How Do I Solve for x and y Using Simultaneous Equations?

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Homework Help Overview

The discussion revolves around solving a system of simultaneous equations involving a linear equation and a quadratic equation. The equations presented are 2x - 3y = 1 and x² - 2xy - 3y² = -3.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to substitute the linear equation into the quadratic to derive new equations in terms of a single variable. They express uncertainty about whether to continue substituting to create additional quadratics. Other participants question the clarity of the equations and suggest focusing on solving the derived quadratic.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's approach. Some guidance has been offered regarding solving the quadratic equation, but there is no explicit consensus on the best method to proceed.

Contextual Notes

The original poster mentions time constraints related to a test scenario, indicating a concern about the efficiency of their approach. There is also a reference to an equation that was not initially included in the discussion, which may affect clarity.

MadmanMurray
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Crazy Simultaneous Equation ?!?

Homework Statement



Solve for x and y
1)2x - 3y = 1
2)x2 - 2xy - 3y2 = -3

Homework Equations


The Attempt at a Solution


I substituted the linear equation into the quadratic and got a quadratic with only y in it:
3.) 15y2 - 2y = 13

I then substituted the linear into eq3 and got a quadratic with only x in it:
4.) 5x2 - 6x = 9

Do I need substitute that linear into these quadratics twice more so I get 2 simultaneous quadratics with only x in them and 2 quadratics with only y in them?

I've been on this question for about half an hour now but this is a sample question from a test so there's no way I could spend that long on a single question on a test. I must be doing it wrong.
 
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where is eqn 3?
You say it has only y in it why aren't you solving it then?
 


Its a quadratic if has y^2 and y. I'll edit the post and put in all the equations I got
 


3.) 15y2 - 2y = 13
Looks good! The rest is easy - just use one of your techniques for solving a quadratic equation to find the number(s) for y.
 


Ah yea I didn't even think of factorizing or anything like that. Thanks.
 

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