How do you solve the system $9x^2+4y^2=36$ and $-x+y=-4$ ?

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How do you solve the system $9x^2+4y^2=36$ and $-x+y=-4$ ?

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Answer 1 · 2016-09-30T01:57:07

Substitute $x - 4$ for y in $9x^2 + 4y^2 = 36$ to discover that there are no real roots for the resulting quadratic, therefore, the line does not intersect with the ellipse.

Explanation:

Given:
$y = x - 4$
$9x^2 + 4y^2 = 36$

$9x^2 + 4(x - 4)^2 = 36$

$9x^2 + 4(x^2 - 8x + 16) = 36$

$13x^2 - 32x + 64 = 36$

$13x^2 - 32x + 28 = 0$

$b^2 - 4(a)(c) = (-32)^2 - 4(13)(28) = -432$

There are no real roots, therefore, the line does not intersect with the ellipse

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How do you solve the system $9x^2+4y^2=36$ and $-x+y=-4$ ?

1 Answer

Explanation:

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Science

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Humanities

... and beyond

How do you solve the system 9x^2+4y^2=369x^2+4y^2=36 and -x+y=-4-x+y=-4?

1 Answer

Explanation:

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How do you solve the system $9x^2+4y^2=36$ and $-x+y=-4$ ?