when you have a fraction + x
you can bring in x into the fraction

no there aren't multiple solutions

this is the only one

only place where you'd find multiple solutions would be in a quadratic equation

I'm talking about equations

if a = b and b = c them a = c
so
a = b
And
a = c

lol that's not how it works

I remember thinking that years ago

you can try to prove it algebraically

(a/b)+c = (a+ac)/c

here's a simplified answer

that's a law of equality

look up equivalence relation

transitivity

it'd only work if a b and c would be the same value

Is a requirement of equality

either way there's only one solution to this problem

il do it again see if I messed up

use paper

text math is awful

X = h/3(a+b)
X/h = 1/3(a+b)
h/X = 3(a+b)
(h/3X)-b = a
a = (h/3X)-b

🤔

I see what you’re doing wrong

Just multiply out the denominator first

Not the numerator

I have no idea what gwence was doing

il try what you said

She distributed first

Then removed the denominator

X = h/3(a+b)
3X(a+b) = h
a+b = h/3X
a = (h/3X)-b

there you go

oh hey it's quicker

same answer tho

wait I see where you went wrong

moving (a+B) to the Left hand side

did you divide

you didn't

I multiplies both sides by that value

h(a+b) needs to be moved to the LHS by dividing h(a+b) to remove (a+b)

you would be multiplying (a+b) * (a+B)