• Welcome to Religious Forums, a friendly forum to discuss all religions in a friendly surrounding.

    Your voice is missing! You will need to register to get access to the following site features:
    • Reply to discussions and create your own threads.
    • Our modern chat room. No add-ons or extensions required, just login and start chatting!
    • Access to private conversations with other members.

    We hope to see you as a part of our community soon!

Math Question

Debater Slayer

Vipassana
Staff member
Premium Member
Alternatively

5 - 12x = 8 - 7x - [6/(3 (2+5³) + 5x)]

5 - 12x = 8 - 7x - [6/(3 (127) + 5x)]

5 - 12x = 8 - 7x - 6/(381 + 5x)

5(381 + 5x) - 12x(381 + 5x) = 8(381 + 5x) - 7x(381 + 5x) - 6

1905 + 25x - 4572x - 60x^2 = 3048 + 40x - 2667x + 35x^2 - 6

1905 -4547x - 60x^2 = 3042 - 2627x +35x^2

1905 - 3042 - 4547x + 2627x - 60x^2 - 35x^2 = 0

-1137 - 1920x -95x^2 = 0

1137 + 1920x + 95x^2 = 0

Then I gave up :)

You have to put the equation in the correct formula:

aX² + bX + c = 0

Which would be 95X² + 1920X + 1137 = 0

Then factorize them (not sure if it's possible with these particular numbers, though).

5-12x = 8-7x-[6/3(2+5^3)+5x]

5-12x+7x = 8-[6/3(127)+5x]

5-5x = 8-[254+5x]

5-8-5x = -254+5x

-3-5x = -254+5x

-5x = -254+3+5x

-10x = -251

x = 25.1

Knowing me there's a glaring fault in there somewhere, but *shrugs* [/2 cents].

There. :D It should have been -254-5X (the -ve sign would change everything inside the bracket into a negative). Fell for the same trick before myself. XD
 
Last edited:

9Westy9

Sceptic, Libertarian, Egalitarian
Premium Member
You have to put the equation in the correct formula:

aX² + bX + c = 0

Which would be 95X² + 1920X + 1137 = 0

Then factorize them (not sure if it's possible with these particular numbers, though).

I know. I just couldn't be bothered to work it out xD
 

9Westy9

Sceptic, Libertarian, Egalitarian
Premium Member
5-12x = 8-7x-[6/3(2+5^3)+5x]

5-12x+7x = 8-[6/3(127)+5x]

5-5x = 8-[254+5x]

5-8-5x = -254+5x

-3-5x = -254+5x

-5x = -254+3+5x

-10x = -251

x = 25.1

Knowing me there's a glaring fault in there somewhere, but *shrugs* [/2 cents].

found it. Always the negative signs.
 

Phasmid

Mr Invisible
I see so if you have, for example, -[10+4x] you effectively have -1[10+4x] and since you have to expand the brackets by multiplying all the internal terms by the external ones you get -10-4x?
 

Debater Slayer

Vipassana
Staff member
Premium Member
I see so if you have, for example, -[10+4x] you effectively have -1[10+4x] and since you have to expand the brackets by multiplying all the internal terms by the external ones you get -10-4x?

Basically, yes. The 10+4X inside can't stay that way with the +ve sign still in between the two terms after eliminating the brackets.
 

cablescavenger

Well-Known Member
Alright, this is very similar to what I've been doing lately, so I'll try to explain it as simply as possible:

In equations like this, you have to follow the order of operation BODMAS (Brackets Orders Division Multiplication Addition Subtraction) as shown here. First deal with any unsolved brackets, then exponents/roots, then divisions, then multiplications, then additions, then finally, subtractions.

Now I'll solve this for you as an example:

5 - 12X = 8 - 7X - [6/3 (2+5³) + 5X]

5 - 12X = 8 - 7X - [2 (2+125) + 5X]

5 - 12X = 8 - 7X - [4 + 250 + 5X]

Now shift all the constants to the L.H.S (or left hand side), and all the variables to the right hand side (or R.H.S):

5 - 8 = -7X + 12X - [4 + 250 + 5X]

-3 = 5X - [254 + 5X]

Now here's how you eliminate those brackets:

-3 = 5X - 254 - 5X (the previous negative sign changed the signs into -ve ones for the entire bracket function).

