Rewrite the expression $\\frac{cx+d}{(x+a)(x+b)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{B}{x+b}$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{({c}x+{d})/((x+{a})(x+{b}))} .\\]
", "advice": "To express \\[ \\simplify{({c}x+{d})/((x+{a})(x+{b}))} \\] as partial fractions, we want to set this equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:
\n\\[ \\simplify{({c}x+{d})/((x+{a})(x+{b}))} = \\simplify{A/(x+{a}) + B/(x+{b})}.\\]
\nTo find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives
\n\\[ \\simplify{{c}x+{d}=A(x+{b})+B(x+{a})}.\\]
\nThere are 2 methods of finding $A$ and $B$. The first is to choose suitable values for $x$ which will eliminate one of the terms, and the other is to compare the coefficients of each side of the equation. We will cover both methods here.
\nMethod 1:
\nTo find $A$, we can eliminate $B$ by setting $\\simplify{x={-a}}$:
\n\\[ \\simplify{{d-c*a}=A{b-a}} \\implies \\simplify[fractionNumbers]{A={(d-c*a)/(b-a)}}.\\]
\nSimilarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:
\n\\[ \\simplify{{d-c*b}=B{a-b}} \\implies \\simplify[fractionNumbers]{B={(d-c*b)/(a-b)}}.\\]
\nTherefore,
\n{check}
\nMethod 2:
\nBy comparing the coefficients of the $x$-terms and the constant terms we can form a pair of simultaneous equations to find $A$ and $B$.
\n\\[ \\begin{split} \\simplify{{c}x+{d}} &\\,= \\simplify{A(x+{b})+B(x+{a})} \\\\ &\\,= \\simplify{A*x+{b}A+B*x +{a}B} \\\\ &\\,=\\simplify{(A+B)x +{b}A+{a}B} . \\end{split} \\]
\n\\[ \\begin{split}&(x):\\quad \\var{c} &\\,= \\simplify{A+B} \\\\ &(c):\\quad \\var{d} &\\,= \\simplify{{b}A+{a}B} .\\end{split} \\]
\nHence,
\n\\[A=\\simplify[fractionNumbers]{{Asol}},\\,B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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