Integrated math 3
Sal solves (-2x+4)/(x-1)=3/(x+1)-1.
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- 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?(22 votes)
- 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(4 votes)
- Can somebody help me figure out what I did wrong? I got
x = -3but 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!(8 votes)
- 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
0 = (x + 3)(x - 2)
Use zero product rule to split the factors & solve:
x + 3 = 0creates
x = -3
x -2 = 0creates
x = 2
Hope this helps.(23 votes)
- Can't you just reduce 3/x+1 and -1 to a common denominator and make the problem really easy?
-(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!
You multiply x^2-1 on both sides of the equation
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.(11 votes)
- 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.(1 vote)
- 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?(3 votes)
- 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.(4 votes)
- 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)
--- +4 = ---
--------- = ------
-------- = ------
------- = -------
After this step I lost z=-3 as a solution
------ = ------
4(z+2) = z+5
4z+8 = z+5
- 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.(7 votes)
- At5:00, he should have divided both sides by -1, right?(2 votes)
- 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(4 votes)
- 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.(2 votes)
- 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.(3 votes)
- 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.(5 votes)
- How to find cube root of 3x to exponent 3?(2 votes)
- 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.(2 votes)
- I understood the video until1:47and 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(1 vote)
- 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.(4 votes)
- [Voiceover] So we have a nice, little equation here that has some rational expressions in it and like always, pause the video and see if you can figure out which x's satisfy this equation. All right. Let's work through it together. Now, when I see things at the denominator like this, my instinct is to try to not have denominators like this. And so what we could do is to get rid of this x minus one in the denominator on the left-hand side. We can multiply both sides of the equation times x minus one, x minus one. We're gonna multiply both sides by x minus one. And once again, the whole point of doing that is so that we get rid of this x minus one in the denominator right over here. And then to get rid of this x plus one in the denominator over here, we can multiply both sides of the equation times x plus one. So, x plus one. Multiply both sides times x plus one. And so what is that going to give us? Well, on the left-hand side, that is going to, x minus one divided by x minus one is just gonna be one for the x's where that's defined, for x not being equal to one. And so, we're gonna have x plus one times negative two x plus four. So let me write that down. So we have x plus, I'm gonna need some space, so let me make sure I don't write too big. x plus one times negative two x, negative two x plus four is going to be equal to. Now, if we multiply both of these times three over x plus one, the x plus one is going to cancel with the x plus one and we're gonna be left with three times x minus one. So that is going to be three x minus three, three x minus three, and then minus, minus one times both of these. So one times x minus one times x plus one. So, minus one times x minus one times x plus one. All I did is I multiplied, took the x minus one times x plus one, multiplied times each of these terms. When I multiply it times this first term, the x plus one and the x plus one canceled, so I just have to multiply three times x minus one. And then for the second term, I just multiply times both of these. And now you might recognize, if you have something x plus one times x minus one, that's going to be x squared minus one. So I could rewrite all of this right over here as being equal to... as being equal to x squared minus one, and once again, that's because this is the same thing as x squared minus one. And since I'm subtracting an x squared minus one, actually, let me just, I don't wanna do too much on one step so let's go to the next step. So, I could multiply this out. So I could multiply x times negative two x which would give us negative two x squared. x times four, which is going to give us plus four x and I could multiply one times negative two x. I'm gonna subtract two x. And then one times four, which is gonna be plus four, and then that is going to be equal to, that is going to be equal to, we have three x minus three and then we can distribute this negative sign. So we can say minus x squared plus one. And over here, we can simplify it a little bit. This is going to be, that is four x minus two x is going to be, I'm gonna try to write over, yep, four x minus two x, so that would be two x. And so this is simplified to, let's see, well this is, we have a negative three and a one, so those two together are going to be equal to, subtracting at two. So we can rewrite everything as, do the neutral color now, negative two x squared plus two x plus four is equal to negative x squared plus three x minus two. Now we can try to get all of this business onto the right-hand side, so let's subtract it from both sides. So we're gonna subtract, or we'll say we'll add x squared to both sides, add x squared, that gets rid of this white negative x squared. We subtract three x from both sides, subtract three x from both sides, add two to both sides, add two, and we will be left with, we are going to end up with. See, negative two x squared plus x squared is negative x squared. Two x minus three x is negative x. And then four plus two is six, is going to be equal to, well that's going to cancel that, that, that, is equal to zero. I don't like having this negative on the x squared, so let's multiply both sides times negative one. And so if I do that, so if I just take the negative on both sides, so I just multiply that times negative one. Same thing, it's taking the negative on both sides. I'm gonna get positive x squared plus x minus six is equal to zero. And we're making some good progress here. So we can factor this. And actually, let me just do it right over here so that we can see the original problem. So, if I were to factor this, what two numbers, their product is negative six, they're gonna have different signs since their product is negative. And they add up to one, the coefficient on the first degree term. Well, positive three and negative two work, so I can rewrite this as x plus three times x minus two is equal to zero. Did I do that write? Yeah. Three times negative two is negative six. Three x minus two x is positive x. All right. So, I just wrote this in this quadratic in factored form. And so the way that you get this equaling to zero is if either one of those equals zero. x plus three equals zero or x minus two is equal to zero. Well, this is going to happen if you subtract three from both sides. That's gonna happen if x is equal to negative three or over here, if you add two to both sides, x is equal to two. So either one of these will satisfy, but we wanna be careful. We wanna make sure that our original equation isn't going to be undefined for either one of these. And negative three does not make either of the denominators equal to zero, so that's cool. And positive two does not make either of the denominators equal to zero. So it looks like we're in good shape. There's two solutions to that equation. If one of them made any of the denominators equal to zero, then they would have been extraneous solutions. It would have been solutions for some of our intermediate steps, but not for the actual original equations with the expressions as they're written. But this, we can feel good about because neither of these make any of these denominators equal to zero.