How would Veeky Forums solve this?

How would Veeky Forums solve this?

this is so trivial my first line would be the solution.

10/3 = 7/x
70/21 = 70/10x
x=2.1

[math] \displaystyle

\begin{cases}
z - 1 = 8 \\
z + 1 = 3
\end{cases}

\\ 2z = 11
\\ z = 5.5
[/math]

Actually fucking kill yourself and get off this board

Edgelord

nice troll

>not using [math]\LaTeX[/math]
Actually fucking kill yourself and get off this board

...

you forgot 4. miltiplier

4* [(z-1)/3] = 8/(z+1)

>4/3 != 8/6
lmao ok

let the fourth be with you

...

God damn it...

You forgot to add a constant to both sides in step 12.

>he doesn't know Cauchy's integral theorem

I can tell you're in high school

He got the wrong answer.

>he looks at the answer as a value in the Real-number Set

Get real, pre-call faggalo

z = 5 all of you have downs

Try plugging that back in for z.

half the answer, brainlet

z=+5,-5.

(z-1)(z+1)=24
z^2-25=0
z=+5, -5

4/3 = 8/6 true

or

-6/3 = -8/4 true

okay?

Seriously....I hope OP is a troll, because atleast I know that the majority of answers below OP are trying to troll.


I think it is time to implement some sort of test before being able to enter this board.

Engineer here.
Let the LHS and RHS be f(z) and g(z), respectivly. Let h(z) = f(z) - g(z), and use some numerical method to find the roots of h(z).

cross-multiply, then difference of squares:
[math]\frac{z-1}{3} = \frac{8}{z+1}\\
\implies (z-1)(z+1) = 24\\
\implies z^2 - 1 = 24\\
\implies z^2 = 25\\
\implies z = \pm 5[/math]