How do you get 5?

Question

The problem is

1 / x - 3 + 1 / x + 3 = 10/x^(2-9)

the book says the answer is 5 and I've been working on this problem for 20 minute (sad) and I haven't gotten 5 yet. Help please.

sorry. I meant to say ... how do we get x = 5

Answer 1

1/(x - 3) + 1/( x + 3) = 10 /(x^2-9)
take lcm as (x-3)(x+3)
(x+3 +x-3 ) /(x-3)(x+3) = 10/(x^2-9)
2x / (x^2-9) = 10 /(x^2-9)
coz (a+b)(a-b) =a^2-b^2
so 2x = 10
x=5

Answer 2

I'm going to assume you mean this:

1/(x - 3) + 1/(x + 3) = 10/(x^2 - 9)

The added brackets emphasize what is in the denominator; otherwise your question looks ambiguous.

First, factor the denominator on the right hand side.

1/(x - 3) + 1/(x + 3) = 10/ [ (x - 3)(x + 3) ]

Look at the greatest common denominator; it is (x - 3)(x + 3). If you multiply the equation by (x - 3)(x + 3), you should eliminate all fractions, because all denominators cancel out. This gives us

(x + 3) + (x - 3) = 10

Now, solve this as per normal algebra.

x + 3 + x - 3 = 10
2x = 10
x = 5

Answer 3

you've got the wrong problem

i think the problem is 1/x-3 + 1/x+3 = 10/x^2 - 9 not x^(2-9)

if that is the case: the answer of 5 is correct

solution:

by LCD
[x+3 + x-3]/x^2-9 = 10/x^2-9

2x = 10(x^2-9)/x^2-9
eliminate x^2-9

x = 5

Answer 4

1/2 + 1/8 =? 10/ (25 - 9)

5/8 =? 10/16= 5/8

yes so 5 is indeed an answer.

1 /(x - 3)+ 1 /(x + 3)= 10/(x^2-9)

1 /(x - 3)+ 1 /(x + 3)= 10/{ (x-3)(x+3) }

(x+3) /{(x+3)(x - 3)}+ (x-3) /{(x-3)(x + 3)}= 10/{ (x-3)(x+3) }
2x/{(x-3)(x+3)} = 10/{ (x-3)(x+3) }

2x = 10 => x = 5.

Answer 5

1/(x - 3) + 1/( x + 3) = 10 /(x^2-9)
take (x-3)(x+3)
(x+3 +x-3 ) /(x-3)(x+3) = 10/(x^2-9)
2x / (x^2-9) = 10 /(x^2-9)
coz (a+b)(a-b) =a^2-b^2
so 2x = 10 so
x=5

Answer 6

firstly please DO be careful with brackets because that can yeild different answers accordingly

multiply both sides by (x-3)

1 + (x-3)/(x+3) = 10(x-3)/(x-3)(x+3) = 10/(x+3)

then multiply both sides by (x+3)

(x + 3)+(x - 3) = 10

2x = 10

x = 10/2 = 5