Solve the following simultaneous equations for $x$ and $y$. Input your answers as fractions or integers, not as decimals.
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", "showStrings": false, "strings": ["."]}, "vsetRange": [0, 1], "showCorrectAnswer": true, "failureRate": 1, "answerSimplification": "std", "showFeedbackIcon": true, "unitTests": [], "variableReplacementStrategy": "originalfirst", "checkingAccuracy": 0.001, "variableReplacements": [], "checkVariableNames": false}], "showCorrectAnswer": true, "showFeedbackIcon": true, "unitTests": [], "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "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.
\n"}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "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.
"}, "variablesTest": {"condition": "{a}<>{a1} and {b}<>{b1}", "maxRuns": 100}, "extensions": [], "tags": [], "ungrouped_variables": ["a", "c", "b", "that", "this", "fromorto", "s1", "s6", "a1", "aort", "a2", "b1", "b2", "sc", "sb", "sa", "sc1", "c1"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "functions": {}, "advice": "\\[ \\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 either of the equations (1) and (2), then solving for $y$ 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.