Question

Help me find the answer to this problem? I got
x = -4, x = -5 did I do it right?

Answer 1

No something is wrong..

Answer 2

I am just really having a hard time trying to figure out how to solve this. Ive been trying for over a day now. ._.

Answer 3

\(\frac{x+3}{x+5} + \frac{x}{6} = \frac{5}{6}\)

multiply the LCD 6(x+5) through out the equation

\(6(x+5) [\frac{x+3}{x+5} + \frac{x}{6}] = 6(x+5) [\frac{5}{6}]\)
\(\)

\(6(x+5)*\frac{x+3}{x+5} + 6(x+5) * \frac{x}{6}] = 6(x+5) *\frac{5}{6}\)

\(6\cancel{(x+5)}*\frac{x+3}{\cancel{x+5}} + \cancel{6}(x+5) * \frac{x}{\cancel{6}} = \cancel{6}(x+5) *\frac{5}{\cancel{6}}\)


\(6*(x+3) + (x+5) * x = (x+5) * 5\)


\(6x + 18 + x^2+5x = 5x + 25\)

\(x^2 + 6x - 7 = 0\)

Answer 4

did you get the same quadratic ?

Answer 5

I think you went by long method I show you simpler..

Answer 6

Divide x/6 both the sides first..

Answer 7

\[\frac{x+3}{x+5} = \frac{5-x}{6} \implies 6x + 18 = (x+5)(5-x)\]

\[6x + 18 = (x+5)(5-x) \implies 6x + 18 = -(x^2 - 25) \implies x^2 + 6x - 7 = 0\]