Basic Algebra Question (2 Viewers)

[Katsumi]

Hi All,

I've been touching up on my algebra in preparation for university and have found a knowledge gap due to me never learning a concept.

Could someone please explain to me why -125/-8 became positive? Image below

Thanks in advance

 

[Shadowdude]

factor out -1 from both -125 and -8 and cancel it out

 

[Green Yoda]

Hi All,

I've been touching up on my algebra in preparation for university and have found a knowledge gap due to me never learning a concept.

Could someone please explain to me why -125/-8 became positive? Image below

Thanks in advance

Firstly, When you had -125/-8 on LHS the minus and minus cancel out to become positive. Its basically -1/-1 multiplied by 125/8 and when you have two of the same numbers on the numerator and the denominator they yield to = 1.

Now you have 125/8=1/r^3
Multiply 8 to the other side
∴ 125=8/r^3
Multiply r^3 to both sides
∴125*r^3=8
Divide 125 to both sides
∴ r^3=8/125

 

[Katsumi]

I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

So fundamentally when you have r^-3 = -125/-8 you complete the following process

1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
-1/r^-3 = -125/-8

2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
1/r^3 = 125/8

3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
8/r^3 = 125

4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

Operation One
8 = 125 * r^3

Operation Two
r^3 = 8/125

5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

r = 2/5

Is this the correct way of thinking or is there a concept behind it.

 

[Green Yoda]

I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

So fundamentally when you have r^-3 = -125/-8 you complete the following process

1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
-1/r^-3 = -125/-8

2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
1/r^3 = 125/8

3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
8/r^3 = 125

4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

Operation One
8 = 125 * r^3

Operation Two
r^3 = 8/125

5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

r = 2/5

Is this the correct way of thinking or is there a concept behind it.

The negative in the indices cannot be cancelled. So it is -1/r^-3=125/8 and its only RHS where the negative 'goes away'

 

[Katsumi]

The negative in the indices cannot be cancelled. So it is 1/r^-3=125/8

If that's the case how did it become positive later on? (r^-3 became r^3 at some point of the equation)

 

[InteGrand]

I failed so many tests in high school because of that. If i'm thinking in the right way below then my problem was believing that 1/x^y automatically converted into just x^y without any further simplification needed.

So fundamentally when you have r^-3 = -125/-8 you complete the following process

1. As r has a negative coefficient you first convert it into a fraction so that it can become positive
-1/r^-3 = -125/-8

2. As both sides of the equation are negative - the negative exponents cancel out and the equation stays balanced
1/r^3 = 125/8

3. You then need to separate r^3 from the 1 which it is dividing. You do this by multiplying across the 8.
8/r^3 = 125

4. You then need to isolate variable r. As the coefficients 8 and 125 are on different sides of the equation this can be done in 2 operations.

Operation One
8 = 125 * r^3

Operation Two
r^3 = 8/125

5. Now that variable r has been isolated we can find the value by taking the cube root of both sides of the equation.

r = 2/5

Is this the correct way of thinking or is there a concept behind it.

 

[InteGrand]

If that's the case how did it become positive later on? (r^-3 became r^3 at some point of the equation)

The -125/(-8) is actually a positive quantity and can immediately be simplified to 125/8 (negative signs on top and bottom "cancel out"). So the original equation is equivalent to r^-3 = 125/8.

 

[Katsumi]

....... does that mean that a negative number divided by a negative number is a positive number

you're shitting me if that's the case

 

[InteGrand]

....... does that mean that a negative number divided by a negative number is a positive number

Correct, a negative divided by a negative is a positive.

(Also a negative times a negative is a positive.)

 

[Green Yoda]

....... does that mean that a negative number divided by a negative number is a positive number

you're shitting me if that's the case

That is correct, -125/-8 is the same as -1/-1 multiplied by 125/8.. where -1/-1 =1 so 125/8 multiplied by 1 is 125/8.

 

[Katsumi]

I never knew that..... I sat General Math for 2 years and was never told that even once.....

I finally understand basic algebra.

Thanks guys; very much appreciated

 

[He-Mann]

I never knew that..... I sat General Math for 2 years and was never told that even once.....

I finally understand basic algebra.

Thanks guys; very much appreciated

Why is negative divide by negative equal to a positive?

 

[Mathew587]

Why is negative divide by negative equal to a positive?

Firstly, you should know that anything divided by itself is 1 so that's one way i.e.
5/5 = 1 and -10/-10 =1

Another way is factorisation i.e.
check the attachement cos i cant latex

 

[He-Mann]

Firstly, you should know that anything divided by itself is 1 so that's one way i.e.
5/5 = 1 and -10/-10 =1

Another way is factorisation i.e.
check the attachement cos i cant latex
View attachment 33829

I highlighted the word 'understand' because I demanded a concrete understanding, not abstract.

 

[Mathew587]

there's three approaches in the link. i'm sure that you undestand atleast one of them

 

[InteGrand]

If you know a bit of abstract algebra, you can read the top answer here about why (-x)(-y) = xy in any ring (though in that case you may have already thought about this and had it answered before): http://math.stackexchange.com/questions/878836/product-of-two-negative-numbers-is-positive .

 

[He-Mann]

View attachment 33831

there's three approaches in the link. i'm sure that you undestand atleast one of them

Still abstract. Can you provide a concrete example demonstrating this idea? i.e. incorporating realistic objects like monkey noses, hexagonal-shaped dishes, dollar coins covered in acetic acid, etc

 

[Mathew587]

Let x= 1 dollar coin covered in aceitic acid
Let y= he-mann's shitty sarcasm
-x/-x = -1(x)/-1(x) = x/x = 1
-x/-y = -1(x)/-1(y) = x/y

 

[He-Mann]

I'll give you a basic example.

Let x = Matthew587's debt to He-Mann = 2017 apologetic letters typesetted in LaTeX and must include a picture of a golden coin covered in acetic acid.
Let y = 2017 letters sent to He-Mann.

Then, x/y = 2017/2017 = 1.

That is, your debt can be paid in one payment!

Here, we are working with positives. Try incorporating negatives and try to make sense out of it.