### A good answer might be:

-1 (1, 2, -3)^{T} - (2 + 1)(1, 0, 1)^{T} =
( -1, -2, 3)^{T} - 3(1, 0, 1)^{T} =
( -1, -2, 3)^{T} + (-3, 0, -3)^{T} = (-4, -2, 0 )^{T}

# More Algebra!

If there is an unknown or two in the mix,
you can use algebra in the usual way to solve for them.
Just be careful that each move you make results in a valid
expression.
For example,
solve the following for a and y:

a(1, y)^{T} + 2(0, 5 )^{T} = (2, 20)^{T}

You might say that this cannot be done, since there
are two unknowns (a and y) but only one equation.
However:

a(1, y)^{T} + 2(0, 5)^{T} = (2, 20)^{T}
(a, ay)^{T} + (0, 10)^{T} = (2, 20)^{T}
(a, ay+10)^{T} = (2, 20)

Corresponding elements
must be equal.
So:

a = 2
ay + 10 = 20
2y = 10
y = 5

This last manuever is a very common trick,
called *equating corresponding elements*.

### QUESTION 6:

Here is an opportunity to practice.
Find x and y :

4( -1, y)^{T} + 2( 3x, 10)^{T} = (8, 24)^{T}