Is there an easier way to solve this rational equation? Reddit r/askmath

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Күн бұрын

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@bprpmathbasics 2 ай бұрын

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@jyotsanabenpanchal7271 2 ай бұрын

Plz make a vid on stolz Cesaro theorem

@kennethgee2004 2 ай бұрын

so this might be a case of get a common denominator on each side instead of the entire thing. We can then use the fact a rational expression is equal when its cross product are equal. This should make everything in terms of a quadratic.

@Ninja20704 2 ай бұрын

It may be slightly easier to combine the fractions on the LHS and RHS individually and then cross multiply. It would be easier on the algebra in terms of simplifying the equation afterwards.

@incircIe 2 ай бұрын

Yeah I also figured that out 😅

@OptimusPhillip 2 ай бұрын

That's what I did

@florianbasier 2 ай бұрын

exactly. I didn't want to bulldoze my way into a cubic equation from line 1, and the symmetry helped a lot since you get -x on both side's numerators. which means that x=0 or that the denominators are equal, and you are down to a binomial which simplifies into a very simple linear one.

@quantumbuddha777 2 ай бұрын

That's what I did too. Kept it pretty simple.

@yahoo532511 2 ай бұрын

Try to get the common denominator of each side, and multiply by(-1): 2(X+1)-(X+2)/(X+1)(X+2) = 4(X+3)-3(X+4)/(X+3)(X+4) and that is : X/(X+1)(X+2) = X/(X+3)(X+4) There's an obvious answer for this eq. as X=0 # If X is NOT 0, then (X+1)(X+2)=(X+3)(X+4) → 3X+2=7X+12 therefore X= -5/2 #

@Misteribel 2 ай бұрын

Yeah, that's how I started out. It simplified quite cleanly and quickly. Nice!

@ianfowler9340 2 ай бұрын

It's not really any better, but let y = x+2. You will then get a -y + 2 in the num of both sides. This gives y = 2 ==> x = 0. Cancel the numerators and cross multiply: 3y + 2 = -y ==> x = -5/2. As you said, too bad the middle is 2.5. If you had an whole number average then this method would be much cleaner.

@ausaramun 2 ай бұрын

If we take the mean of 1, 2, 3, and 4, we end up with 10/4.

@henryucha2705 2 ай бұрын

Another was is to distribute -1 two times to both sides of the equation to make each numerator 1 and -1 I.e 1/(x+1)-1 -2/(x+2) -1 For the left hand side and do the same for the right hand side Resulting to -1/x+1+1/x+2 =-1/x+3+1/x+4 Substitute y=x+1 We get -1/y+1/y+1=-1/y+2+1/y+3 Again this results to -1/y(y+1)=-1/(y+2)(y+3) (Y+2)(y+3)=y(y+1) Y²+5y+6=y²+y 4y=-6 Y=-3/2 X+1=-3/2 X=-5/2

@GillesF31 7 күн бұрын

May I suggest using substitution variables which lightens the calculations and number of steps without skipping any? 1/(x + 1) - 2/(x + 2) = 3/(x + 3) - 4/(x + 4) = ? set: • a = x + 1 • b = x + 2 • c = x + 3 • d = x + 4 1/a - 2/b = 3/c - 4/d b/(ab) - 2a/(ab) = 3d/(cd) - 4c/(cd) (b - 2a)/(ab) = (3d - 4c)/(cd) note: • b - 2a = (x + 2) - (2x + 2) = -x • 3d - 4c = (3x + 12) - (4x + 12) = -x • ab = (x + 1)·(x + 2) = x² + 3x + 2 • cd = (x + 3)·(x + 4) = x² + 7x + 12 => = -x/(x² + 3x + 2) = -x/(x² + 7x + 12) = -x·(x² + 7x + 12) = -x·(x² + 3x + 2) = -x³ - 7x² - 12x = -x³ - 3x² - 2x = -x³ - 7x² - 12x + x³ + 3x² + 2x = 0 = -x³ + x³ - 7x² + 3x² - 12x + 2x = 0 = -4x² - 10x = 0 = -2x·(2x + 5) = 0 => • solution #1: -2x = 0 => x = 0 • solution #2: 2x + 5 => x = -5/2 --- /// final results: ■ x = 0 ■ x = -5/2 🙂

@nicholasscott3287 2 ай бұрын

If anyone cares, the result of substituting x=-2.5 into the equation is 3 1/3...=3 1/3.... On the left hand side you get -2/3+4 and on the right hand side you get 6-(2 2/3).

