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Solving a Linear and a Non-linear system of equations
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\nInput the \\(y\\) value that corresponds to the previous answer. \\(y = \\) [[1]]
\n\nInput the lesser of the two \\(x\\) values that satisfies both equations. \\(x = \\) [[2]]
\nInput the \\(y\\) value that corresponds to the previous answer. \\(y = \\) [[3]]
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\n\\(\\var{a1}x+\\var{b1}y=\\var{r1}\\)
\nand
\n\\(\\var{c1}x^2+\\var{d1}y^2=\\var{r2}\\)
\nThere are two solutions for \\(x\\) that satisfy both of these equations and for each \\(x\\) value there exists a corresponding \\(y\\) value that forms a solution pair.
", "variablesTest": {"condition": "(2*{a1}*{d1}*{r1})^2>4*({b1}^2*{c1}+{d1}*{a1}^2)*({d1}*{r1}^2-{b1}^2*{r2})", "maxRuns": 100}, "advice": "To solve a system that involves a linear equation and a non-linear equation we must use the substitution method.
\n\\(\\var{a1}x+\\var{b1}y=\\var{r1}\\)
\n\\(\\var{c1}x^2+\\var{d1}y^2=\\var{r2}\\)
\nThe first equation is a linear equaton, we use this to write one variable in terms of the other.
\nFor example, we could make y the subject of this equation
\n\\(y=\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\)
\nWe can then insert this for every \\(y\\) in the non-linear equation to get
\n\\(\\var{c1}x^2+\\var{d1}*\\left(\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\right)^2=\\var{r2}\\)
\n\\(\\var{c1}x^2+\\var{d1}*\\frac{(\\var{r1}-\\var{a1}x)^2}{\\var{b1}^2}=\\var{r2}\\)
\nMultiplying across by \\(\\simplify{{b1}^2}\\) gives
\n\\(\\simplify{{c1}*{b1}^2}x^2+\\var{d1}(\\var{r1}-\\var{a1}x)^2-\\simplify{{r2}*{b1}^2}=0\\)
\n\\(\\simplify{{c1}*{b1}^2}x^2+\\var{d1}\\left(\\simplify{{r1}^2}-\\simplify{2*{r1}*{a1}}x+\\simplify{{a1}^2}x^2\\right)-\\simplify{{r2}*{b1}^2}=0\\)
\nGathering the like terms together gives
\n\\(\\simplify{({c1}*{b1}^2+{d1}*{a1}^2)x^2-(2*{a1}*{d1}*{r1})x+({d1}*{r1}^2-{r2}*{b1}^2)}=0\\)
\nThis is a quadratic equation.
\nRecall the formula for solving a quadratic equation of the form \\(ax^2+bx+c=0\\) is given by
\n\\(x=\\frac{-b\\pm \\sqrt{b^2-4ac}}{2a}\\)
\nIn this example \\(a = \\simplify{({c1}*{b1}^2+{d1}*{a1}^2)}, b = \\simplify{-(2*{a1}*{d1}*{r1})}\\) and \\(c = \\simplify{{d1}*{r1}^2-{r2}*{b1}^2}\\)
\nOnce we have each \\(x\\) value we insert it into \\(y=\\frac{\\var{r1}-\\var{a1}x}{\\var{b1}}\\) to find the corresponding \\(y\\) value.
", "extensions": [], "name": "Solving a Linear and a Non-linear system of equations", "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}]}], "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Xiaodan Leng", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2146/"}]}