You have two equations here:
- \( 3x + 5y = 2 \)
- \( y = \frac{2}{3}x - \frac{14}{3} \)
To solve this system of equations, you can either substitute the second equation into the first or set them equal to each other. I will substitute the second equation into the first.
Starting with the first equation:
\[ 3x + 5y = 2 \]
Substituting \( y \) from the second equation:
\[ 3x + 5\left(\frac{2}{3}x - \frac{14}{3}\right) = 2 \]
Now, distribute the \( 5 \):
\[ 3x + \frac{10}{3}x - \frac{70}{3} = 2 \]
To eliminate the fractions, multiply the entire equation by \( 3 \):
\[ 3(3x) + 3\left(\frac{10}{3}x\right) - 3\left(\frac{70}{3}\right) = 3(2) \]
This simplifies to:
\[ 9x + 10x - 70 = 6 \]
Combine like terms:
\[ 19x - 70 = 6 \]
Now, add \( 70 \) to both sides:
\[ 19x = 76 \]
Now, divide by \( 19 \):
\[ x = 4 \]
Now we can substitute \( x = 4 \) back into the second equation to find \( y \):
\[ y = \frac{2}{3}(4) - \frac{14}{3} \]
This simplifies to:
\[ y = \frac{8}{3} - \frac{14}{3} = \frac{-6}{3} = -2 \]
So, the solution to the system of equations is:
\[ \boxed{(4, -2)} \]