@ElderEagle42 2 ай бұрын

Who else wants this guy as you math teacher?

@orenfivel6247 2 ай бұрын

Do the overkill: since x=0 is a sol, multiply and divive by x and do partial fraction decomposition to each a/(x(x-a))

@iabervon 2 ай бұрын

The answer to the original question is that, in order to divide by -x, you need to consider whether x is 0. If it is, you've found the x=0 solution, so you should write that solution down. If it's not, you can divide by it, and you get x=-5/2. In this case, both options are possible, so you get those two solutions. In general, whenever you want to cancel something from both sides, and x value that makes that something 0 is a solution that you need to recognize, but you can cancel the something once you've addressed the fact that you're now only looking for other solutions besides the ones you've found.

@koenth2359 2 ай бұрын

I just went ((x+2)-2(x+1))/(x+1)(x+2) = (3(x+4)-4(x+3))/(x+3)(x+4) -x/(x^2+3x+2) = -x /(x^2+7x+12) x=0 or x^2+3x+2 = x^2+7x+12 x=0 or 0=4x+10 x=0 or x=-5/2

@gibbogle 2 ай бұрын

Ditto.

@Metaverse-d9f 2 ай бұрын

use (+x-x) on top of every term, simplify to x/(x+1)(x+2)=x/(x+3)(x+4), say x=0 is a sol. and solve the rest linear equation(though it looks like a quadratic)

@YTUBE508 2 ай бұрын

When you have -x(x+3)(x+4) = -x(x+1)(x+2), could you not divide by -x on both sides and then multiply out the two binomials on both sides? I know technically since x = 0 is a solution you are dividing by 0, but if you note down x ≠ 0, would you then be allowed to divide by -x?

@DotRabbit 2 ай бұрын

I wrote it very complicatedly like [1/(x+1)] - [2/(x+2)] = [3/(x+3)] - [4/(x+4)] then i transposed the negative in the two sides we get [1/(x+1)]+[4/(x+4)] = [3/(x+3)] + [2/(x+2)] then i added them up and got (5x+8)/(x^2+5x+4) = (5x+12)/(x²+5x+8) then i transposed the denominators into LHS and RHS sides then i got 5x³+33x²+70x+48 = 5x³+37x²+80x+48 then we get 4x²= -10x then 4x²+10x = 0 2x(2x+5)=0 2x = 0 ; x =0 2x+5=0 ;2x= -5; x= -5/2

@FFSytstoptryingtobetwitter 2 ай бұрын

That first step looks like it'd be intimidating for someone new to solving rational equations. I think a more inviting technique is to multiply each of the 4 fractions by 1 written in a careful way, since each of the 4 steps is very easy and this is a common technique in all sorts of math. I'd maybe go as far as to say that this problem is meant to illustrate that technique. So you'd multiply 1/(x+1) by (x+2)/(x+2) and do 2/(x+2) times (x+1)/(x+1). Since we're already assuming neither x+1 nor x+2 is 0 for this equation to make sense, each of these is just multiplying by 1 so it won't change our inequality. We do the same with 3/(x+3) times (x+4)/(x+4) and and 4/(x+4) * (x+3)/(x+3) on the right side. Then we can simplify each side before we multiply by the common denominator (x+1)(x+2)(x+3)(x+4). It just makes all the arithmetic much much simpler and we arrive at the same quadratic you solve in the back half of your video.

@nicholasscott3287 2 ай бұрын

There are also solutions of 0 = 0 at + or - infinity which can be easily seen by taking the limit.

@bobh6728 2 ай бұрын

The only real solutions are the 0 and -5/2. Infinity is not a real number. Including the infinities is usually called the extended real numbers, which is very similar to taking limits as you said.

@meirhdjduehehd 2 ай бұрын

Can you please explain the meaning of implies (=>) , implies and implied by () and iff ( if and only if )in mathematics Please.

