Solve \( p x \) ad \( y \) \( 3 y=-2 x+12 \) and \( 4 y=2 x+2 \)

To solve the system of equations given by \(3y = -2x + 12\) and \(4y = 2x + 2\), we can start by rewriting the equations in slope-intercept form \(y = mx + b\). From the first equation, we have: \[ y = -\frac{2}{3}x + 4 \] From the second equation, we can express it as: \[ y = \frac{1}{2}x + \frac{1}{2} \] Now, we can set the two equations equal to each other because both are equal to \(y\): \[ -\frac{2}{3}x + 4 = \frac{1}{2}x + \frac{1}{2} \] To eliminate the fractions, we can find a common denominator, which is 6: Multiplying everything by 6 gives us: \[ -4x + 24 = 3x + 3 \] Now we can solve for \(x\): \[ 24 - 3 = 4x + 3x \\ 21 = 7x \\ x = 3 \] Now that we have \(x\), we can substitute \(x = 3\) back into either original equation to find \(y\). Using the first equation: \[ 3y = -2(3) + 12 \\ 3y = -6 + 12 \\ 3y = 6 \\ y = 2 \] Thus, the solution is: \( x = 3 \) and \( y = 2 \). So, the values of \(p\) and \(x\) are \(p = 3\) and \(y = 2\).