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Inverse signs on both sides of an equation [Confusion]

neurohype

New member
May 31, 2012
6
[SOLVED] Inverse signs on both sides of an equation [Confusion]

Hey everyone,

I've just reinserted myself in maths after so many years, done a lot of review but I sometimes fall on small "glitches" between my test answers and the suggested answers in my textbook.

I've been getting confused with a new "trick":

Right now I'm at getting solution-pairs to solve systems of equations, graph them after getting y = mx + b , etc (sorry if my terminology is still shaky) and on previous pages (in the textbook) I noticed that you could cancel out like a -y and make it y (positive) by inverting all signs on the other side of the equation.

Everything seemed to work okay and I was all hyped about this new trick until I found the detailed answer (step by step) to be different and break that rule I thought I had learned. I'll copy the last two steps from what the answer section gives me :

Fig. 1

\(\displaystyle y = \frac{-2x}{-3} - \frac{14}{-3}\)

then becomes:

\(\displaystyle y = \frac{2x}{3} + \frac{14}{3}\)

My confusion :

Why does the right side of the equation invert all signs when the left one did not (stayed positive Y). Of course, that's what we want at the end, a positive y, but I don't understand why it remains positive while the rest all inverts (if "invert" is the correct term).

Here's an example among others that led me to believe BOTH SIDES would be affected when inverting (afterall, it's an equation)

Example 1:

\(\displaystyle -y = -2x - 5\)

becomes...

\(\displaystyle y = 2x + 5\)

Here, you invert -y to +(1)y for convenience and subsequently invert all signs on the other side of the equation. My problem lies in the fact Fig 1 (first equation) has y remain positive y but yet all signs change on the other side.

I'd hugely appreciate any explanation and/or hints.
 
Last edited:

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Hi neurohype (Wave),

This is a great question. You are correct that it's a nice technique to change $-y$ to $y$ by inverting the sign of all terms, however in the first example you gave that is not what is going on. What is going on is applying the fact that a double negative is a positive.

Let's look at \(\displaystyle \frac{-2x}{-3}\). You could think of this as \(\displaystyle \frac{(-1)2x}{(-1)3}=\frac{2x}{3}\).

Now for the second part: \(\displaystyle -\frac{14}{-3}\). You can think of subtraction as adding a negative.

Put another way, \(\displaystyle -\frac{14}{-3}=+(-1)\frac{14}{-3}=\frac{(-1)14}{(-1)3}=\frac{14}{3}\).

Does that help any? :)
 

neurohype

New member
May 31, 2012
6
What is going on is applying the fact that a double negative is a positive.
Wow! I am speechless. Okay, I know this might sound like I'm overreacting, but I've done 8 hours straight yesterday just so I could re-immerse myself good and there's so many "small rules" that I have to learn, many of which are not mentioned in my books because it's assumed I already know them having completed my high school almost 15 years ago. So I confused my new learned trick with the fact (a fact I hadn't yet observed but which make perfect sense, thanks alot for that one!) that double negatives become positive.

Thanks for also using the (-1)'s in front of the numerators/denominators, I had learned about that two days ago but still wasn't sure about that "minus 14 fourteen over negative 3", if it was the same as "negative fourteen over negative 3".

Does that help any? :)
It helped my GREATLY! So much I'm enabling smilies for this post : :)

Thanks!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Glad to help and you've found the right place for self-study. :) I think it's awesome that you are trying to improve your skills again. It's never too late to keep going and it's never wrong to start with the basics.

Hope to see you around more and look forward to your questions.
 

neurohype

New member
May 31, 2012
6
Hope to see you around more and look forward to your questions
Thanks! I'm under the impression I might often have questions that pop up, though.

And yah, I'm glad I finally found the courage to getting myself back to it. I guess it's the thirties, I have to go back to uni and since I did my high school with easy maths (don't know how it works in Russia but here in Canada it's Math 416 & Math 514) while what courses I need for the other 4 prerequisites to software engineering are Math 436 & 536, well it came back to haunt me now. I didn't end up a rockstar but at least I have made a living coding/programming but now I want more control over my employment than random contracts here and there.

I said "haunt me", but I did find some pleasure in some things like the useful tricks mentioned in this thread and more to come. I do admit I'm fairly intimidated by what's to come (math-wise, I'm very comfortable with code already). Just the titles of some threads make me feel like I'm lightyears behind. Hope it all works well.

Personal question: did you find higher math (Calculus I & II, etc) intimidating at first, or am I innately retarded on the math level to find it all very foreign?
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
I'm from the US actually, just spent a whlie in Russia. :)

When I took calculus I was in high school and a full on math nerd, but it took quite a while to transition to the new ideas - that is for sure. You need a very solid background in all the algebra techniques you've seen throughout the years and feel comfortable with limits before doing derivatives and integrals. It's not just you, trust me. It's a different way of thinking about math so transition time is to be expected.
 

neurohype

New member
May 31, 2012
6
Sounds reassuring!

Anyway, I'm planning on really investing myself in this so it might be less frightening than I expect. Thanks for your encouragement, very much appreciated.