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[Basic Algebra] Math/Wolfram
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You are currently reading a thread in /wsr/ - Worksafe Requests

Thread replies: 30
Thread images: 2
Hello. Preferably I'd like a Wolfram Pro ID so I can just reference what I need, but some simple guidance would also be appreciated.

I need to solve the attached. Most of the problems are similar, so if I can have one explained in steps, I think I can do the rest.

I'd use the textbook my wallet is empty.
>>
You can reduce the denominator of the left expression by a factor of two. Now the denominators are equal and you can add the numerators then just cross multiply and solve.
>>
>>49334
I don't know what you mean.
>>
Forgot to add the answer to check your work against, m=5 I'll be on for the next couple of minutes if you have any questions
>>
>>49341
That doesn't help me, I still don't know where to even start. I don't know how to reduce it, or even what to reduce.
>>
2/(8m+16)= 2/(2(4m+8))=1/(4m+8)

So
6/(4m+8) + 2/(8m+16)

6/(4m+8) +1/(4m+8)

7/(4m+8)=1/4
>>
So If your other problems are similar you should look at the denominators (bottom part of the fraction) and try to make them equal each other by reducing or multiplying as in this one you could have gone another route of multiplying the right expression by 2/2
>>
Just realized in all my comments got my lefts and rights confused, sorry
>>
File: wsr.png (721 B, 118x45) Image search: [iqdb] [SauceNao] [Google]
wsr.png
721 B, 118x45
I've got another one I don't really get, but it's different from the first.

I need to find all values of x for which the rational expression is undefined.
>>
>>49366
2x+3 = 0

5x^2+7x-6 =/= 0

solve it
>>
>>49368
Do I set the entire expression =0 or =/=0, or, as you write, numerator to 0 and denominator to =/=0? How can I set two parts of the same expression to different things? Is that not a rule?
>>
>>49368
Denominator can't be equal to 0 in ANY CASE. I don't know how teacher did taught you, but this is absolutely legit. Oh, I thought that you need to solve an inequality...

Fuck the numerator, calculate only
5x^2+7x-6 =/= 0
>>
>>49370
In this case how do I isolate X?
>>
>>49370
And I forgot to mention "Why?". Because denominator can't be 0 and these values when denominator is 0 aren't legit
>>
>>49371
Simple quadratic equation, just put everywhere =/= instead of =. And these numbers will be the ones which are not suitable for equation. So the answer will be all numbers except these two x's
>>
>>49373
Oh, hangover kicks in... It says undefined value... Then the only answer will be these two x's
>>
when I set denominator to =/=0 and solved, and plugged in quadratic equation to solve for x, I found x=0.395445115 and 1.795445115. I've only plugged in the first value so far, and when I solved I got -2.45=/=0.
>>
>>49375
unless I was supposed to include numerator this time too, in which case I have -1.547302135... both are still unusual to me.
>>
>>49376
alright buddy, see if this helps. with >>49366

whenever a fraction is something over zero (y/0 for example) its undefined. Thus, for the example we are finding all values where the denominator is = to zero. You can do this a variety of ways

First, you could use the quadratic equation to find roots for the denominator. This will neatly solve all your answers in one application.

Otherwise, we can factor the denominator, as its not a non-reduceable quadratic
I belive (5x -3) (x +2) gives the expression.

Then we solve EACH of these quantities for =0 (5x-3 = 0 and x + 2 = 0) this is because if either is 0, the whole denominator is 0, and thus the entire expression is undefined.
>>
when I solve for x using
5x+3=0 and x+2=0 {3/5;-2}
and plug in my answers for X to check, I get 7/22=0 and 4 and 1/5=0, respectively. This isn't right, isn't it?
>>
>>49394
ya fucked up a bit. look at that first answer. its clearly not 3/5, but its quite close (-3/5, remember, you're subtracting 3)

for -2 you should get zero. (7*-2 = -14 ; -14-6 = -20; -2^2=4, 4*5=20, 20-20 = 0, numerator is some value /0, therefore undefined)
>>
>>49399
ah, i fucked up a bit here. It is 3/5 (non negative) and it does result in some value over 0, and is thus undefined
>>
>>49402
also, because its a quadratic expression (5x^2 +7x-6), it can only have two real roots, therefore these are the only cases where the entire rational expression can be undefined.
>>
>>49330
Ok look
6
-
4m+8

6 is the numerator, (4m+8) is the denominator. Fractions with the same denominator can we added and subtracted. Fractions can cancel down if both numerator and denominator are divisible by the same number.

So look at the next fraction
2
-
8m+16

2/2=1
8m is 8*m, 8/2=4, so 8m/2=4m
16/2=8

so
2
-
8m+16

is reduced to

1
-
4m+8

you now add it to the previous fractions
so
[6 OVER 4m+8] PLUS [1 OVER 4m+8] EQUALS
7
-
4m+8

so now

[7 OVER 4m+8] EQUALS [1 OVER 4]

Multiply both sides by 4 to get rid of the denominator 4 on the right-hand side.

[7 OVER 4m+8] TIMES 4 = [28 OVER 4m+8]
[1 OVER 4] TIMES 4 = 1

so now [28 OVER 4m+8] = 1
[28 OVER 4m+8] can be cancelled down, divide numerator and denominator by 4

now equals

7
-= 1
m+2

now multiply both sides by (m+2) to get rid of the denominator on left side

so

7 = m+2

subtract 2 from both sides

m = 5

that is your answer
>>
>>49330
Holy fuck, if you're struggling with basic shit speak to your teacher/lecturer/tutor whatever, you're missing basic logic that is far easier explained in person in a way YOU can understand rather than getting use to solve random examples for you in the most awkward way.
>>
Math Idiot here. No this is not bait.

I'm taking an online astronomy class for fun and am working my way through an equation. I am having trouble with how

z a(t) = 1 - a(t)

can become

(1+z) a(t) = 1

The lecture says "move this [a(t] over" but when I do

start z a(t) = 1 - a(t)
subtract a(t) from both sides: z a(t) a(t) = 1
simplify: z 2a(t) = 1

Which is not correct.

Can someone please walk me through this? I am so close to the end and so sick of this equation. Thank you.
>>
>>49517
>start z a(t) = 1 - a(t)
>subtract a(t) from both sides: z a(t) a(t) = 1
>simplify: z 2a(t) = 1

you are retarded
>>
>>49517
here is how it is:

z*a*t = 1 - a*t
(add a*t to both sides)
z*a*t + a*t = 1

a*t is equal to 1*a*t
now by simplifying it
z*a*t + a*t = 1
becomes
(z+1)*(a*t) = 1

in algebra, we do not usually leave in the multiplication sign as this makes it needlessly confusing, algebra is manipulating TERMS. A TERM is anything separated by a + or -. It doesn't matter about multiplication or division.

it's easier if you do it without parentheses or * or x signs.

zat = 1 - at
zat + at = 1
(z+1)(at) = 1
>>
>>49525

Thanks for trying.

a(t) is a scale factor as a function of time in astronomy and seems to be treated as a single calculated value which is why I never split it.

>>49518

Thanks for pointing out my typo.
>>
>>49542
>z a(t) = 1 - a(t)

This means z times a(t) = ...
The multiplication sign is omitted if it is not needed.

Add a(t) to both sides

z a(t) + a(t) = 1

You can read the second summand as 1 times a(t).

Now you count how many times a(t) is present in the left side. z+1. This gives

(z+1) a(t) = 1.
Thread replies: 30
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