Question

3l-2m+3n=0
l-2m-n=0
find l,m & n...

Answer 1

3l - 2m + 3n = 0
l - 2m - n = 0
2l + 4n = 0
l = -2n

-3n - 2m = 0
m = -2/3 n

Answer 2

\[3l-2m+3n=0\]\[l-2m-n=0\]
Subtracting both equations,\[2l+4n=0\]\[l=-2n\]
Replacing into the second equation,\[(-2n)-2m-n=0\]\[-2m=3n\]\[m=-\frac{3}{2}n\]
Replacing m and l into the first equation,
\[3(-2n)-2(-\frac{3}{2}n)+3n=0\]\[-6n+3n+3n=0\]
Here's the interesting part,\[n=\mathbb{R} \]n can be any real number, yet l and m depend on the value of n.\[l=-2n\]\[m=-\frac{3}{2}n\]

Answer 3

can it be done by cross multiplication?