What algebraic property can I use here?

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Homework Help Overview

The discussion revolves around an algebraic equation presented in a general chemistry context, specifically the equation 5 = [(x)(4+x)] / (4-x). Participants are exploring the algebraic properties and techniques that can be applied to solve for x, while expressing uncertainty about the methods used in class.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the concept of multiplying by the reciprocal and balancing equations. There are questions about the validity of the solution x = 5 and whether it satisfies the original equation. Some participants inquire about the rules of multiplication and addition in the context of solving equations.

Discussion Status

The discussion is ongoing, with participants questioning the assumptions made about the solution and exploring different algebraic properties. Some guidance has been offered regarding the rules of operations in equations, but there is no consensus on the approach to take or the correctness of the proposed solution.

Contextual Notes

Participants note that the problem is situated within a general chemistry class, which may influence their understanding of the algebra involved. There is also mention of uncertainty regarding the professor's methods and the steps taken in class.

megaboy123

Homework Statement


[/B]
example problem: 5 = [(x)(4+x)] / (4-x)

answer: 5

Homework Equations


Unsure what to use.

3. The Attempt at a Solution
Not sure what my professor did, but I thought that if i multiply by the reciprocal of something, I have to balance by multiplying the other side as well. It seems like everything just canceled out and the only thing left was x = 5. I would just like to know what algebraic property or technique can be used to find x BESIDES quadratic formula.
 
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megaboy123 said:
Not sure what my professor did,...
This is the purpose of taking notes in class.
... but I thought that if i multiply by the reciprocal of something, I have to balance by multiplying the other side as well.
Yes.
It seems like everything just canceled out and the only thing left was x = 5.
Do you mean that is your vague recollection of what the prof did or that is what happened when you did it?
I would just like to know what algebraic property or technique can be used to find x BESIDES quadratic formula.
...what's wrong with the usual rules for multiplication and addition?
What was the topic of the class?

Have you tried substituting the stated solution x=5 into the equation to see if it really is the solution?
(I am guessing that "answer:5 means x=5 will make the expression true...)
 
Last edited:
its for a general chemistry class, everything makes sense (as far as chemistry goes) up until that last step right before the answer. Not sure what you mean by rules for multiplication and addition.
 
I am guessing that "answer:5" means that putting x=5 into the expression 5=[x(4+x)]/(4-x) will make it true.
Have you tried this to see?

Not sure what you mean by rules for multiplication and addition.
Do you know how to add and how to multiply?
 
megaboy123 said:
Not sure what you mean by rules for multiplication and addition.
If you add the same quantity to both sides of an equation, the new equation will have the same solutions as the original equation.
If you multiply both sides of an equation by the same nonzero quantity, the new equation will have the same solutions as the original equation.
 
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megaboy123 said:

Homework Statement


[/B]
example problem: 5 = [(x)(4+x)] / (4-x)

answer: 5

Homework Equations


Unsure what to use.

3. The Attempt at a Solution
Not sure what my professor did, but I thought that if i multiply by the reciprocal of something, I have to balance by multiplying the other side as well. It seems like everything just canceled out and the only thing left was x = 5. I would just like to know what algebraic property or technique can be used to find x BESIDES quadratic formula.

Are you solving an equation of the form
$$
\frac{x(4+x)}{4-x} = 5 ?
$$
If so, x=5 has nothing to do with it: there are two solutions, which are obtainable from the quadratic solution formula: x ≈ -10.84 and x ≈ 1.84.
 
Last edited:

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