SRL Archive - Mathematics

Hey guys, just doing some linear algebra in math, I kinda missed the one example the prof did so I'm unsure of how to complete this (probably very easy)

[x, 3] = [2, x+y]
[4, y] = x[2, 3]

Solve for x and y.

The coordinate format is throwing me off, any thoughts?

Are you sure those are coordinates and not interval notation?

Iv'e never seen anything like this before so you're probably in college or something but my best guess is x=2 and y=1.

The only thing you're going to be able to do is assign x, and y, to a number, and also make the [x, 3] the same sum as [2, x+y], i believe. I'm not too terribly sure, but, we'll see.

X=2, and Y=1
There are several ways to approch this problem. each line should be looked at as
[a,b]=[a,b]
[c,d]=[c,d]
where you have 2 variables, x and y which are related to both sets.
the first row
[x, 3] = [2, x+y] is saying that x=2, and 3=x+y
the second row
[4, y] = x[2, 3] is really [4,y]=[2x,3x] and says that 4=2x, and y=3x
so the only number for x which holds true for all case is 2, and the only number which holds true for y is 1.

There may be cases where you have a problem that the only solutions could be
x=2y, and y=0.5x

such as the following
[x,2y]=[2y,x]
x[4,4(y^2)]=[8y,(2(x^2)y)]

disregard, my math is flawed!

From: EtherFreak
X=2, and Y=1
There are several ways to approch this problem. each line should be looked at as
[a,b]=[a,b]
[c,d]=[c,d]
where you have 2 variables, x and y which are related to both sets.
the first row
[x, 3] = [2, x+y] is saying that x=2, and 3=x+y
the second row
[4, y] = x[2, 3] is really [4,y]=[2x,3x] and says that 4=2x, and y=3x
so the only number for x which holds true for all case is 2, and the only number which holds true for y is 1.

There may be cases where you have a problem that the only solutions could be
x=2y, and y=0.5x
such as the following
[x,2y]=[2y,x]
x[4,4(y^2)]=[8y,(2(x^2)y)]

The first eqn confirmed that x=2, y=1.
In the 2nd eqn, x=2, but 3x=y => y=6!

Hence the 2 eqns cannot be simultaneous? The 2 eqn must be independent so there is a unique (x,y) for both eqn?

From: riwu
The first eqn confirmed that x=2, y=1.
In the 2nd eqn, x=2, but 3x=y => y=6!

Hence the 2 eqns cannot be simultaneous? The 2 eqn must be independent so there is a unique (x,y) for both eqn?

Ah, ya, my math has a flaw, thats what happens when I try to make these problems on the fly. The thing I was trying to prove is there are some cases where they are not solvable because x=y, no mater the value you put in for x or y.

From: EtherFreak
Ah, ya, my math has a flaw, thats what happens when I try to make these problems on the fly. The thing I was trying to prove is there are some cases where they are not solvable because x=y, no mater the value you put in for x or y.

In that case, it means that x,y has no unique values as all real values of x,y satisfy the 2 eqn (you can replace x,y with any no. and the eqn will be equivalent).

In this case, if the eqns are meant to be simultaneous, then there is no solution for y (contradiction).

Ah thank you for the responses! I believe I misinterpreted the question, and those 2 equations are separate. Thank you EtherFreak

[x 3] = [2 (x+y)]

x = 2
y = 1


[4 y] = x[2 3]

x = 2
y = 3/2

Just equating the components of the vectors;
(i.) [x, 3] = [2, x+y] => x = 2, and 3 = x + y => 3 = 2 +y => y = 1,
(ii.) [4, y] = x[2, 3] = [2x, 3x] => 4 = 2x, and y = 3x => x = 4/2 = 2, and then y = 3x = 3*2 = 6.

it can be solved as a matrix i believe