Now cancel out the 5X along with the -5X and you get:

-3 = -254

Shift all the constants once again to the L.H.S:

-3 + 254 = 0

Therefore you get: 251 = 0

Therefore the answer is refused or false.

Yes, I'm not bluffing. :D And to be even more sure of it, just copy & paste your equation here for the solution (which will be false too ;)).

Hope this helped. :)

That looks good to me :drool:
 

Debater Slayer

Vipassana
Staff member
Premium Member
Odd, I don't think I've ever been given a question that results in 0 = anything.

The formula aX² + bX + c = 0 is generally used in solving quadratic equations. In any given equation, you can reduce something to = 0, even if it's not a quadratic one. Example:

2 + 1 = 3

In order to equate it to zero, we shift the constant we have in this case into one side with the others and change its sign:

2 + 1 - 3 = 0

Following that up:

3 - 3 = 0

0 = 0.
 

Phasmid

Mr Invisible
The formula aX² + bX + c = 0 is generally used in solving quadratic equations. In any given equation, you can reduce something to = 0, even if it's not a quadratic one. Example:

2 + 1 = 3

In order to equate it to zero, we shift the constant we have in this case into one side with the others and change its sign:

2 + 1 - 3 = 0

Following that up:

3 - 3 = 0

0 = 0.

I meant specifically that an integer other than zero = 0.
 

Koldo

Outstanding Member
5(381 + 5x) - 12x(381 + 5x) = 8(381 + 5x) - 7x(381 + 5x) - 6

1905 + 25x - 4572x - 60x^2 = 3048 + 40x - 2667x + 35x^2 - 6

1905 -4547x - 60x^2 = 3042 - 2627x +35x^2

Correction:

1905 + 25x - 4572x - 60x^2 = 3048 + 40x - 2667x - 35x^2 - 6
1905 -3048 + 6 +25x - 4572x -40x + 2667x -60x^2 + 35x^2 = 0
-1137 - 1920x - 25x^2=0
25x^2 + 1920x + 1137 = 0
 

LegionOnomaMoi

Veteran Member
Premium Member
Eek. You can't do this.
6/(381+5x) is NOT equal to 6/381 + 6/5x.
I will explain why. You may correct me if , for some reason, i am wrong.

In addition to (not instead of) an example, sometimes rewriting the expression also helps to understand the issue.
6/(381+5x)=6 * 1/(381+ 5x)= 6 * (1/381+ 1/5x)= 6(1/381) + 6(1/5x).

One of my favorite singular variable calculus textbooks (Calculus by Spivak), designed for students with some background in calculus (e.g., a high school course), begins with a chapter on addition, subtraction, multiplication, and other basic operations. He also specifically states that the chapter "is not a review of old material" despite the fact that everyone who has or will read the book would have been exposed to addition years and years ago. The reason it's so great is because it covers the concepts behind the operations, and uses simple ones to define others. Another book I encourage students to buy as a supplement is The Calculus Direct which also starts with arithmetic, and explains variables and why certain operations are "allowed" or "not allowed" using vectors (graphically displayed as line segments on a number line with all the numbers other than 0 removed).
 

Shadow Wolf

Certified People sTabber
The answer is an unreal number, the zero with a / through it.
[6/3(2+5^3)+5x]
that is six divided by three. I don't know if there is a way to insert a division symbol.
Has anyone seen Poly?
 

Draka

Wonder Woman
If it is 6 divided by 3 then times 2 plus 5 cubed then plus 5x, then the answer earlier of x=25.1 would seem to be right. If it is 6 divided by the whole equation of 3(2+5^3)+5x then it appears to be 25x^2 + 1920x + 1137 = 0 At least those are the end results I'm getting.
 

Koldo

Outstanding Member
The answer is an unreal number, the zero with a / through it.
[6/3(2+5^3)+5x]
that is six divided by three. I don't know if there is a way to insert a division symbol.
Has anyone seen Poly?

