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Find the $x$ and $y$ values that satisfy both of the following equations. That is, find the point of intersection of the two curves.
\n$y$ | \n$=$ | \n$\\simplify{{grad}x+{yint}}$ | \n$(1)$ | \n
$y$ | \n$=$ | \n$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$ | \n$(2)$ | \n
$x=$ [[0]], $y=$ [[1]]
\n", "scripts": {}, "steps": [{"showCorrectAnswer": true, "variableReplacements": [], "prompt": "There are many ways to solve these equations simultaneously. Here is one method.
\n$y$ | \n$=$ | \n$\\simplify{{grad}x+{yint}}$ | \n$(1)$ | \n
$y$ | \n$=$ | \n$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$ | \n$(2)$ | \n
Substitute the expression for $y$ given in $(1)$ into $(2)$:
\\[\\simplify{{grad}x+{yint} =x^2+{quadxcoeff}x+{quadccoeff}}\\]
Since we have a quadratic here we get everything onto one side:
\\[0=\\simplify{x^2+{sroots}x+{proots}}\\]
There are various ways to solve a quadratic, in this particular case we can factorise the quadratic:
\n\\[(\\simplify{x-{root1}})(\\simplify{x-{root2}})=0\\]
\nTherefore, $x=\\var{root1}$.
\n
Now we know the $x$ value we can determine the corresponding $y$ value by substituting $x=\\var{root1}$ into either equation $(1)$ or $(2)$, below we substitute into $(1)$:
$y$ | \n$=$ | \n$\\simplify[!collectnumbers]{{grad}({root1})+{yint}}$ | \n
\n | $=$ | \n$\\var{ansyvalue}$ | \n
Therefore the values that satisfy equations $(1)$ and $(2)$ are $x=\\var{root1}$ and $y=\\var{ansyvalue}$.
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