- #1

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## Homework Statement

x+2y=3

x^2-4y^2=-33

## Homework Equations

## The Attempt at a Solution

x-2y=root-33?

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- Thread starter d3v3st4t10n
- Start date

- #1

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x+2y=3

x^2-4y^2=-33

x-2y=root-33?

- #2

HallsofIvy

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Probably it is simplest to solve the first equation for x, replace the "x" in the second equation by that expression.

- #3

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Try this:

x+2y=3

=> x=3-2y

**<< detailed solution hints deleted by berkeman >>**

x+2y=3

=> x=3-2y

Last edited by a moderator:

- #4

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and u have the a+b eqn. with you..

Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.

- #5

HallsofIvy

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Oh, that's cute.

- #6

Mentallic

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It is always easiest (imo) to make x or y the subject in the linear equation: i.e. the first equation becomes [tex]x=3-2y[/tex] or [tex]y=\frac{3-x}{2}[/tex] and then substitute either of these values into the quadratic equation so that becomes: [tex](3-2y)^2-4y^2=-33[/tex] or if you made y the subject: [tex]x^2-4(\frac{3-x}{2})^2=-33[/tex]

Oh and remember you can't take the square root of a negative value.

- #7

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x+2y=3

x^2-4y^2=-33

-33 = (x+2y)(x-2y)

got it now?

x^2-4y^2=-33

-33 = (x+2y)(x-2y)

got it now?

- #8

Mentallic

Homework Helper

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- #9

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I know, I wouldn't have found it until I was halfway through the problem. It was a nice catch.

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