For this case we must solve the following equation:
[tex]3x- \frac {1} {5} = x + \frac {1} {3}[/tex]
Subtracting "x" from both sides of the equation we have:
[tex]3x-x- \frac {1} {5} = \frac {1} {3}\\2x- \frac {1} {5} = \frac {1} {3}[/tex]
Adding[tex]\frac {1} {5}[/tex]to both sides of the equation we have:
[tex]2x = \frac {1} {3} + \frac {1} {5}\\2x = \frac {5 + 3} {15}\\2x = \frac {8} {15}[/tex]
We divide between 2 on both sides of the equation:
[tex]x = \frac {\frac {8} {15}} {2}\\x = \frac {8} {15 * 2}\\x = \frac {8} {30}\\x = \frac {4} {15}[/tex]
Thus, the solution of the equation is:
[tex]x = \frac {4} {15}[/tex]
Answer:
[tex]x = \frac {4} {15}[/tex]
