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# Equations with rational expressions (example 2)

Sal solves (-2x+4)/(x-1)=3/(x+1)-1.

## Want to join the conversation?

• 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?
• 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
• 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--------   =   --------   -1x - 1          x + 12( -x + 2)       3----------   =   --------   -1  x - 1          x + 12( -x + 2)       3          x + 1----------   =   --------   -   -------- x - 1           x + 1        x + 12( -x + 2)       -x + 2----------   =   -------- x - 1           x + 12( -x + 2)( x + 1 )  =  ( -x + 2 )( x - 1 )2( -x + 2)( x + 1 )     ( -x + 2 )( x - 1 )--------------------  =  ------------------   -x + 2                -x + 22( x + 1)  =  x - 12x + 2  =  x - 12x - x + 2 + 1  =  x - x - 1 + 1x + 3  =  0x  =  -3``

Sorry about the weird notation. Thanks for the help!
• When you divided by (-x + 2), you lost the 2nd solution. You changed a quadratic into a linear equation. To avoid this, never divide an equation by a value containing the variable.

Instead of doing that division, multiply out each side:
`2( -x + 2)( x + 1 ) = ( -x + 2 )( x - 1 )`
`2( -x^2 + x + 2 ) = -x^2 + 3x - 2 `
` -2x^2 +2 x + 4 = -x^2 + 3x - 2 `
Move all terms to the same side: ` 0 = x^2 + x - 6 `
Factor: `0 = (x + 3)(x - 2)`
Use zero product rule to split the factors & solve:
`x + 3 = 0` creates `x = -3`
`x -2 = 0` creates `x = 2`

Hope this helps.
• Can't you just reduce 3/x+1 and -1 to a common denominator and make the problem really easy?
-2(x-2)/x-1=2-x/x+1
-(x-2 )and 2-x cancel out.
Of course, when you cancel them out, you need to consider the situation when x=2. You may put x=2 into the original equation and finds that x=2 does work!
2/x-1=1/x+1
You multiply x^2-1 on both sides of the equation
2x+2=x-1
x=-3
The result comes out which is x=2/-3
Nice and neat.
I think solving the problem in this way is much easier than Sal's method.
• This is an easier method, but Sal is just explaining how to solve the problem the tradional way. Your method of solving these types of problems are faster but needs more thought.
• 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?
• 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.
• 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
(1 vote)
• 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.
• At , he should have divided both sides by -1, right?
• 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
• Didn't he make a mistake a ? He multiplied the (x+1) so he could remove the fraction, but then puts its again in the equation
• Rewind to and watch from there a few times to understand. The answer to your question is at .
(1 vote)
• 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.
• 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.
• Can someone please explain what extraneous solutions are and, why they can't be one of our actual solutions?
(1 vote)
• 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.