// Numbas version: exam_results_page_options {"name": "Partial Fractions 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "ungrouped_variables": ["a", "c", "b", "d", "nb", "a1", "a2", "s1", "a_", "c_", "b_", "d_", "nb_", "a1_", "a2_", "s1_"], "name": "Partial Fractions 1", "tags": ["algebra", "algebraic fractions", "algebraic manipulation", "combining algebraic fractions", "common denominator"], "preamble": {"css": "", "js": ""}, "advice": "

a)

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We use partial fractions to find $A$ and $B$ such that: 
\\[ \\simplify[std]{({a*a2+c*a1}*x+{c*b+a*d})/(({a1}x +{b})*({a2}x+{d}))} \\;\\;\\;=\\simplify[std]{ A/({a1}x+{b})+B/({a2}x+{d})}\\]

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Dividing both sides of the equation by $\\displaystyle \\simplify[std]{1/( ({a1}x+{b})({a2}x+{d}) )}\\;\\;$ we obtain:

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$\\simplify[std]{A*({a2}x+{d})+B*({a1}x+{b}) = {a*a2+c*a1}*x+{a*d+c*b}} \\Rightarrow \\simplify[std]{({a2}A+{a1}B)*x+{d}*A+{b}*B={a*a2+c*a1}*x+{a*d+c*b}}$

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Identifying coefficients:

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Constant term: $\\simplify[std]{ {d}*A+{b}*B={a*d+c*b} }$

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Coefficent $x$: $ \\simplify[std]{ {a2}A+{a1}B = {a*a2+c*a1} }$ 

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On solving these equations we obtain $A = \\var{a}$ and $B=\\var{c}$

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Which gives:\\[ \\simplify[std]{({a*a2+c*a1}*x+{c*b+a*d})/(({a1}x +{b})*({a2}x+{d}))}\\;\\;= \\simplify[std]{{a}/({a1}x+{b})+{c}/({a2}x+{d})}\\]

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b)

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We use partial fractions to find $A$ and $B$ such that: 
\\[ \\simplify[std]{({a_*a2_+c_*a1_}*x+{c_*b_+a_*d_})/(({a1_}x +{b_})*({a2_}x+{d_}))} \\;\\;\\;=\\simplify[std]{ A/({a1_}x+{b_})+B/({a2_}x+{d_})}\\]

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Dividing both sides of the equation by $\\displaystyle \\simplify[std]{1/( ({a1_}x+{b_})({a2_}x+{d_}) )}\\;\\;$ we obtain:

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$\\simplify[std]{A*({a2_}x+{d_})+B*({a1_}x+{b_}) = {a_*a2_+c_*a1_}*x+{a_*d_+c_*b_}} \\Rightarrow \\simplify[std]{({a2_}A+{a1_}B)*x+{d_}*A+{b_}*B={a_*a2_+c_*a1_}*x+{a_*d_+c_*b_}}$

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Identifying coefficients:

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Constant term: $\\simplify[std]{ {d_}*A+{b_}*B={a_*d_+c_*b_} }$

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Coefficent $x$: $ \\simplify[std]{ {a2_}A+{a1_}B = {a_*a2_+c_*a1_} }$ 

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On solving these equations we obtain $A = \\var{a_}$ and $B=\\var{c_}$

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Which gives:\\[ \\simplify[std]{({a_*a2_+c_*a1_}*x+{c_*b_+a_*d_})/(({a1_}x +{b_})*({a2_}x+{d_}))}\\;\\;= \\simplify[std]{{a_}/({a1_}x+{b_})+{c_}/({a2_}x+{d_})}\\]

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Split \\[\\simplify{({a*a2 +  c*a1} * x + {a * d +  c * b})/ (({a1}*x + {b}) * ({a2}*x + {d}))}\\] into partial fractions.

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Input the partial fractions here: [[0]].

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Input as the sum of partial fractions.

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Split \\[\\simplify{({a_*a2_ +  c_*a1_} * x + {a_ * d_ +  c_ * b_})/ (({a1_}*x + {b_}) * ({a2_}*x + {d_}))}\\] into partial fractions.

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Input the partial fractions here: [[0]].

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Input as the sum of partial fractions.

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5/08/2012:

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Added tags.

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Added description.

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Changed to two questions, for the numerator and denomimator, rather than one as difficult to trap student input for this example. Still some ambiguity however.

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12/08/2012:

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Back to one input of a fraction and trapped input in Forbidden Strings.

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Used the except feature of ranges to get non-degenerate examples.

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Checked calculation.OK.

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Improved display in content areas.

\n \t\t \n \t\t", "description": "

Split $\\displaystyle \\frac{ax+b}{(cx + d)(px+q)}$ into partial fractions.

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