Could I get some help?

Question

x/5x+10+x-3/x+2=7/5
I am getting mixed answers every time I work the problem.

Answer 1

Use parentheses to make things clearer.
Assuming this is what you meant:
x/(5x+10) + (x-3)/(x+2) = 7/5
Note that 5x+10 = 5(x+2)
Multiply everything by 5(x+2) to get rid of the fractions
5(x+2)(x/(5x+10)) + 5(x+2)(x-3)/(x+2) = 5(x+2)(7/5)
x + 5(x-3) = (x+2)(7)
x + 5x - 15 = 7x + 14
6x - 15 = 7x + 14
Subtract 6x from each side
-15 = x + 14
Subtract 14 from each side
-29 = x

Answer 2

Just realized you implied some parenthesis in there...

x / (5x + 10) + (x-3) / (x+2) = 7/5

5x + 10 = 5 (x+2) so...

x / (5 * (x+2) ) + (x-3) / (x + 2) = 7/5

Let's first get rid of the fraction on the right by multiplying by 5

x / (x+2) + 5*(x-3)/ (x+2) = 7

Then multiply by (x+2) to get rid of fractions on the left.

x + 5*(x-3) = 7* (x+2)

Now multiply out the terms.

x + 5x - 15 = 7x + 14

6x - 15 = 7x + 14

-29 = x


Move terms around...

Answer 3

if i consider your ecuation like you wrote
in this case you obtain a cuadratic equation
ax^2+bx-c=0
and a formula for solve this kind of equation
it is (+/-)square root from(b^2-4ac)/2a
the idea is towo obtain a perfect binome
because we know that (x+b/2a)^2 is x^2+2x(b/2a)+(b/2a)^2
making the necesary operation your equation became
x^2+54x-3=0
quadratiac equation have always 2 solutions real or complex
in this case your solution are10 and 64
in
due to editing limitacion i recomand you to visit this:
http://www.enchantedlearning.com/math/algebra/quadratic/
or this
http://www.themathpage.com/aPreCalc/quadratic-equation.htm

Answer 4

x = (15/59) Try editing your question by putting parentheses. That is probably why you are missing it. This is the answer without parentheses though.

Answer 5

EDIT

oh. you're missing parentheses. thanks for wasting 5 minutes of my life working out a quadratic formula, sweetie.