27 \\times \\frac{3}{8}$\n $x = -\\frac{81}{8}$\n $x = -10 \\frac{1}{8}$\n\n4. **Vul de gevonden $x$-waarde in één van de formules in om $y$ te vinden:**\n We gebruiken de formule van lijn $k: y = x + 24$.\n Vul $x = -10 \\frac{1}{8}$ in:\n $y = -10 \\frac{1}{8} + 24$\n $y = 24 - 10 \\frac{1}{8}$\n Om dit te berekenen, kun je 24 schrijven als $23 \\frac{8}{8}$:\n $y = 23 \\frac{8}{8} - 10 \\frac{1}{8}$\n $y = (23 - 10) + (\\frac{8}{8} - \\frac{1}{8})$\n $y = 13 + \\frac{7}{8}$\n $y = 13 \\frac{7}{8}$\n\n5. **Noteer het snijpunt:**\n Het snijpunt van lijn $k$ en lijn $l$ is $(-10 \\frac{1}{8}, 13 \\frac{7}{8})$.",["D",1760054451197],["D",1789082174021],"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],"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],"compositeLecture","hasSummary","sectionUuid","bc217a82-e768-4475-8ab1-7c1f8687368e","chapterUuid","e89cf41b-7840-45a5-9054-eba836360718","chapterTitle","1 Vergelijkingen","content",[246,247,248,249,250,251],"book",{"_38":147,"_40":148,"_58":149,"_56":150,"_49":50,"_151":152,"_153":154,"_155":50,"_156":157,"_158":159,"_160":161,"_162":163,"_164":52,"_165":166,"_167":168,"_169":170,"_171":172,"_173":174,"_175":176},"userHasSelectedBook","alerts",[],"myBooksLinkToAddBook","success","df0d1f99-1bb0-41d5-b594-6ea25daea422","flex-boek-deel-1-vwo-bovenbouw-121",["D",1769522826723],["D",1757506548648],"chapterPrefix","","configurationStep","FINISHED","digitalBookIsIndexed","methodSlug","moderne-wiskunde","type","EXTERNAL","methodTitle","Moderne 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Periodieke functies","hasJoJoLists","sections",[260,261,262,263,264,265],"hasLecture","hasPractice",{"_38":296,"_40":297,"_198":298,"_269":299,"_271":300,"_273":50,"_258":52,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":52},{"_38":291,"_40":292,"_198":293,"_269":294,"_271":295,"_273":50,"_258":52,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":52},{"_38":286,"_40":287,"_198":288,"_269":289,"_271":290,"_273":50,"_258":52,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":52},{"_38":281,"_40":282,"_198":283,"_269":284,"_271":285,"_273":50,"_258":52,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":52},{"_38":276,"_40":277,"_198":278,"_269":279,"_271":280,"_273":50,"_258":52,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":52},{"_38":266,"_40":267,"_198":268,"_269":270,"_271":272,"_273":50,"_258":50,"_131":52,"_274":-5,"_255":50,"_259":52,"_275":50},"9c998203-a46a-4907-9452-a75ec2acca27","6-vergelijkingen-oplossen","Vergelijkingen oplossen","positionName","6.6","fullTitle","6.6 Vergelijkingen oplossen","checked","numberOfTasks","hasAuthoredExercises","42d868a2-463f-4072-8d70-45db645ac57c","5-algemene-vorm-van-een-sinusoide","Algemene vorm van een sinusoïde","6.5","6.5 Algemene vorm van een sinusoïde","f5ba1518-92cb-48d5-8bcf-5295b26c88da","4-transformaties","Transformaties","6.4","6.4 Transformaties","c12c38bf-079a-4573-9817-32279e4ec11b","3-cosinusfunctie","Cosinusfunctie","6.3","6.3 Cosinusfunctie","9a02bb10-e3a2-4bb7-baf3-bb04a3209058","2-sinusfunctie","Sinusfunctie","6.2","6.2 Sinusfunctie","b47fa817-005f-400e-a54b-b47f5329a66e","1-radialen","Radialen","6.1","6.1 Radialen","25973d1a-d9cd-4c1f-83b3-d9a416ab3c70","5-afgeleide-functies","5 Afgeleide 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