hard math help

[sharp dew]

Alrighty so

When you're solving linear equations

You want to get an x term on one side

And a constant term on the other

Eg.
5x=10

Then you can easily solve for x

I'll do an example problem for you and you can try these ones yourself

Lets take 5x+10=7+8x

Let's try to move all x terms to one side

The right in this case

We minus 5x on both sides

To get rid of 5x on the left side

'Moving' it to the right

So
5x+10=7+8x
Becomes
10=7+8x-5x

And we can simplify that to
10=7+3x

Then we can move all the constants to the right by minus 7 on both sides

We obtain 3x=3

Then we divide by 3 on both sides to find x

x=1

@narrow flint Anything you want to clarify

[narrow flint]

thanks

this guy the goat

patience level is absurd

i use cramer

[sharp dew]

What's a cramer

[thorny swan]

It’s a method called Cramer’s rule

It’s used to solve linear equations much faster than elimination or substitution

[sharp dew]

Oh matrix stuff

[thorny swan]

Yea

[sharp dew]

For some reason our syallbus doesn't touch matrices

Like we learn basic integration before matrices 😭

[thorny swan]

I learned it here

Intro to algebra

I don’t know much about matrices

Just learned it while solving this problem

[thorny swan]

Ohh

Which country you in?

[sharp dew]

Singapore

[thorny swan]

Ohhk

[sharp dew]

Technically you can just generalise this to
$x=\frac{c_2-c_1}{m_1-m_2}$

[boreal apex]

denzio321

[sharp dew]

Oh nvm this is different

[thorny swan]

Different from what?

[sharp dew]

Cramer's methods is for system of 2 linear equations

[sharp dew]

This one is a general solution for
$m_1x+c_1=m_2x+c_2$

[boreal apex]

denzio321

[thorny swan]

Yes