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There are many ways to solve these equations simultaneously. Here is one method.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify{{grad}x+{yint}}$               $(1)$
$y$$=$$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$               $(2)$
\n

Substitute the expression for $y$ given in $(1)$ into $(2)$:
\\[\\simplify{{grad}x+{yint} =x^2+{quadxcoeff}x+{quadccoeff}}\\]

\n

Since we have a quadratic here we get everything onto one side:
\\[0=\\simplify{x^2+{sroots}x+{proots}}\\]

\n

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\\]

\n

Therefore, $x=\\var{root1}$.

\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 $(1)$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify[!collectnumbers]{{grad}({root1})+{yint}}$
$=$$\\var{ansyvalue}$
\n

Therefore the values that satisfy equations $(1)$ and $(2)$ are $x=\\var{root1}$ and $y=\\var{ansyvalue}$.

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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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$y$$=$$\\simplify{{grad}x+{yint}}$               $(1)$
$y$$=$$\\simplify{x^2+{quadxcoeff}x+{quadccoeff}}$               $(2)$
\n

$x=$ [[0]],   $y=$ [[1]]

\n

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