// Numbas version: finer_feedback_settings
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Der svaret ikke er et helt tall skal du svare med brøk. Eksempel: Bruk 1/2 i stedet for 0,5, 3/2 i stedet for 1,5 osv.
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", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Vi løser på samme måte som enkle ulikheter, men pass på å gjøre det samme på alle sider av ulikhetstegnene. Målet er å få $x$ alene i midten.
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", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Vi løser på samme måte som enkle ulikheter, men pass på å gjøre det samme på alle sider av ulikhetstegnene. Målet er å få $x$ alene i midten.Solving inequalities is similar to solving equations, ensure you do the same thing to all sides. Note that the operations we do are to get $x$ by itself.
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\n\n | \n | \n | \n | \n | \n
\n\n$\\var{left_bc*c}$ | \n$<$ | \n$\\simplify{{a}x+{b}}$ | \n$<$ | \n$\\var{right_bc*c}$ | \n
\n\n | \n | \n | \n | \n | \n
\n\n$\\simplify[basic]{{left_bc*c}-{b}}$ | \n$<$ | \n$\\simplify[basic]{{a}x+{b}-{b}}$ | \n $<$ | \n$\\simplify[basic]{{right_bc*c}-{b}}$ | \n
\n\n | \n | \n | \n | \n | \n
\n\n$\\var{left_bc*c-b}$ | \n$<$ | \n$\\var{a}x$ | \n$<$ | \n$\\var{right_bc*c-b}$ | \n
\n\n | \n | \n | \n | \n | \n
\n\n$\\displaystyle\\frac{\\var{left_bc*c-b}}{\\var{a}}$ | \n$<$ | \n$\\displaystyle\\frac{\\var{a}x}{\\var{a}}$ | \n$<$ | \n$\\displaystyle\\frac{\\var{right_bc*c-b}}{\\var{a}}$ | \n
\n\n | \n | \n | \n | \n | \n
\n\n$\\displaystyle\\simplify{{left_bc*c-b}/{a}}$ | \n$<$ | \n$x$ | \n$<$ | \n$\\displaystyle\\simplify{{right_bc*c-b}/{a}}$ | \n
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