How to Solve for Variables in Equations with Multiple Variables and Equations?

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SUMMARY

The discussion focuses on solving for four variables (a, b, c, d, e) in a system of equations derived from the equation \(\frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{dx+e}{x^2}\). The participants identify that there are two equations available: \(c=0\) and \(b+e=0\), along with \(a+d=0\). The challenge lies in determining the values of the variables given the insufficient number of equations relative to the number of variables.

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  • Understanding of algebraic manipulation and simplification.
  • Familiarity with polynomial equations and their properties.
  • Knowledge of systems of equations and methods for solving them.
  • Basic calculus concepts, particularly limits and continuity.
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  • Study methods for solving systems of linear equations, such as substitution and elimination.
  • Learn about polynomial long division and its applications in algebra.
  • Explore the concept of degrees of freedom in systems of equations.
  • Investigate the implications of underdetermined systems in algebraic contexts.
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Students and educators in mathematics, particularly those focused on algebra and calculus, as well as anyone involved in solving complex equations with multiple variables.

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[tex]\frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{dx+e}{x^2}\\[/tex]
[tex]ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\\[/tex]
[tex]c=0\\[/tex]
[tex]b+e=0\\[/tex]
[tex]a+d=0\\[/tex]

i got 4 variables with 2 equations

what to do??
 
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proto said:
[tex]\frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{dx+e}{x^2}\\[/tex]
[tex]ax^4+bx^3+cx^2+dx^4+ex^3=x^4(a+d)+x^3(b+e)+cx^2=1\\[/tex]
[tex]c=0\\[/tex]
[tex]b+e=0\\[/tex]
[tex]a+d=0\\[/tex]

i got 4 variables with 2 equations

what to do??

[tex]\frac{1}{x^3(x^2+1)}=\frac{ax^2+bx+c}{x^3}+\frac{dx+e}{x^2+1}\\[/tex]
 

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