For this case we must solve the following equation:
![3x- \frac {1} {5} = x + \frac {1} {3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/auroqlbdukvktb3k7ea1dcz7v4g2yjseza.png)
Subtracting "x" from both sides of the equation we have:
![3x-x- \frac {1} {5} = \frac {1} {3}\\2x- \frac {1} {5} = \frac {1} {3}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lm758u5wsalu222401fdzer4hiq5p8cymq.png)
Adding
to both sides of the equation we have:
![2x = \frac {1} {3} + \frac {1} {5}\\2x = \frac {5 + 3} {15}\\2x = \frac {8} {15}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/78tjkdfj5gs26v7iiq19v6k8fprlnl8x1d.png)
We divide between 2 on both sides of the equation:
![x = \frac {\frac {8} {15}} {2}\\x = \frac {8} {15 * 2}\\x = \frac {8} {30}\\x = \frac {4} {15}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/awozij2a31d15fr8ueacs9ly4p8yyg2o1u.png)
Thus, the solution of the equation is:
![x = \frac {4} {15}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/apsvha63a7rlcdlso420uboe5r2mdoaioy.png)
Answer:
![x = \frac {4} {15}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/apsvha63a7rlcdlso420uboe5r2mdoaioy.png)