two math problems that i need help on ..?

Question

Could someone please help me with these two problems please thanks so much ...

1. (x + 1) / (x + 31) = 1 / (x + 3)

A) x = -7, 4
B) No Solution
C) x = -7, -4
D) x = 7, -4


2. Find the sum and express in simplest form
x^2 / (x^2 - 9) + (2x -15) / (x^2 -9)

Answer 1

1. (x + 1) / (x + 31) = 1 / (x + 3)
(x + 1)(x + 3) = x + 31
x² + 4x + 3 = x + 31
x² + 3x - 28 = 0
(x + 7)(x - 4) = 0
x = -7
x = 4

2. x^2 / (x^2 - 9) + (2x -15) / (x^2 -9) =
(x² + 2x - 15) / (x² - 9) =
[(x + 5)(x - 3)] / [(x - 3)(x + 3)] =
(x + 5) / (x + 3).

Answer 2

I am in Algebra 2!!! Each line below is a step!!

#1
Cross multiply - (x+1)(x+3)=(x+31)
Multiply (x+1)(x+3) - (x^2+4x+3)
Bring the (x+31) over to the other side
Subtract the (x+31) from the (x^2+4x+3) the 31 becomes a minus because it is switching sides - (x^2+3x-28)
Factor - (x-4)(x+7)
So x equals - 4 and -7
The answer is A

#2
Factor both (x^2-9) - (x+3)(x-3)
The problem is now - x^2/(x+3)(x-3)+2x-15/(x+3)(x-3)
There is a common denominator
The problem is now - x^2+2x-15/(x+3)(x-3)
Factor the top - (x+5)(x-3)
The problem is now - (x+5)(x-3)/(x+3)(x-3)
The (x-3) cancels on the top and the bottom!
The answer is (x+5)/(x+3)!


Well good luck!!!

My e*mail address is [e-mail] if you need more help!!!

Answer 3

1. (x + 1) / (x + 31) = 1 / (x + 3)

A) (-7 +1)/(-7+31) = 1/(-7+3)
-6/24 = -1/4 True!
(4+1)/(4+31) = 1/(4+3)
5/35 = 1/7 True!

2. x^2 / (x^2 - 9) + (2x -15) / (x^2 -9)
= (x^2+2x - 15)/(x^2 - 9)
= (x^2+5x - 3x - 15)/(x^2 - 3^2)
= (x(x+5) - 3(x+5)) /( (x-3)(x+3))
= (x-3)(x+5) / ((x-3)(x+3))
= (x+5)/(x+3)

Answer 4

1 (a)
2 Combine to obtain (x^2+2x-15)/(denom)
(x+5)(x-3)/(denom)
Note the denom is the difference of 2 squares, so it factors (x-3)(x+3).
The (x-3) in num. and denom. cancel out, leaving x+5 / x+3

Answer 5

1.a you have the answer
2.(x*2+2x-15)/x*2-9=
((x+5)(x-3))/((x-3)(x+3))=
(x+5)/(x+3)=
(x+3+2)/(x+3)=
((x+3)/(x+3))+(2/(x+3))=
1+((2)/(x+3))