1. Basic Algebra Question

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

2. Re: Basic Algebra Question

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

3. Re: Basic Algebra Question

Originally Posted by 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

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

4. Re: Basic Algebra Question

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.

5. Re: Basic Algebra Question

Originally Posted by 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.
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'

6. Re: Basic Algebra Question

Originally Posted by Rathin
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)

7. Re: Basic Algebra Question

Originally Posted by 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.
$\noindent The first step isn't correct. The correct step is: r^{-3} = \frac{1}{r^{3}} (shouldn't be \frac{-1}{r^{-3}}). So we have \frac{1}{r^{3}} = \frac{-125}{-8}. This becomes \frac{1}{r^3} =\frac{125}{8} (in the fraction \frac{-125}{-8}, we can cancel negative signs: this fraction is equal to \frac{125}{8}). Now reciprocate (invert) both sides: r^{3} = \frac{8}{125}. Take the cube root of both sides now: r = \frac{2}{5}.$

8. Re: Basic Algebra Question

Originally Posted by Katsumi
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.

9. Re: Basic Algebra Question

Originally Posted by InteGrand
$\noindent The first step isn't correct. The correct step is: r^{-3} = \frac{1}{r^{3}} (shouldn't be \frac{-1}{r^{-3}}). So we have \frac{1}{r^{3}} = \frac{-125}{-8}. This becomes \frac{1}{r^3} =\frac{125}{8} (in the fraction \frac{-125}{-8}, we can cancel negative signs: this fraction is equal to \frac{125}{8}). Now reciprocate (invert) both sides: r^{3} = \frac{8}{125}. Take the cube root of both sides now: r = \frac{2}{5}.$
....... 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

10. Re: Basic Algebra Question

Originally Posted by Katsumi
....... 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.)

11. Re: Basic Algebra Question

Originally Posted by 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
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.

12. Re: Basic Algebra Question

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

13. Re: Basic Algebra Question

Originally Posted by 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
Why is negative divide by negative equal to a positive?

14. Re: Basic Algebra Question

Originally Posted by He-Mann
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
Capture.PNG

15. Re: Basic Algebra Question

Originally Posted by Mathew587
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
Capture.PNG
I highlighted the word 'understand' because I demanded a concrete understanding, not abstract.

16. Re: Basic Algebra Question

CodeCogsEqn.gif

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

18. Re: Basic Algebra Question

Originally Posted by Mathew587
CodeCogsEqn.gif

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

19. Re: Basic Algebra Question

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

20. Re: Basic Algebra Question

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.

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