The equation you provided, \(2y = 3x - 2 = 0\), seems to be a bit confusing due to the double equal sign. To clarify, let's break it down into two parts:
- \(2y = 3x - 2\)
- \(3x - 2 = 0\)
Part 1: Solve for \(y\)
Starting with the first equation, we can isolate \(y\):
\[ 2y = 3x - 2 \]
Dividing both sides by 2:
\[ y = \frac{3x - 2}{2} \]
Part 2: Solve for \(x\)
From the second part of the equation, we can set \(3x - 2 = 0\) and solve for \(x\):
\[ 3x - 2 = 0 \]
Adding 2 to both sides:
\[ 3x = 2 \]
Dividing by 3:
\[ x = \frac{2}{3} \]
Final Values
Now you have \(x = \frac{2}{3}\) and can substitute this value back into the first equation to find \(y\):
\[ y = \frac{3\left(\frac{2}{3}\right) - 2}{2} \]
Calculating further:
\[ y = \frac{2 - 2}{2} = \frac{0}{2} = 0 \]
Summary
The solution to the equations is:
\[ \left(x, y\right) = \left(\frac{2}{3}, 0\right) \]