A series of algebra problems...?

Question

Can I get some help with these please?

(9/5)/(2/3)

[(2x)/(x + 7)] + [(4)/(x + 4)]

[(x + 5)/(x + 2)] + 4

[(2x - 2)/(x - 8)] + x + 5

(-2b - 9)/(b2 + 7b + 12) = b/(b + 3) + 2/(b + 4)

1/(y - 4) - 2/(y - 8) = -1/(y + 6)

If you're just gonna tell me to do my own work, then get your crappy 2 points elsewhere. Furthermore, if I could do the work on my own, I wouldn't have to ask you people, would I?

Answer 1

(9/5)/(2/3) = 9/5 * 3/2 = 27/10

[(2x)/(x + 7)] + [(4)/(x + 4)] = (2x)(x+4)/(x+7)(x+4) + 4(x+7)/(x+7)(x+4)
= (2x^2+8x)/(x+7)(x+4) + (4x+28)/(x+7)(x+4)
= (2x^2+12x+28)/(x+7)(x+4)
= 2(x^2+6x+14)/(x+7)(x+4)

[(x + 5)/(x + 2)] + 4 = (x + 5)/(x + 2) + 4(x+2)/(x+2)
= [(x+5) + (4x + 8)] / (x+2)
= (5x + 13) / (x+2)

[(2x - 2)/(x - 8)] + x + 5 = (2x-2)/(x-8) + x(x-8)/(x-8) + 5(x-8)/(x-8)
= (2x-2)/(x - 8) + (x^2-8x)/(x - 8) + (5x-40)/(x-8)
= (x^2 -x-42)/(x-8)
= [(x-7)(x+6)]/(x-8)

(-2b - 9)/(b2 + 7b + 12) = b/(b + 3) + 2/(b + 4)
(-2b-9)/(b+3)(b+4) = b(b+4)/(b + 3)(b + 4) + 2(b+3)/(b + 3)(b + 4)
(-2b - 9)/(b + 3)(b + 4) = (b^2+4b+2b+6)/(b + 3)(b + 4)
(-2b - 9) = (b^2+6b+6)
0 = b^2 +8b + 15
0 = (b+5)(b+3)
b = either -5 or -3

1/(y - 4) - 2/(y - 8) = -1/(y + 6)
(y-8)/(y-4)(y-8) -2(y-4)/(y-4)(y-8) = -1/(y + 6)
(y-8-2y+8)/(y-4)(y-8) = -1/(y + 6)
-y/(y-4)(y-8) = -1/(y + 6)
-y(y+6) = -1(y-4)(y-8)
-y^2-6y = -y^2 +12y -32 <- There was a mistake here
6y = 32
y = 16/3

Answer 2

That's a pretty tall order. Let's see
1. When you divide by a fraction, it's the same as multiplying by the fraction's reverse. So (9/5)/(2/3) is the same as (9/5)*(3/2) which = (9*3)/(5*2).

2. Find a common denominator by multiplying the first bracket by (x+4)/(x+4), and the second bracket by (x+7)/(x+7). (Remember that if the numerator and denominator are the same, you're essentially multiplying by 1).
[(2x)(x+4)/(x+7)(x+4)] + [(4)(x+7)/(x+4)(x+7)] or
[(2x^2)+8x]/[(x+4)(x+7)] =
[(2x^2)+8x]/[x^2+7x+4x+28] =
[(2x^2)+8x]/[x^2+11x+28]

3. Multiply both pieces by (x+2)/(x+2)

4. Multiply all three pieces by (x-8)/(x-8)

5. b2+7b+12 = (b+3)(b+4), therefore
(-2b-9)/(b+3)(b+4) = b/(b+3) +2/(b+4)
[(-2b-9)/(b+3)(b+4)] - 2/(b+4) = b/(b+3)
[(-2b-9)/(b+3)(b+4)] - [2/(b+4)] - [b/(b+3)] = 0
multiply the 2nd bracket by b+3/b+3 and the third bracket by (b+4)/(b+4) to get a common denominator for all.

6. [1/(y - 4)] - [2/(y - 8)] +[1/(y + 6)] = 0
Multiple the first bracket by (y-8)(y+6) then
Multiple the second bracket by (y-4)(y+6) then
Multiple the third bracket by (y-8)(y-4) to get a common denominator.

Answer 3

If you arent able to do anything, not even the first one, then begin asking about this one. Ask something specific.

Try to do something and I will help you.

Nikkis answer is mistaken. Sorry, Nicky.

And the last ones answers are mistaken too.

I will explain to you the first one.

When you divide two fractions, you have to invert the second one.

So: (9/5)/(2/3) = (9/5))(3/2)

When you multiply 2 fractions, you have to multiply the numerators and the denominators.

Ana

Answer 4

algebra was hard for me to. but not to scare you but geometry is way harder. ok so do whats in the parenthises inside first then go out ot the other ones.
example:
[(2x)/(x + 7)] + [(4)/(x + 4)]
{2x/7x} + {4/4}
since / means devide just do 2 divided by 7 and 4 divide by 4. that how i would do it. maby call someone for help?

Answer 5

I didn't get answer for last 2,the answers r 270,6x(x+6),5x+13/x+2,(x-7)(x+6)/x-8

Answer 6

= (9/5) / (2/3)
= (9*3) / (2*5)
= ( 27 ) / ( 10 )
= 2.7

Answer 7

sorry im too tired to answer! But it looks a bit difficult!

Answer 8

I could do these for you, but do your own homework.. you'll never learn anything this way

Answer 9

sorry but i didn't get the question, are u suppose to simplify or develop of factorise?????

Answer 10

dont you have a book ?