algebraic fraction question?

Question

how would i solve
6x / x + 1 - 5 / x + 3 = 3

thank you

Answer 1

I assume this is what you meant.

[6x / (x + 1)] - [5 / (x + 3)] = 3

You need a common denominator...
which in this case is (x + 1)(x + 3)

so...
6x / (x + 1) becomes
6x(x+3) / (x+1)(x+3)

and
5 / (x + 3) becomes
5(x+1) / (x+1)(x+3)

now you have:
[6x(x+3) - 5(x+1)] / (x+1)(x+3) = 3
[6x^2 + 18x - 5x + 5] / [x^2 + x + 3x + 3] = 3
(6x^2 + 13x + 5) / (x^2 + 4x + 3) = 3
cross multiply....
6x^2 + 13x + 5 = 3(x^2 + 4x + 3)
6x^2 + 13x + 5 = 3x^2 + 12x + 9
3x^2 + x - 4 = 0
(3x + 4)(x - 1) = 0

so...
3x + 4 = 0 ----> x = -4/3
x - 1 = 0 ----> x = 1

Answer 2

It's not clear where those division signs are supposed to be, but I'm going to assume you mean
(6x)/(x+1) - 5/(x+3) = 3

In this case, multiplying both sides by (x+1) will get rid of the denominator of the first fraction. Then multiplying both sides by (x+3) will clear the second one:
6x - 5(x+1)/(x+3) = 3(x+1)
6x(x+3) - 5(x+1) = 3(x+1)

Multiply everything out and combine like terms. You should end up with something in the form of ax^2 + bx + c = 0. Then solve this using factoring, completing the square or the quadratic formula.

Answer 3

multiply everything by the first denominator x + 1
6x - 5(x + 1)/(x + 3) = 3(x+1)
Then multiply by x + 3
6x(x+3) - 5(x+1) = 3(x+1)(x+3)
Distribute
6x^2 + 18x - 5x - 5 = 3 (x^2 + 4x + 3)
6x^2 + 13x - 5 = 3x^2 + 12x + 9
Group like terms
3x^2 + x - 14 = 0
Factor
(3x + 7)(x - 2) = 0
So 3x + 7 = 0 or x -2 = 0
So the answers are
x = -7/3 or 2

Answer 4

Solve the "x" variable in the equation.

(6x/x+1) - (5/x+3) = 3

First: eliminate fractions - multiply the denominators with each term.

(x+1)(x+3)(6x/x+1)-(x+1)(x+3)(5/x+3)=(x+1)(x+3)(3)

Sec: cross cancel "like" terms & combine the rest.

(x+3)(6x)-(x+1)(5)=(x+1)(x+3)(3)
6x^2+18x-(5x+5) = (x^2+3x+x+3)(3)
6x^2+18x-5x-5 = 3x^2+9x+3+9
6x^2+13x-5 = 3x^2+12x+9

*Combine "like" terms - subtract 3x^2, subtract 12x & subtract 9 from both sides (when you move a term to the opposite side, always use the opposite sign).

6x^2+13x-5-3x^2-12x-9 = 3x^2-3x^2+12x-12x+9-9
3x^2+x-14 = 0

Third: factor the expression & you should get...

(x-2)(3x+7) = 0

Now, set the parenthesis to "0' & solve the "x" variables.

a. x-2 = 0
x = 2

b. 3x+7=0
3x = -7
3x/3 = -7/3
x = -7/3

Solutions: 2, -7/3

Answer 5

if 6x/x + 1 - 5/x + 3 = 3
then 6 + 1 - 5/x + 3 = 3
=> 7 - 5/x + 3 = 3
=> - 5/x + 10 = 3
=> - 5/x = 3 - 10
=> -5/x = -7
=> 5/x = 7
=> x = 5/7

Answer 6

you shoud use parentheses becasue your equation as written in confusing buy I think you meant to say

6x/(x+1) -5/(x+3) = 3

clear fractions by multiplying through by (x+1)*(x+3) to get
6x(x+3) -5(x+1) = 3(x+1)(x+3)

6x^2 + 18x -5x -5 = 3(x^2+4x+3)
6x^2+13x-5=3x^2+12x+9
3x^2+x-14 = 0
(3x+7)(x-2) = 0
x1 = -7/3
x2 = 2

Answer 7

6x/x + 1 - 5/x + 3 = 3
6 + 1 + 5/x + 3 = 3
5/x + 10 = 3
5/x = 3 -10
5/x = -7
5 = -7x
x = 5/-7

Answer 8

6x.(x + 3) - 5.(x + 1) = 3.(x + 1).(x + 3)
6x² + 18x - 5x - 5 = 3x² + 12x + 9
3x² + x - 14 = 0
(3x + 7).(x - 2) = 0
x = - 7/3 , x = 2

Answer 9

6x / x is the same as 6

6 + 1 equals 7

7 - 5/x + 3 equals 4 - 5/x

so you have
4 - 5/x = 3
subtract 3 from both sides to get 1 - 5/x = 0

add 5/x to both sides to get 1 = 5/x

5/x = 1 when x = 5

Answer 10

6x / x + 1 - 5 / x + 3 = 3
6x(x+3-5(x+1)=3(x+1)(x+3)
6x^2+18x-5x-5=3x^2+12x+9
3x^2+x-14=0
3x^2+7x-6x-14=0
x(3x+7)-2(3x+7)=0.
(3x+7)(x-2=0
x=2,-7/3 answer