I am so close to figuering them out?

Question

I need some help with these problems.
subtract 7x/10-x/10

I think that you set this one up like this 7x-x/10 and get either 7/10 or 6x/10 but I am not sure.

the other problem is 9-b^2/b^2-b-6 (write in simplest form.)
I know with this one I must factor but I am not sure how to factor 9-b^2
can I get some help?

Answer 1

(7x)/10 - x/10

To solve this, since we have a common denominator, we can merge the fractions as one.

(7x - x)/10

Group like terms,

(6x)/10

Which reduces to

(3x)/5

(9 - b^2) / (b^2 - b - 6)

First, let's use the "negative one" technique on the numerator; how this works is factoring out a (-1), causing the terms to switch.

(-1)(b^2 - 9) / (b^2 - b - 6)

Now, factor each quadratic. The top factors as a difference of squares.

[ (-1)(b - 3)(b + 3) ] / [ (b - 3)(b + 2) ]

Note the common term that cancels out (b - 3).

[ (-1)(b + 3) ] (b + 2)

Answer 2

Since the denominator is common, yes, you can say that it is equal to (7x-x)/10, which is 6x/10 which is 3x/5.

2) Remember that (a^2-b^2) = (a+b)(a-b), so....

9-b^2 = (3+b)(3-b)

Also b^2-b-6 = (b-3)(b+2)

If we look at the whole equation, we see that (3-b)/(b-3) = -1

So, it all equals -(b+3)/(b+2)

Answer 3

Question 1
7x/10 - x/10 = 6.x/10 = (3/5).x
NB if you let x/10 = A, question becomes:-
7A - A = 6A = 6.(x/10) = (3/5).x as above.

Question 2
= (3 - b).(3 + b) / (b - 3).(b + 2)
= - 1(b - 3).(3 + b) / (b - 3).(b + 2)
= - (b + 3) / (b + 2)

Answer 4

do you mean 7x - x/10 = 70x/10 - x/10 = 69x/10
or 7x/(10-x/10) = 70x/(100-x) ?

i assume you mean
y = ( 9 - b^2 ) / ( b^2 - b - 6)
= [(3-b)(3+b)] / [(b-3)(b+2) ]
= - (b+3)/ (b+2)

Answer 5

the first one is 6x/10 = 3x/5

9-b^2/b^2-b-6
(3-b)(3+b)/(b-3)(b+2)
(3-b)(-3-b)/(3-b)(b+2)
-(3+b)/(b+2)