- #1

Strafespar

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## Homework Statement

It seemed simple at first but, how would you solve x+y=12 also given xy=35. I can only seem to solve it by trial and error. Any solutions, I think I may be missing something that's all.

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- Thread starter Strafespar
- Start date

- #1

Strafespar

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It seemed simple at first but, how would you solve x+y=12 also given xy=35. I can only seem to solve it by trial and error. Any solutions, I think I may be missing something that's all.

- #2

Redsummers

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Have you tried solving it by substitution?

- #3

hotvette

Homework Helper

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- #4

Strafespar

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xy=35 =y-y,x-x? since im subtracting by the same number, as opposed to a simple problem like 2x+6y=54

2x+3y=45 I can use simple substitution to figure out 3y=9, and then substitute the answer into one of the equations. I'm sorry I am posting this on a physics forum, it was just the only question I had in mind.

- #5

NatAvery

- 1

- 0

You then substitute your new y into xy=35.

Hope that helps.

- #6

Strafespar

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- #7

Redsummers

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You have two equations:

x+y = 12

xy = 35

Now, to do a substitution you will need e.g. to set the x in terms of y, i.e.:

x = 12 - y

Now, putting this equation into the other:

(12 - y)y = -y^2 +12y = 35

Solving for y:

[tex]y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

where a=-1, b=12, and c=-35

And that's definitely not leading you to an imaginary number solution, so it's perfectly doable in [tex]\Re[/tex]. It's not that

- #8

Strafespar

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- #9

Redsummers

- 163

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Yeah, you gotta use that formula, is like one of the commandments in linear equation religion. You're welcome ^^

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