Just by 3?
Then:
5-12x=8-7x-[6/3(2+5^3)+5x]
5-12x=8-7x-[6/3(2+125)+5x]
5-12x=8-7x-[2(127)+5x]
5-12x=8-7x-[254+5x]
5-12x=8-7x-254-5x
-12x+7x+5x=8-254-5
0x=-251 OR 0=-251

Case solved.
Weird result? Yep. But that is how it is.
 
Last edited:

methylatedghosts

Can't brain. Has dumb.
I have a math problem on my homework and there are two parts I'm not sure what to do with.
5-12x=8-7x-[6/3(2+5^3)+5x]
What I'm not sure how to do is do I multiple 5 to the third, then that number times 3, or 3*5, and then that number (15) to the third? I'm also not sure how to simplify with a division sign.

Ok, I'll have a go... it's been a while, using BEDMAS...

5-12x=8-7x-[6/3(2+5^3)+5x]
Firstly, you have brackets within brackets. Solve within brackets first, as though it's a separate equation. So first to solve is
6/3(2+5^3)+5x
There is a bracket in that equation, so we need to solve for that one first, leaving
2+5^3
Now that brackets have been eliminated, solve the Exponents before Addition (again, refer to bEdmAs
5^3 = 125
2+125=127
So here we've solved the first bracket, so we can replace 2+5^3 with 127:
6/3*(127)+5x
Now, with that, the first thing to do, according to BEDMAS is Division, so.... 6/3 = 2
Giving us 2(127)+5x
Then, it's multiplication
2*127=254
Giving us 254+5x
As we have no value for x, this is as far as we can go with this portion, so put that back into original equation...

5-12x=8-7x-(254+5x)

Now, to multiply out (the section -(254+5x) can also read -1*(254+5x))
5-12x=8-7x-254-5x

This is as far as I'm able to take it... maybe someone else can go further, but I have a hunch that you're not going to be able to solve for x. The problem is the 5x. If that x hadn't been there, I think it'd be solvable. I don't think there's any number you could replace x with, and the sides of the equation remain both positive, or both negative, to begin with. I think any number would end up being something like (super-simplified example) x = -x, which is of course, impossible.

EDIT: Alternatively, if it had read -5x, the - outside of the larger bracket would have made that into a positive, and made the equation solvable
 
Last edited:

Koldo

Outstanding Member
Now, to multiply out (the section -(254+5x) can also read -1*(254+5x))
5-12x=8-7x-254-5x

This is as far as I'm able to take it... maybe someone else can go further, but I have a hunch that you're not going to be able to solve for x.

The next step is to take the x's to one side, and the pure numbers to the other, such as every x to the left and every pure number to the right. Paying attention to invert their signals when performing the exchange of sides.

-12x+5x+7x=8-254-5

Basic math here, and the result is:

0x = -251

Considering that any number multiplied by 0 is equal to 0, then it is same as saying that:

0 = -251
 

methylatedghosts

Can't brain. Has dumb.
The next step is to take the x's to one side, and the pure numbers to the other, such as every x to the left and every pure number to the right. Paying attention to invert their signals when performing the exchange of sides.

-12x+5x+7x=8-254-5

Basic math here, and the result is:

0x = -251

Considering that any number multiplied by 0 is equal to 0, then it is same as saying that:

0 = -251

Of course...

So really x cannot be said to be a specific number, yes? So x = ∞ ?
 

Shadow Wolf

Certified People sTabber
So really x cannot be said to be a specific number, yes? So x = ∞ ?
X ≠ ∞ though, as when you add 12x to -12x it comes out to 0, which makes the answer written as Ø.

I do have to admit though, I did not realize that for the 3(2+5³) that you could add whats in the parenthesis first, then multiply the number by 3. I was using the distributive property for that part, which didn't make too much of a difference other than adding an extra step.
 

Debater Slayer

Vipassana
Staff member
Premium Member
The answer is an unreal number, the zero with a / through it.
[6/3(2+5^3)+5x]
that is six divided by three. I don't know if there is a way to insert a division symbol.
Has anyone seen Poly?

You mean like this ÷? :D

Here you'll find all the maths symbols you need.

Still curious though, what's the right answer to that equation according to your teacher/textbook?
 
Top