Several questions on dividing algrbraic fractions.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Write the following as a single fraction.
\n\\[\\frac{\\var{a}}{\\var{b}}\\div\\frac{\\var{d}}{\\var{c}}\\]
", "advice": "Dividing algebraic fractions is the same as dividing any other fractions. We flip the second fraction and then multiply them together. This means we have,
\n\\[\\frac{\\var{a}}{\\var{b}}\\div\\frac{\\var{d}}{\\var{c}}=\\frac{\\var{a}}{\\var{b}}\\times\\frac{\\var{c}}{\\var{d}}=\\frac{\\var{a}\\times\\var{c}}{\\var{b}\\times\\var{d}}=\\simplify{{a}{c}/({b}{d})}.\\]
\n\nUse this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "Evaluate the following division, giving the fraction in its simplest form.
\n\\[\\frac{\\var{a}}{\\simplify[unitPower]{y^{n2}}}\\div\\frac{\\var{b}}{\\simplify[unitPower]{y^{n1}}}\\]
", "advice": "To divide two fractions you just flip the second fraction and then multiply the numerators and multiply the denominators. This means we have,
\n\\[\\frac{\\var{a}}{\\simplify[unitPower]{y^{n2}}}\\div\\frac{\\var{b}}{\\simplify[unitPower]{y^{n1}}}=\\frac{\\var{a}}{\\simplify[unitPower]{y^{n2}}}\\times\\frac{\\simplify[unitPower]{y^{n1}}}{\\var{b}}=\\frac{\\var{a}\\times{\\simplify[unitPower]{y^{n1}}}}{\\simplify[unitPower]{y^{n2}}\\times\\var{b}}=\\simplify[unitPower,unitDenominator,unitFactor]{({a/gcd_ab}y^{n1-n2})/({b/gcd_ab})}\\]
\n\n
Use this link to find some resources which will help you revise this topic.
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", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Expand and simplify $\\displaystyle{\\var{LeftMul}\\div\\var{RightDiv}}.$
", "advice": "Remember that dividing two fractions is the same as flipping the second fraction and then multiplying. Therefore, our division can be re-written as
\n\\[\\var{LeftMul}\\div\\var{RightDiv}=\\var{LeftMul}\\times\\var{RightMul}.\\]
\nBefore we multiply the fractions together first lets check if we can do any cancellation. Notice that $\\var{RightMulBottom}$ has a factor of $\\var{Num}$ so we can cancel this straight away.
\nWe also have a factor of $x$ in both $\\var{QuadCoeff[0]}x^2+\\var{QuadCoeff[1]}x$ and $\\var{RightMulTop}$ so we're now left with multiplying
\n\\[\\frac1{\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]}}\\times\\frac{\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}}{\\var{Lin2Coeff[0]}x+\\var{Lin2Coeff[1]}}.\\]
\nWe're not necesserily done with cancellation though! To make sure that a fraction with a quadratic is simplified we have to factorise it to make sure there are no linear factors we can cancel. In this case we have
\\[\\simplify[all,expandBrackets]{(x+{Lin1Coeff})*({QuadCoeff[0]}x+{QuadCoeff[1]})}={(x+\\var{Lin1Coeff})(\\var{QuadCoeff[0]}x+\\var{QuadCoeff[1]})}.\\]
This gives us one last factor to cancel and then we can finally multiply what's left of each fraction to give us a final answer of
\n\\[\\var{ans}.\\]
\nUse this link to find some resources which will help you revise this topic.
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