I don't know if linear algebra was the right sub context but here it goes...
how do I compute (f-h)(4) if f(x)=2x+5 and h(x)=7-x/3?

Question

I don't know if linear algebra was the right sub context but here it goes...
how do I compute (f-h)(4) if f(x)=2x+5 and h(x)=7-x/3?

anonymous · Answer

(f-h)(x)=2x+5 -(7-x/3)
(f-h)(4)=2*4+5 -(7-4/3)

anonymous · Answer

so it's 2/3?

anonymous · Answer

22/3

anonymous · Answer

Ok.  I do the stuff in the parenthesis first and come up with 2*4+5(-3/3), 2*4 is 8, 8+5=13(-3/3) which =-39/3 and that =-13.  What did I do wrong?

anonymous · Answer

Why are you multiplying anything against (-3/3)? 
$$(2)(4) +5-(7-4/3)$$
$$8+5 -(21-4)/3$$
$$13-17/3$$
$$39/3-17/3=22/3$$

anonymous · Answer

remember that 5-5 doesn't mean (5)(-5).

anonymous · Answer

I forgot to do the 7*3 to make it 21/3 and -4/3... so I just subtracted -7-4 and divided that by 3.  Thanks!  I get so flustered with fractions that I make stupid mistakes!

anonymous · Answer

Yeah, it happens a lot. Negative signs are everyone's bane.