Math Question

[Shadow Wolf]

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.

 

[Shadow Wolf]

I do have the answer, but it's those steps that I'm not sure if I have them correct.
What I done was:
-[6/3(2+5^3)+5x]
-[6/6+375+5x]
-1-375-5x

 

[Gharib]

wow good thread, haven't done any math for years, i'll give it a try

 

[Gharib]

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

the red bit is 6 over 3 yeah

 

[Gharib]

hope what i have is right:

5-12x=8-7x-[6/3(2+5^3)+5x]
5-12x = 8-7x-[254+5x]
5-12x=8-7x-254-5x
5-8+254=12x-5x
251=7x

 

[Gharib]

oops my bad now i see what you mean, let me think a little more to make sure i get it right this time

 

[Gharib]

you can't multiply the 3 under the 6 with the numbers in the round brackets () because the 6/3 is already being multiplied with them, to change the 3 from under the 6 you need to get it on the other side of the equation that is the only time you can change one number from being divided into it being multiplied. your equation should look like this:

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

i'm guessing you don't have a formula on how that stuff works, it would make it easier.

 

[LegionOnomaMoi]

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.

One problem with how you've written the equation is the division symbol. Order of operations in algebra depends on methods of grouping. If an entire expression is written under the fraction/division line (i.e., 6/[an algebraic expresion]), then everything operation under the division line has precedence over the division itself. More explicitly, if I write 8/2 +3, then this is equivalent to 4 +3. However,
8
________
2+3

where that line represents division means that the expression is equivalent to 8/5, because addition has precedence. So it's hard to answer your question without knowing whether the entire expression 3(2+5^3)+5x is written under 6/.
If it is, then yes, you would distribute the 3 and get (6 + 75). Exponents take precedence over multiplication unless specific grouping symbols (brackets, parentheses), are used, so you don't multiply 3*5 until you have raised 5 to the 3rd.

It's helpful if you keep in mind what all these operations mean. You can define virtually all algebraic operations in terms of addition, or an inverse of an operation defined by addition. 5^3 means 5*5*5 which means (5+5+5+5+5)+(5+5+5+5+5)+(5+5+5+5+5). Division is the inverse of multiplication. a/b=c is the number b such that a=bc.

This becomes important when trying to figure out how you can manipulate algebraic equations. For example, the left side of your equation reads 5-12x. The reason this is NOT the same as -2x is because multiplication is a shorthand for addition. 12x means adding the amount "x" represents to itself 12 times. However, you don't know what this amount is. Whatever it is, there is a big difference between adding 12 of this amount "x", and adding -2 of "x."

 

[Koldo]

If it is, then yes, you would distribute the 3 and get (6 + 75).

???

It's helpful if you keep in mind what all these operations mean. You can define virtually all algebraic operations in terms of addition, or an inverse of an operation defined by addition. 5^3 means 5*5*5 which means (5+5+5+5+5)+(5+5+5+5+5)+(5+5+5+5+5).

Actually, your explanation is way off.
That is because 5^3=125, and NOT 75.
5*5=25
25*5=125

 

[LegionOnomaMoi]

???



Actually, your explanation is way off.
That is because 5^3=125, and NOT 75.

You are absolutely correct. I was probably concentrating on the three (25*3). I've been up for over 24 hours so these kind of slips tend to happen (it really messed me up on an exam in linear algebra some time ago).

But thankfully calculators can correct the type of mistake I made. The bigger issue (I think) in the OP concerns order of operations and how to determine these. I have tutored/taught high school kids since I was an undergrad, and while they are all familiar with PEMDAS (order of operations), they get confused when addition takes precedence over division despite the abscence of grouping symbols.

I find that in addition to explaining grouping "rules," some students benefit from understanding why they exist in the first place and what the operations involved in high school/college algebra (as opposed to abstract algebras, boolean algebra, and so on) really mean. That way, if one forgets a particular rule (like whether one "adds" exponents with the same base when the exponent is outside of parentheses or multiplies them), one can derive the rule.

 

[Gharib]

One problem with how you've written the equation is the division symbol. Order of operations in algebra depends on methods of grouping. If an entire expression is written under the fraction/division line (i.e., 6/[an algebraic expresion]), then everything operation under the division line has precedence over the division itself. More explicitly, if I write 8/2 +3, then this is equivalent to 4 +3. However,
8
________
2+3

where that line represents division means that the expression is equivalent to 8/5, because addition has precedence. So it's hard to answer your question without knowing whether the entire expression 3(2+5^3)+5x is written under 6/.
If it is, then yes, you would distribute the 3 and get (6 + 75). Exponents take precedence over multiplication unless specific grouping symbols (brackets, parentheses), are used, so you don't multiply 3*5 until you have raised 5 to the 3rd.

