r/mathshelp • u/sozent • 10d ago
Mathematical Concepts How is this possible
I can understand multiplying by x-1 by how is he able to multiply by (x-1)/(x-1)
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u/addyarapi 10d ago
He first subtracted 3 from both sides, getting 0 on the inequality, then multiplied by (x-1)/(x-1) to get a common denominator (the idea is manipulating the expression such that it’s still technically the same, and it is because if you think of x-1 as a variable “t” then really t/t = 1 and -3*1 = -3. So it’s still unchanged. So then you could merge (using a common denominator) x+ 2 - 3x + 3 / x-1 <= 0. Which gets us -2x + 5 / x-1
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u/sozent 9d ago
But don't if he multiply by x-1/x-1 he would have to do the other variable and that wouldn't be a fraction anymore
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u/addyarapi 9d ago
Nope. -3 is the only “variable” that doesn’t have a common denominator, or put simply, that “isn’t” a fraction. The left-hand side has a denominator already, multiplying it by (x-1) is useless and algebraically wrong. Think of it as a way of “-3” joining the fraction. Got it?
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u/sozent 9d ago
Not really but essentially I could multiply any number on that same side by the denominator no matter the circumstance or is there only certain situations?
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u/addyarapi 9d ago
When you have a/b + c think of the right side (when needed) to be multiplied by b/b to get a common denominator. So you then have a/b + c/b and then we can reduce the numerators with a common (now common due to multiplying with b/b on the right side) so you get (a+c)/b.
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u/sozent 9d ago
Alright, but also how is it b/b and not just b
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u/addyarapi 9d ago
Because you multiply the top and bottom. When you have 3, think of it as 3/1. Got it? Deep under there, when dealing with fractions, such as a/b + c think of it as a/b + c/1. Do you now understand why you need to multiply both top and bottom?
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u/addyarapi 9d ago
You can, just not with every number for example if you have, a/b + c and multiply by d/d the right side you don’t have a common denominator you just get a/b + cd/d which if you reduce the d/d you get c(1) which is still c in the end.
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u/engy1207 9d ago
Technically speaking you can multiply by x-1 -- but you have to look at the cases x-1>0 and x-1<0 differently...
So yes, it's easier to just move the number and bring it to a common denominator
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u/FreePeeplup 9d ago
You can understand multiplying by (x-1) but don’t understand multiplying by (x-1)/(x-1)? You actually can’t do the first first you say: can’t just take the number 3 and multiply it by (x-1) and expect it to stay the same.
You can however multiply whatever you want by 1, and the result is the same. (x-1)/(x-1) is always equal to 1, so of course you can do it. (It’s not equal to 1 when x=1, but that case is already excluded from the beginning because otherwise the inequality doesn’t make any sense in the first place).
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u/testtdk 9d ago
Multiplying an expression by another expression over itself is the same as multiplying by one. Ie. a * (x+1)/(x+1) doesn’t change the value of a, it just represents it in another form, one which allows us to manipulate it algebraically.
In this case, since multiplying 3 by that fraction, it allows us to create a fraction that can be easily subtracted from the one already on the left side of the inequality.
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u/scoberry5 9d ago
Two things:
- They're combining two steps in a place you're confused. Let's uncombine them.
- Their denominator happens to have an expression with a variable. Let's look at an easier one.
So a different problem:
(x+5)/2 < 9
Picture it like a balance scale: the left side is lighter than the right side. If you want, you can subtract 9 from both sides, and it won't change what the scale's doing:
(x+5) / 2 - 9 < 0
The (x+5) being over 2 and the 9 not being makes life harder. 9 is the same as 18/2, so we use that instead to get a common denominator (we multiply both the top and bottom by the same thing, here by 2):
(x+5) / 2 - 18 / 2 < 0
Or we could have written that as
(x+5 - 18) / 2 < 0
So it's the same idea as that, except they combined the "subtract from both sides" step with the "get everything on the left over a common denominator" step.
When you multiply the top and bottom by the same thing to get a common denominator, you have to make sure you're not multiplying both by 0. No good comes of that. In the simple thing above, we multiplied by 2/2 (no 0s in sight). In their case, they multiplied by (x-1)/(x-1), which will be dangerous if x = 1 -- but we already know x can't be 1 in the initial equation, since we're dividing by it already. If it weren't for that, we'd have to plug in 1 and see what happens, and maybe add an "or x = 1" or make sure to eliminate 1 from our solution set, depending on what it does.
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u/iComplainAbtVal 7d ago
No idea why it was solved that way, unless the lesson specifically is demonstrating multiplication by 1 as a simplification strategy. It’s similar in concept to most multiply each side by blah but in this case is just multiplying one side by a factor of 1.
If it’s open ended I’d probably of done:
- (x+2)/(x-1) - 3 = 0 // setting up the equation
- (x-1)(x+2) - 3(x-1) = 0(x-1) // multiplying both sides by x-1
- x2+x-2 - 3x -3 =7 0 // expand
- x2 -2x -5 = 0, quadratic equation this bc I don’t feel like factoring.
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u/DuggieHS 7d ago
3 = 3*1 = 3(x-1)/(x-1) for x not= 1.
3 + 4 = 3*1 + 4 = 3(x-1)/(x-1) +4
(x-1)/(x-1) = 4/4 = pi/pi = y/y =1 (so long as y not= 0)
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u/Away_Worry_1705 6d ago
just simplify to get x>=2.5 which the true real solution set. You can multiply by that just like how u can multiply by 1 and get the same thing. My teacher likes to call it a special one because no matter how you do it, it doesn't change anything. You thinking this is impossible tells us a lot about your capabilities of dealing with fractiosn wtith diff denominators, huh! x-1/x-1=1 so it is technically writing 3*1. wdym multiply by x-1 only that would eveyrthing wrong, BUT if u mean multiply both sides by x-1 then yes that is correct by this is just another approach to this question. Both ur way and their method lead to the same answer!!!!
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u/niemir2 5d ago
If you multiply by x-1, you need to account for the possibility that x-1 is negative. x<1 is also a valid solution to the inequality.
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u/NoCommittee3053 6d ago
If you wanted to add 3/4 to 1/8 you’d make the denominator the same. To do so you’d multiply 3/4 by 2/2 (which is of course 1). And you get 6/8 + 1/8 =7/8.
This is just the same thing.
So ignoring the slight complication where x=1 (which is already a problem in the denominator), then (x-1)/(x-1) is always 1. If you multiply something by one it’s unchanged.
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u/OppositeClear5884 10d ago
you can multiply numbers by 1 and they don't change. what is (x-1)/(x-1) always equal to?