The vectors
are linearly independent if the ONLY solution to the following equation:
is
Write the set of points from -6 to 0 but excluding -4 and 0 as a union of intervals
First we take the interval -6 to 0. In that -4 and 0 are excluded.
So we split the interval -6 to 0.
Start with -6 and go up to -4. -4 is excluded so we break at -4. Also we use parenthesis for -4.
Interval becomes [-6,-4) . It says -6 included but -4 excluded.
Next interval starts at -4 and ends at 0. -4 and 0 are excluded so we use parenthesis not square brackets
(-4,0)
Now we take union of both intervals
[-6,-4) U (-4,0) --- Interval from -6 to 0 but excluding -4 and 0
This is a ratio problem; the ratio of the length to width is constant (and therefore equal):
4 /6 = 15 / x
Now, with a ratio, we may do any allowable algebra operation: cross-multiply, invert both sides, multiply or divide both sides by the same amount, etc.
Let's cross-multiply:
4x = (15)(6)
x = 90/4
x = 22.5 in.
Solve the equation:
5 3
x – —— = – ——
7 7
5
Add —— to both sides, so you get x isolated at the left-hand side:
7
5 5 3 5
x – —— + —— = – —— + ——
7 7 7 7
3 5
x = – —— + ——
7 7
Now, add those fractions at the right-hand side of the equation:
– 3 + 5
x = —————
7
2
x = —— <——— this is the answer (first option: 2/7).
7
I hope this helps. =)
