You should be able to solve this

My sytem works to, you cunt!

1. x2
2. +11
3. x2
4. +11
...

112

>1*2 = 1

Literally what. Please explain.

I am able to solve it but too lazy to actually do it.

I don't hate kurisuTIIINA as a character but every time I see her on this centrafrican mask engraving picture book I get an aneurysm

x+y=8
x+z=13
y+w=8
z-w=6

y+z=14
x-y=-1
2x=7
x=3.5
y=8-3.5=4.5
z=14-4.5=9.5
w=8-4.5=3.5

you didn't show how you got 2x=7

Add the 2 previous terms right above 2x=7

lol, you are right :D

And that, my son, is why you fail.

38

poor little brainlet

left base 7
right base 3

answer is 112

3.5+4.5=8
+ +
9.5-3.5=6
= =
13 8

I did brute force but in retrospect its just basic math and I'm just a retard

x+y=8
x+z=13
z-x=6
Solving for z first because I fucking feel like it
x+z=13
z-x=6
Combine them
2z=19
z=9.5

Now solve for x
9.5+x=13
x=3.5
Now y
3.5+y=8
y=4.5

It only took me like a minute longer for brute force though.

20 = 20 by definition

42

01=10*0+(0+1)
10=10*2+(1+0)
11=10*2+(1+1)
20=10*4+(2+0)=42
xy=10*2x+(x+y)

Would you care to show 2 of these?

This solution literally makes no sense.

>his solution doesn't work for numbers of more than two digits

lol

see

It's in base 21 you retard

The solution is trivial and left to the reader as an exercise.

Y'all muthafuckers need gauss

[eqn]
\begin{pmatrix}
a & b & c & d
\end{pmatrix}
*
\begin{pmatrix}
1 & 0 & 0 & 1\\
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 1 & -1 & 0
\end{pmatrix}
=
\begin{pmatrix}
8 & 8 & 6 & 13
\end{pmatrix}
[/eqn]

[eqn]

\left.\begin{matrix}
1 & 0 & 0 & 1\\
1 & 1 & 0 & 0\\
0 & 0 & 1 & 1\\
0 & 1 & -1 & 0
\end{matrix}\right|

\begin{matrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{matrix}
[/eqn]

[eqn]

\left.\begin{matrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{matrix}\right|

\begin{matrix}
0.5 & 0.5 & -0.5 & 0.5\\
-0.5 & 0.5 & 0.5 & -0.5\\
-0.5 & 0.5 & 0.5 & -0.5\\
0.5 & -0.5 & 0.5 & 0.5
\end{matrix}
[/eqn]

[eqn]

\begin{pmatrix}
8 & 8 & 6 & 13
\end{pmatrix}
*
\begin{pmatrix}
0.5 & 0.5 & -0.5 & 0.5\\
-0.5 & 0.5 & 0.5 & -0.5\\
-0.5 & 0.5 & 0.5 & -0.5\\
0.5 & -0.5 & 0.5 & 0.5
\end{pmatrix}
=
\begin{pmatrix}
4.5 & 3.5 & 8.5 & 2.5
\end{pmatrix}
[/eqn]

top left = 3.5, top right = 4.5, bottom left = 9.5, bottom right = 3.5

>fucked up the final equation in latex

fugg

you still get the idea

What the actual fuck, who taught you to do it that way?
[eqn]\left(\left.\begin{matrix} 1 & 1 & 0 & 0\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & -1 \end{matrix}\right |\begin{matrix} 8 \\ 13 \\ 8 \\ 6 \end{matrix}\right) \sim \left(\left.\begin{matrix} 1 & 0 & 0 & -1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & -1 \end{matrix}\right |\begin{matrix} 0 \\ 8 \\ 13 \\ 6 \end{matrix}\right) [/eqn]
[eqn]\sim \left(\left.\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{matrix}\right |\begin{matrix} 7/2 \\ 9/2 \\ 19/2 \\ 7/2 \end{matrix}\right) [/eqn]

+9 01 | 1 + 20
+1 10 | 21 + 1
+9 11 | 22 + 20
+1 20 | 42 + 1
etc.
If the exponential value isn't going to change it should work like this.

Sorry but why did you need x-y=-1 and how did you get it? From y+z=14 I was able to find 2x=7 easily and the rest is done, but why did you need x-y=-1?