Question 1

I don't get how he multiplied ( 3/(x+1) -1 )_by (x -1)(x+1). By what drawn out math does that make sense to multiple the first term 3/(x+1) by both (x-1)(x+1) at the same time? is it reverse distribution? I didn't realize you could do reverse distribution with two binomials at the same time. I always thought it was only ex: (b+5)(b-5) and you multiply b+5 by b and b+5 by -5. Am i missing some simple math here? Why is he able to distribute two binomials into one binomial's first term "3/(x+1)" . at the same time?

Accepted Answer

Sal is able to multiply [(3/x+1) -1] by  (x-1)(x+1) because the (x+1) 's cancel out, leaving the (2-x)*(x-1), which can then be simplified. The reason why he multiplied by multiple binomials is to get rid of *all* fractions. Hope this helped! :D

Question 2

Can somebody help me figure out what I did wrong? I got `x = -3` but I didn't get `x = 2`.
This is how I worked the problem: 
```
-2x + 4       3
--------   =   --------   -1
x - 1          x + 1

2( -x + 2)       3
----------   =   --------   -1
  x - 1          x + 1

2( -x + 2)       3          x + 1
----------   =   --------   -   --------
 x - 1           x + 1        x + 1

2( -x + 2)       -x + 2
----------   =   --------
 x - 1           x + 1

2( -x + 2)( x + 1 )  =  ( -x + 2 )( x - 1 )

2( -x + 2)( x + 1 )     ( -x + 2 )( x - 1 )
--------------------  =  ------------------
   -x + 2                -x + 2

2( x + 1)  =  x - 1

2x + 2  =  x - 1

2x - x + 2 + 1  =  x - x - 1 + 1

x + 3  =  0
x  =  -3
```

Sorry about the weird notation. Thanks for the help!

Accepted Answer

It's been a year, but I'm curious about this. Why do you say "never" divide by a value containing the variable? What about all the times we have to do exactly that? Does it just depend on what the question is asking us to do?

For example, if we're asked to simplify then we want to find common factors and divide them out, but here it asks us to find the zeros, so we can't without potentially removing one?

Question 3

We can't divide by zero. Question: We also can't take the square root of a negative number. We solve that by introducing the concept of imaginary numbers, _i_. Is there not a way to solve the inability to divide by zero with a similar concept/solution?

Accepted Answer

Good thinking, but it just doesn't work. There is a problem with defining something to represent dividing by 0.

We can define i^2=−1 and add it to our realm of math, and still have all of the rules of algebra work. And as a bonus, complex numbers are super nice to use (after all, they can be thought of as simply transformations from the number line to an imaginary plane).

However, we cannot define n/0 and still have even the simplest of our rules apply when n is a nonzero number. To see this at the very basic level, let's say that n/0 is defined as z.

n/0 = z, then multiply both sides by 0, we get

n = z∗0, or

n = 0, which is inconsistent.

Now, let's define z so that z*0 = n.

z*0 = z(0+0) based on basic rules of algebra, then distribute

z*0 = z*0 + z*0, then subtract z*0 from both sides

0 = z*0, which is inconsistent

There is something called the Riemann sphere that sets z/0 to infinity for any value other than 0, but it is still causes some problems (negative vs. positive infinity, the problem with 0/0, doesn't fit some rules of algebra, etc). Read into it if you are interested.

Question 4

At 5:00, he should have *divided* both sides by -1, right?

Accepted Answer

Multiplying and dividing by -1 (or positive 1) is the same.  That kind of number has a special name called an identity.  with addition and subtraction 0 is the identity because adding or subtracting 0 is the same

Question 5

If you are multiplying 0 by -1, it makes -0 doesn't it? But -0 can't exist, right? So what is the logic behind -0? Because 0 = nothing, -0 = -nothing?

I'm so curious to find out if -0 isn't equal to 0.

Accepted Answer

Yes, in some sense 0 and -0 are the same.  There is only one 0 on a number line.  It is considered a neutral number (neither positive not negative).

Remember:  -0 is still -1(0).  Anything times 0 = 0.  So, -0, becomes 0 once you fully complete the multiplication.

Question 6

How to find cube root of 3x to exponent 3?

Accepted Answer

Cuberoot[3x^3] = x cuberoot(3)
For something to come out of the cuberoot, it needs an exponent of 3 or a multiple of 3 like 6, 9, 12, etc.

Sinces x^3 has an exponent of 3, the x comes out.  The 3 has an exponent of 1.  So it says in the radical.

Hope this helps.

Question 7

I hope someone can help me understand where I went wrong below. I seem to have lost one of the solutions. Thanks, Scott ( and if there is a better way than using plain text to enter questions, like screen shots, please enlighten me)
-8             z+3
--- +4     =   ---
z+5            z+2

-8+4(z+5)      z+3
---------        = ------ 
  (z+5)            z+2

-8+4z+20       z+3
  --------        = ------
   (z+5)            z+2

4(z+3)        z+3
-------     = -------
(z+5)         z+2

After this step I lost z=-3 as a solution

4         1
------ = ------
(z+5)     z+2

4(z+2) = z+5

4z+8 = z+5

3z=-3

z=-1

Accepted Answer

You lost the z=-3 solution because it looks like you divided your equation by z+3.  This destroys a solution.  Never divide by a variable when you are solving an equation because it causes you to lose potential solutions.
Hope this helps.

Question 8

Can someone please explain what extraneous solutions are and, why they can't be one of our actual solutions?

Accepted Answer

An extraneous solution is an invalid solution created when you solve the equation - you have no math errors, but the solution doesn't actually satisfy the equation.  It either makes the 2 sides unequal, or it creates an undefined state.

Tational equations (equations with fractions) where you have a variable in any denominator have the potential for having extraneous solutions.  The value you find may cause a denominator to = 0, which means it is creating an undefined state.  Undefined is not a valid state for the equation to be in.  So, the solution is invalid.  And, we call it an extraneous solution.

Radical equations are another example where extraneous solutions occur.  The process of applying an exponent to both sides of the equation to eliminate the radical(s) can introduce extra solutions that don't actually satisfy the equation.

Hope this helps.

Question 9

I understood the video until 1:47 and I started to get lost.. I understood how the (x+1) is going to cancel the (x+1) and the problem is left with 3-1(x-1) but then 3x-3 why does he subtract the 1 with both (x+1) and (x-1)? Can it just be  solved like 3-1 (x+1) then 2(x-1) 2x+2 ...? could someone explain this please and thank you

Accepted Answer

On the right side, the (x+1)(x-1) must be distributed across the [3/(x+1) - 1]
3/(x+1) * (x+1)(x-1) = 3(x-1)
But -1(x+1)(x-1), nothing cancels out.  The -1 has no denominator other than 1.

I think you are treating as if the -1 is -1/(x+1).  But it isn't.
Hope this helps.

Algebra (all content)

Course: Algebra (all content) > Unit 13

Equations with rational expressions (example 2)

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