Originally Posted by elvinas
\displaystyle \begin{align*} \frac{x - 1}{3} &= \frac{x + 5}{x - 1} \\ (x - 1)^2 &= 3(x + 5) \\ x^2 - 2x + 1 &= 3x + 15 \\ x^2 - 5x - 14 &= 0 \\ (x + 2)(x - 7) &= 0 \\ x + 2 = 0 \textrm{ or } x - 7 &= 0 \\ x = -2 \textrm{ or }x &= 7 \end{align*}

Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0

Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0

Originally Posted by elvinas
Thank You very much but in Line 3 isnt 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0
Definitely not. When moving everything to the LHS, we need to subtract 3x from both sides. -2x - 3x = -5x.

Thank You very much but in Line 3 isn't 3x-2x is 1x? when you add them together i think it should look like this x(to the second power)+1x-14=0