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Mathematics 8 Online
OpenStudy (compassionate):

Quick question

OpenStudy (zzr0ck3r):

quick would be having it in the topic \(\uparrow\)

OpenStudy (compassionate):

\[\frac{1}{x} = \frac{x - 1}8 - \frac{x}{4} \] My first step would be \[\frac{x-1}{8} - \frac{x \times 2}{4 \times 2}\]

OpenStudy (compassionate):

Didn't want to put more than one line of LaTeX in the subject and make it freeze up on me.

OpenStudy (zzr0ck3r):

I would multiply everything by \(8x\)

OpenStudy (astrophysics):

\[\huge \frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\] pretty much with all fractions

OpenStudy (astrophysics):

I like your style zz

OpenStudy (zzr0ck3r):

:)

OpenStudy (compassionate):

\[\frac{4(x - 1) - 8x}{32} \]

OpenStudy (astrophysics):

That works

OpenStudy (astrophysics):

You can now cross multiply with the left side if you like

OpenStudy (compassionate):

\[4x - 4 - 8x/32 = 1/x\]

OpenStudy (compassionate):

I get it. So when adding or subtracting, we can use the ab + bd/ad rule, and when multiplying, we can cross multiply, and when dividing we can take the recipricol?

OpenStudy (astrophysics):

Yeah sort of, but know what kind of algebraic rules are going on, when we cross multiply for example we're just multiplying both sides by 32 and x, so yes reciprocal. :)

OpenStudy (astrophysics):

I mean this rule isn't anything special, it's just as you would do fractions regularly, finding common denominator..etc

OpenStudy (zzr0ck3r):

In general it is a good idea to learn the following trick \[\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\\ abc\dfrac{1}{a}+abc\dfrac{1}{b}=abc\dfrac{1}{c}\\ bc+ac=ab\] When we have fractions, it is easy to clear the denominators when we have an equals sign!

OpenStudy (astrophysics):

Hehe yeah, I love that

OpenStudy (astrophysics):

Especially with partial fractions

OpenStudy (compassionate):

Yes. Clearing fractions. I understand. I'm just brushing up on my skills.

OpenStudy (zzr0ck3r):

word

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