Solving Equivalent Expressions Using Subtraction Method

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Discussion Overview

The discussion centers on determining the equivalence of two algebraic expressions using the subtraction method. Participants explore the steps involved in manipulating the expressions to assess their equivalence, focusing on algebraic techniques and common denominators.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how to determine if two expressions are equivalent using the subtraction method, noting that they must equal zero to be considered equivalent.
  • Another participant points out a potential error in the initial manipulation of the expressions, suggesting that the expressions should be combined over a common denominator.
  • A later reply emphasizes the importance of correctly distributing the negative sign when simplifying expressions within brackets.
  • One participant argues that the expression presented is indeed a difference between the two expressions, just reformulated with a common denominator.
  • Another participant advises on the necessity of multiplying out brackets to collect like terms, indicating that this is essential for proper simplification.
  • One participant suggests that to combine terms over a common denominator, one must express the polynomial as a fraction with the denominator \(x + 1\) before proceeding with the subtraction.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to simplifying the expressions and whether the initial manipulations are valid. There is no consensus on the steps to take or the correctness of the expressions as equivalent.

Contextual Notes

Some participants highlight the importance of careful algebraic manipulation, particularly regarding the distribution of negative signs and the need for a common denominator. There are unresolved steps in the algebraic process that may affect the outcome.

eleventhxhour
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How do you determine if the two expressions are equivalent using the subtraction method?$$\frac{2x^3+7x^2-5x-6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] $$

The textbook says that they're equivalent. I know that, using the subtraction method, they have to equal 0 in order to be equivalent. But, I'm not getting that. This is what I've done so far:

$$2x^2+7x-5-\frac{6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] $$

Then I'm not sure what to do...
 
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eleventhxhour said:
How do you determine if the two expressions are equivalent using the subtraction method?$$\frac{2x^3+7x^2-5x-6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] $$

The textbook says that they're equivalent. I know that, using the subtraction method, they have to equal 0 in order to be equivalent. But, I'm not getting that. This is what I've done so far:

$$2x^2+7x-5-\frac{6}{x+1} - [(2x^2+5x-10) + \frac{4}{x+1}] $$

Then I'm not sure what to do...
Start by putting everything over the denominator $x+1$. You have already slipped up by writing $$2x^2+7x-5-\frac{6}{x+1}$$ as though it is the same as $$\frac{2x^3+7x^2-5x-6}{x+1}$$ (it isn't!). What you should find is that the expression is equal to $$\frac{2x^3+7x^2-5x-6 - \bigl[(2x^2+5x-10)(x+1) + 4\bigr]}{x+1}.$$

The next step (in fact, maybe this should have been the first step) is to get rid of the square brackets, noticing that there is a minus sign in front of them, which changes the sign of every term inside.
 
Opalg said:
Start by putting everything over the denominator $x+1$. You have already slipped up by writing $$2x^2+7x-5-\frac{6}{x+1}$$ as though it is the same as $$\frac{2x^3+7x^2-5x-6}{x+1}$$ (it isn't!). What you should find is that the expression is equal to $$\frac{2x^3+7x^2-5x-6 - \bigl[(2x^2+5x-10)(x+1) + 4\bigr]}{x+1}.$$

The next step (in fact, maybe this should have been the first step) is to get rid of the square brackets, noticing that there is a minus sign in front of them, which changes the sign of every term inside.

This is what I got, but it's not equivalent...
$$\frac{2x^3+7x^2-5x-6}{x+1}-\frac{2x^3+5x^2-10x+4}{x+1}$$
 
eleventhxhour said:
This is what I got, but it's not equivalent...
$$\frac{2x^3+7x^2-5x-6}{x+1}-\frac{2x^3+5x^2-10x+4}{x+1}$$

You're doing it wrong. Forget about the minus sign for a minute and multiply out the brackets and collect 'like' terms.
 
How is it not the subtraction method? Isn't Opalg's expression the difference between the two expressions, just written with a common denominator?
 
I don't think my last post was very clear. My advice to multiply out brackets is to find all terms with the highest power of x, then then the terms with the next highest power and so on. So $(2x^2+5x-10)(x+1)+4$=$2x^3+7x^2-5x-6$ which is what you want
 
I think you are trying to "skip steps".

To get everything over a common denominator, we have to write:

$2x^2 + 5x - 10 = \dfrac{\text{something}}{x + 1}$.

The way to do this is to multiply by $1 = \dfrac{x+1}{x+1}$. That is:

$2x^2 + 5x - 10 = \dfrac{(2x^2 + 5x - 10)(x+1)}{x+1} = \dfrac{(2x^2 + 5x - 10)x + (2x^2 + 5x - 10)(1)}{x+1}$

$= \dfrac{2x^3 + 5x^2 - 10x + 2x^2 + 5x - 10}{x+1} = \dfrac{2x^3 + 7x^2 - 5x - 10}{x+1}$

ONLY NOW can we add the two terms:

$2x^2 + 5x - 10 + \dfrac{4}{x+1} = \dfrac{2x^3 + 7x^2 - 5x - 10}{x+1} + \dfrac{4}{x + 1} = \dfrac{2x^3 + 7x^2 - 5x - 6}{x+1}$.

Now what do you get when you subtract this from $\dfrac{2x^3 + 7x^2 - 5x - 6}{x+1}$?
 

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