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{
"url": "https://numbas.mathcentre.ac.uk/api/questions/194/?format=api",
"name": "Solving equations. (Video)",
"published": true,
"project": "https://numbas.mathcentre.ac.uk/api/projects/3/?format=api",
"author": {
"url": "https://numbas.mathcentre.ac.uk/api/users/6/?format=api",
"profile": "https://numbas.mathcentre.ac.uk/accounts/profile/6/?format=api",
"full_name": "Bill Foster",
"pk": 6,
"avatar": null
},
"edit": "https://numbas.mathcentre.ac.uk/question/194/solving-equations-video/?format=api",
"preview": "https://numbas.mathcentre.ac.uk/question/194/solving-equations-video/preview/?format=api",
"download": "https://numbas.mathcentre.ac.uk/question/194/solving-equations-video.zip?format=api",
"source": "https://numbas.mathcentre.ac.uk/question/194/solving-equations-video.exam?format=api",
"metadata": {
"notes": "\n \t\t<p><strong>5/08/2012:</strong></p>\n \t\t <p>Added more tags.</p>\n \t\t <p>Added description.</p>\n \t\t <p>Checked calculation. OK.</p>\n \t\t",
"description": "<p>Solve for $x$ and $y$: \\[ \\begin{eqnarray} a_1x+b_1y&=&c_1\\\\ a_2x+b_2y&=&c_2 \\end{eqnarray} \\]</p>\n <p>The 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.</p>",
"licence": "Creative Commons Attribution 4.0 International"
},
"status": "ok",
"resources": []
}