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Solve for $x$ and $y$: \\[ \\begin{eqnarray} a_1x+b_1y&=&c_1\\\\ a_2x+b_2y&=&c_2 \\end{eqnarray} \\]
\nThe included video describes a more direct method of solving when, for example, one of the equations gives a variable directly in terms of the other variable.
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"}, "answerSimplification": "std", "variableReplacementStrategy": "originalfirst", "checkVariableNames": false, "showCorrectAnswer": true, "showPreview": true, "answer": "{c*b1-b*c1}/{b1*a-a1*b}", "unitTests": [], "scripts": {}, "vsetRange": [0, 1], "marks": 2, "vsetRangePoints": 5, "valuegenerators": [], "type": "jme", "useCustomName": false, "customName": "", "failureRate": 1, "checkingAccuracy": 0.001, "showFeedbackIcon": true, "extendBaseMarkingAlgorithm": true, "checkingType": "absdiff", "customMarkingAlgorithm": ""}, {"variableReplacements": [], "notallowed": {"showStrings": false, "strings": ["."], "partialCredit": 0, "message": "Input as a fraction or an integer not as a decimal
"}, "answerSimplification": "std", "variableReplacementStrategy": "originalfirst", "checkVariableNames": false, "showCorrectAnswer": true, "showPreview": true, "answer": "{c*a1-a*c1}/{b*a1-a*b1}", "unitTests": [], "scripts": {}, "vsetRange": [0, 1], "marks": 2, "vsetRangePoints": 5, "valuegenerators": [], "type": "jme", "useCustomName": false, "customName": "", "failureRate": 1, "checkingAccuracy": 0.001, "showFeedbackIcon": true, "extendBaseMarkingAlgorithm": true, "checkingType": "absdiff", "customMarkingAlgorithm": ""}], "type": "gapfill", "useCustomName": false, "prompt": "\\[ \\begin{eqnarray} \\simplify[std]{{a}x+{b}y}&=&\\var{c}\\\\ \\simplify[std]{{a1}x+{b1}y}&=&\\var{c1} \\end{eqnarray} \\]
\n$x=\\phantom{{}}$[[0]], $y=\\phantom{{}}$[[1]]
\nInput your answers as fractions or integers, not as decimals.
", "unitTests": [], "customName": "", "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "customMarkingAlgorithm": "", "showCorrectAnswer": true, "sortAnswers": false, "scripts": {}, "extendBaseMarkingAlgorithm": true, "marks": 0}], "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t\\[ \\begin{eqnarray} \\simplify[std]{{a}x+{b}y}&=&\\var{c}&\\mbox{ ........(1)}\\\\ \\simplify[std]{{a1}x+{b1}y}&=&\\var{c1}&\\mbox{ ........(2)} \\end{eqnarray} \\]
To get a solution for $x$ multiply equation (1) by {this} and equation (2) by {that}
This gives:
\\[ \\begin{eqnarray} \\simplify[std]{{a*this}x+{b*this}y}&=&\\var{this*c}&\\mbox{ ........(3)}\\\\ \\simplify[std]{{a1*that}x+{b1*that}y}&=&\\var{that*c1}&\\mbox{ ........(4)} \\end{eqnarray} \\]
Now {aort} (4) {fromorto} equation (3) to get
\\[\\simplify[std]{({a*this}+{s6*a1*that})x={this*c}+{s6*that*c1}}\\]
And so we get the solution for $x$:
\\[x = \\simplify{{c*b1-b*c1}/{b1*a-a1*b}}\\]
Substituting this value into any of the equations (1) and (2) gives:
\\[y = \\simplify{{c*a1-a*c1}/{b*a1-a*b1}}\\]
You can check that these solutions are correct by seeing if they satisfy both equations (1) and (2) by substituting these values into the equations.