It's helpful if you keep in mind what all these operations mean. You can define virtually all algebraic operations in terms of addition, or an inverse of an operation defined by addition. 5^3 means 5*5*5 which means (5+5+5+5+5)+(5+5+5+5+5)+(5+5+5+5+5). Division is the inverse of multiplication. a/b=c is the number b such that a=bc.

This becomes important when trying to figure out how you can manipulate algebraic equations. For example, the left side of your equation reads 5-12x. The reason this is NOT the same as -2x is because multiplication is a shorthand for addition. 12x means adding the amount "x" represents to itself 12 times. However, you don't know what this amount is. Whatever it is, there is a big difference between adding 12 of this amount "x", and adding -2 of "x."

holy smokes, i have no idea what you're saying :faint:. it seriously has been a long time. i need to study math, how did i get so bad. :slap:

 

[Debater Slayer]

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. 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.

 

[Koldo]

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):

This is rather interesting.
I have always been taught to do the inverse.
Not that it matters .

 

[9Westy9]

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.

is it 6/3 * (2+5^3)+5x or 6/(3*(2+5^3)+5x)?

 

[Koldo]

You are absolutely correct. I was probably concentrating on the three (25*3). I've been up for over 24 hours so these kind of slips tend to happen (it really messed me up on an exam in linear algebra some time ago).

Happens to everyone once in a while.

So it's hard to answer your question without knowing whether the entire expression 3(2+5^3)+5x is written under 6/.

By the way, i forgot to quote this part of your post.
I wanted to know the same thing before telling her how to solve the problem.

IF the problem was handed out to her exactly like that then most likely only the 3 is under the 6. Because otherwise the problem should be written as: 5-12x=8-7x-{6/[3(2+5^3)+5x]} if the entire 3(2+5^3)+5x is below 6, or 5-12x=8-7x-{6/[3(2+5^3)]+5x} if 3(2+5^3) is below 6.

BUT, considering the problem could be written differently in its actual form ( which would be highly recommended considering how misleading this sort of thing can be ), it is important to know what is really under the 6.

 

[9Westy9]

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. 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.

based on this I'll assume the answer to my question. Time to get to work then.

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

7x - 12x -6/5x = 8 - 6/381 -5

-6/5x -5x = 3 - 6/381

-6/5x -25/5x = 1143/381 - 6/381

-31/5x = 1137/381

-2362 1/5 x = 1137

-11811x = 5685

-x = 2.08 (2DP)

 

[9Westy9]

based on this I'll assume the answer to my question. Time to get to work then.

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

7x - 12x -6/5x = 8 - 6/381 -5

-6/5x -5x = 3 - 6/381

-6/5x -25/5x = 1143/381 - 6/381

-31/5x = 1137/381

-2362 1/5 x = 1137

-11811x = 5685

-x = 2.08 (2DP)

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

 

[Koldo]

based on this I'll assume the answer to my question. Time to get to work then.

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

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.

A simple example to show this:

6/(4+x)=1
x=2

Simple replacement of x for 2:

6/(4+2) = 1
6/6 = 1

Considering the simple equation above, if i were to do the same thing you did in your post it would lead to :

6/(4+x)=1
6/4 + 6/x = 1

Knowing that x=2, if we replace the x with 2 we end with:

6/4 + 6/2 = 1
3/2 + 6/2 = 1
9/2 = 1
4,5 = 1 :areyoucra

 

[9Westy9]

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.

A simple example to show this:

6/(4+x)=1
x=2

Simple replacement of x for 2:

6/(4+2) = 1
6/6 = 1

Considering the simple equation above, if i were to do the same thing you did in your post it would lead to :

6/(4+x)=1
6/4 + 6/x = 1

Knowing that x=2, if we replace the x with 2 we end with:

6/4 + 6/2 = 1
3/2 + 6/2 = 1
9/2 = 1
4,5 = 1 :areyoucra

oh xD. Thanks for that. I forgot. It's been too long since I've done any involved maths xD

 

[Phasmid]

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].