MHB Fractional Equation Help: Solve 3 Problems

  • Thread starter Thread starter Mikeybr
  • Start date Start date
  • Tags Tags
    fractional
Click For Summary
The discussion centers around three fractional equation problems that the user is struggling to solve. The first equation can be simplified by eliminating fractions through multiplication by the denominators. A suggestion is made to rewrite the first fraction and then multiply through by the denominators to simplify the equation further. The importance of careful notation, especially with parentheses, is emphasized to facilitate better assistance. The thread encourages users to tackle one problem at a time for more effective help.
Mikeybr
Messages
1
Reaction score
0
There are three problems on my homework I can't get quite right.

x/x-1 - 1 = 3/x+1

4/b - 1/b+3 = 3b+2/b^2+2b-3

3r+1/r+3 + 2 =5r-2/r+3
 
Mathematics news on Phys.org
Mikeybr said:
There are three problems on my homework I can't get quite right.

x/x-1 - 1 = 3/x+1

4/b - 1/b+3 = 3b+2/b^2+2b-3

3r+1/r+3 + 2 =5r-2/r+3

Hi Mikeybr. Welcome to MHB! :)

Try posting one question at a time please and show some attempt at it. This helps us help you more efficiently. Also, be careful with parentheses!

I believe #1 is the following:

[math]\frac{x}{x-1}-1=\frac{3}{x+1}[/math]

You could do this a few ways but the idea is to get rid of the fractions somehow. Let's first try to rewrite the first fraction by multiplying everything by $(x-1)$.

[math](x-1) \cdot \frac{x}{x-1} -1 \cdot (x-1) = (x-1) \cdot \frac{3}{x+1}[/math]

The first fraction now has [math]\frac{x-1}{x-1}[/math] in it, which is 1 so that fraction is now just x. The rest of the equation is below.

[math]x - (x-1) = \frac{3(x-1)}{x+1}[/math]

Now we want to change the other fraction, so we can multiply everything by $(x+1)$. Doing that we get:

[math]x(x+1)+(x-1)(x+1)=3(x-1)[/math]

Can you solve from here?
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
16
Views
851
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K