MHB Master Algebraic Fractions: Step-by-Step Guide with 3+ Sum Examples

  • Thread starter Thread starter fordy2707
  • Start date Start date
  • Tags Tags
    Fractions
AI Thread Summary
To solve the expression -6/(s+3) - 4/(s+2) + 3/(s+1) + 2, the first step is to determine the lowest common denominator (LCD), which is the product of the distinct denominators. Each term must be adjusted to have this common denominator by multiplying by the appropriate factors. After adjusting, the expression can be combined into a single rational expression with a common denominator of (s+3)(s+2)(s+1). The final step involves expanding the numerator and combining like terms to simplify the expression. This method mirrors the process of adding fractions, making it manageable for those familiar with algebraic operations.
fordy2707
Messages
24
Reaction score
0
hi all thanks for any help ,Can You solve:

View attachment 5445

I have the answer in my book but I don't know how the answer is reached

please show steps to solve, i can solve 2 sum fractions but not sure where to start with 3 or more sums involved
 

Attachments

  • FullSizeRender.jpg
    FullSizeRender.jpg
    49.4 KB · Views: 106
Mathematics news on Phys.org
We are given the expression:

$$-\frac{6}{s+3}-\frac{4}{s+2}+\frac{3}{s+1}+2$$

And instructed to rewrite this expression as a single rational expression. Before we can do this, we need to determine the lowest common denominator (LCD), and since all denominators are prime with respect to each other, this will simply be the product of them all.

To make each term have this LCD, we need to take the LCD, and divide by denominator each term already has, and then multiply each term by 1 in the form of this quotient divided by itself. For example, the first term already has a denominator of $s+3$, and so we find the quotient:

$$\frac{(s+3)(s+2)(s+1)}{s+3}=(s+2)(s+1)$$

And so we will want to multiply the first term by:

$$1=\frac{(s+2)(s+1)}{(s+2)(s+1)}$$

I am emphasizing that this expression is equal to 1 so that it is clear that in doing so we are not changing the value of that term. Doing the same for the other terms, we will have:

$$-\frac{6}{s+3}\cdot\frac{(s+2)(s+1)}{(s+2)(s+1)}-\frac{4}{s+2}\cdot\frac{(s+3)(s+1)}{(s+3)(s+1)}+\frac{3}{s+1}\cdot\frac{(s+3)(s+2)}{(s+3)(s+2)}+2\cdot\frac{(s+3)(s+2)(s+1)}{(s+3)(s+2)(s+1)}$$

Now, all 4 terms have the same denominator, and we can combine them:

$$\frac{-6(s+2)(s+1)-4(s+3)(s+1)+3(s+3)(s+2)+2(s+3)(s+2)(s+1)}{(s+3)(s+2)(s+1)}$$

And so now we have done as instructed. You may wish to expand and then combine like terms in the numerator. :)
 
Nice one with your help I've worked it out, it's a long calculation to get the simplest form with no brackets but a good challenge
 
All we're doing here is repeating a/b + c/d = (ad + bc)/bd. If you're used to doing this, this should be easy.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top