Answer :
[tex] \frac{2x+1}{x+1} = \frac{14}{3x-1} \ \ \Leftrightarrow\ \ (2x+1)(3x-1)=14(x+1)\ \ and\ \ x\in R-\{-1; \frac{1}{3}\}\\ \\6x^2-2x+3x-1=14x+14\\6x^2-13x-15=0\ \ \Rightarrow\ \ \Delta=(-13)^2-4\cdot6\cdot(-15)=169+360=529\\ \\ \sqrt{\Delta} =2\\ \\x_1= \frac{13-23}{2\cdot6} = \frac{-10}{12}=- \frac{5}{6}\ \in D\ \ \ and\ \ \ x_2= \frac{13+23}{2\cdot6} = \frac{36}{12}=3\ \in D[/tex]
2x+1/x+1=14/3x-1
find their lcm which is (x+1)(3x-1)
(2x+1)(3x-1)=14(x+1)
6x^2+x-1)=14x+14
6x^2-13x-15=0
(6x+5)(x-3)=0
X=-5/6 or x=3
um hope this helps
find their lcm which is (x+1)(3x-1)
(2x+1)(3x-1)=14(x+1)
6x^2+x-1)=14x+14
6x^2-13x-15=0
(6x+5)(x-3)=0
X=-5/6 or x=3
um hope this helps
