Algebra: Solving Equation with DeltaXs

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In summary, the conversation discusses a question about an equation and how it can be simplified using common denominator and multiplication by 1/delta x. The steps to simplify the equation are explained in detail, along with the reasoning behind each step. The person asking the question expresses their gratitude for the help and mentions that they struggle with certain concepts in math, but find them manageable with proper understanding.
  • #1

I watched some of the first video on the calculus lectures someone posted, but had a question about some of the algebra when he factors something out of the equation. I've noticed stuff like this is where I get messed up. If someone could explain how this equation:
[1/(x + deltax) - 1/(x)] / deltax

turns into

(1/deltax)[(x - (x + deltax)) /(x + delta x)(x)]

The "X"s, not the "deltaXs", were written with a little subscript zero to the bottom of them, but I don't know how to do that on my keyboard, my apologies. Also, it may hard to read the way it is written, but hopefully you guys can get the jist of it. Thanks
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  • #2
First, instead of having a numerator and denominator fraction where the numerator itself consists of fractions with their own numerator and need to take the denominator and pull it out (1/delta) as multiplying the whole numerator..., you should have the multiplication of two factors: (1/delat) and the rest.

then, you take the second factor and join them using a common denominator.
  • #3
[tex]\frac{\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}}{\Delta x} = \left(\frac{1}{\Delta x}\right)\left(\frac{1}{x_0 + \Delta x}-\frac{1}{x_0}\right)[/tex]

[tex]= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0}{(x_0+\Delta x)(x_0)} - \frac{x_0+\Delta x}{(x_0+\Delta x)(x_0)}\right) [/tex]

[tex]= \left(\frac{1}{\Delta x}\right)\left( \frac{x_0 - (x_0 + \Delta x)}{(x_0 + \Delta x)(x_0)}\right) = \left(\frac{1}{\Delta x}\right)\left( \frac{-\Delta x}{(x_0 + \Delta x)(x_0)}\right)[/tex]

[tex]= \frac{-1}{(x_0 + \Delta x)(x_0)}[/tex]


in the first step, we note that division by u is the same thing as multiplication by 1/u (here our "u" is "delta x").

next, we multiply each fraction in the difference, by the denominator of the other fraction in the difference, to get a common denominator:

(a/b - c/d) = ((a/b)(1) - (c/d)(1)) = ((a/b)(d/d) - (c/d)(b/b)) = ((ad)/(bd) - (bc)/(bd)) <--common denominator of bd

once we have a common denominator, we can subtract numerators:

(ad)/(bd) - (bc)/(bd) = (ad - bc)/(bd) (this is just the standard rule for subtracting fractions).

once that is done, we cancel the "delta x's".
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  • #4
Oh man, you have my sincerest thanks. That is exactly what I needed to see. The ideas and concepts of most math that I have taken and studied on my own (calculus) do not seem to hard to me. It is just things I do not have down that well, here and there, that cause roadblocks. Like my initial question.
  • #5

I understand that algebra can be confusing and challenging at times. It is important to have a solid understanding of algebra in order to solve equations and understand mathematical concepts, especially in fields such as calculus.

In this equation, we are trying to simplify the expression by factoring. The first thing to note is that the denominator, deltax, is common to both fractions. Therefore, we can factor it out and rewrite the equation as:

1/deltax * (1/(x+deltax) - 1/x)

Next, we can simplify the fractions by finding a common denominator. In this case, the common denominator is (x+deltax)(x). We can rewrite the fractions with this common denominator as:

1/deltax * [(x/(x+deltax)) - (x+deltax)/((x+deltax)(x))]

Now, we can combine the fractions by subtracting the numerators and keeping the common denominator:

1/deltax * [(x - (x+deltax))/((x+deltax)(x))]

Finally, we can simplify the numerator by distributing the negative sign and simplifying the expression to get the final answer:

1/deltax * [(x - x - deltax)/((x+deltax)(x))] = 1/deltax * (-deltax/((x+deltax)(x))) = (-1/(x+deltax)(x))

I hope this explanation helps to clarify how the equation was simplified. Remember to always look for common factors and common denominators when simplifying algebraic expressions. Practice makes perfect, so keep practicing and you will become more comfortable with algebraic manipulations.

1. What is an equation with Δx?

An equation with Δx is an algebraic expression that involves a change in the value of x. It is commonly used in physics and calculus to represent small changes in a variable.

2. How do you solve an equation with Δx?

To solve an equation with Δx, you need to isolate Δx on one side of the equation. This can be done by using inverse operations to cancel out any coefficients or constants that are attached to Δx. Once Δx is isolated, you can solve for its value by applying the same operations to both sides of the equation.

3. Can you provide an example of solving an equation with Δx?

Sure, let's say we have the equation 3Δx + 4 = 16. To solve for Δx, we first need to get rid of the 4 on the left side by subtracting it from both sides: 3Δx = 12. Then, we need to isolate Δx by dividing both sides by 3: Δx = 4. Therefore, the solution to the equation is Δx = 4.

4. Why is it important to use Δx in equations?

Δx is important because it allows us to represent small changes in a variable in a precise and concise way. It is commonly used in physics and calculus to calculate rates of change and to approximate solutions to more complex problems.

5. Are there any other notations for Δx?

Yes, Δx can also be represented as Δt or dx. These notations all represent the same concept of a small change in a variable. In some cases, different notations may be used to represent different types of changes, such as a change in time (Δt) or a change in position (dx).

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