@danielemacheda4493 2 ай бұрын

a==>b (a implies b) means that b is a consequence of a ( you can prove that b is true starting from a). This might be a bit of an advanced concept but note that a doesn't have to be true; in fact you can prove virtually anything starting from a false proposition, even a true one. aNOTb (NOTa implies NOTb) which means that the negation of a implies the negation of b (negation is meant here as the negation of a proposition). Proving this second==> is equivalent to prove b and b==>a are true at the same time. To prove this you must prove both ==>. Let me know if you have any questions.

@gibbogle 2 ай бұрын

That's an easy way? (1(x+2) -2(x+1))/((x+1)(x+2)) = (3(x+4) - 4(x+3))/((x+3)(x+4)) -x/((x+1)(x+2)) = -x/(x+3)(x+4)) therefore x = 0 or (x+1)(x+2) = (x+3)(x+4) x^2 + 3x + 2 = x^2 + 7x + 12 4x = -10, x = -5/2.

@TurtleGod2 2 ай бұрын

Can't you just do (x+3)(x+4)=(x+1)(x+2) and solve from there because you could divide both sides by -x

@ellelawliet8977 2 ай бұрын

I'd Call x+2,5=y and do the sub. It will get an easy semplification

@avinandan2008 2 ай бұрын

1/x+1 - 1/x+2 = 3/x+3 - 4/x+4 => x+2-2x-2/(x+1)(x+2) = 3x+12-4x-12/(x+2)(x+4) => -x/(x^2+7x+12) = -x/(x^2+3x+2) => 1/(x^2+7x+12)= 1/(x^2+3x+2) => -5x=10 => x=-10/4 Thus, x is equal to negative 5 devided by 2. I think this process is much easier than yours!

@RatolokoMemo 2 ай бұрын

I generalized this problem, so if: (n/x+n) - (n+1/x+n+1) = (n+2/x+n+2) - (n+3/x+n+3) the solutions are: x = 0 and x = -(2n+3)/2 in this case n=1 which gives x=0 and x=-5/2, so its just a coincidence the appearance of the next number (5), this just happens when n+4 = 2n+3, which gives n=1

@JikoAmonus 2 ай бұрын

This is very interesting

@ChengxiHu-e1u 2 ай бұрын

I understood all but the first step in which he tranforms the fractions into their denominators, can someone explain?

@emyrronain6983 2 ай бұрын

It is an equation so multiplying both sides doesn't change the equation. If you have multiplication of more than one rational expression, multiplying by multiplication of all the denominators will give you the expression without division (a/b)+(c/d)=f can be written as; [(b)(d)][(a/b)+(c/d)]=f(b)(d)= (d)(a)+(b)(c)=f(b)(d)

@Pseudify 2 ай бұрын

It might make more sense if you do it in smaller steps. Multiply everything by only (x+1) at first. Then multiply by x+2. And then x+3 and finally x+4. He’s just combining all four of those steps into one.

@Gunslinger-us1ek 2 ай бұрын

u can do it using GP 1/(1+n) = 1/n-1/n^2+1/n^3-... input n=x, x/2, x/3, x/4 and add/subtract according to the question. simplify the ugly expression and boom they are equal for x=0

@robertveith6383 2 ай бұрын

Write sentences.

@Gunslinger-us1ek 2 ай бұрын

@@robertveith6383 i dont wanna write

@BladeMasterz9801 2 ай бұрын

Too much algebra. There is a pattern to all of the fractions, which is a/(a+b) = 1/(1+b/a). So you can write both sides as following: 1/(1+1/x) - 1/(1+2/x) = 1/(1+3/x) - 1/(1+4/x) => (1/x)/[(1+1/x)(1+2/x)] = (1/x)/[(1+3/x)(1+4/x)] => (1+1/x)(1+2/x) = (1+3/x)(1+4/x) (*) Recall that (1+a)(1+b) = 1+a+b+ab Then (*) becomes: 1+3/x+2/x² = 1+7/x+12/x² => 4/x+10/x² = 0 => 4/x(1+5/2x) = 0 => x = -5/2. This is a beautiful problem indeed.