## Geometry: Common Core (15th Edition)

$y = 11$ $x = 3$
Theorem 6-8 states that in a quadrilateral that is a parallelogram, its opposite sides are congruent. We can now deduce that $\overline{AB}$ is congruent to $\overline{DC}$, and $\overline{AD}$ is congruent to $\overline{BC}$. Let's set up the two equations reflecting this information: $AB = DC$ $AD = BC$ Let's substitute in what we know: $2y + 2 = 3y - 9$ $3x + 6 = y + 4$ Let's turn our attention to the first equation because we can solve for $y$ in this equation: $2y + 2 = 3y - 9$ Subtract $2$ from each side of the equation to isolate constants on one side of the equation: $2y = 3y - 11$ Subtract $3y$ from each side of the equation to isolate the variable on the left side of the equation: $-y = -11$ Divide each side of the equation by $-1$ to solve for $x$: $y = 11$ Now that we have the value for $y$, we can plug in this value for $y$ into our second equation to find $x$: $3x + 6 = 11 + 4$ Subtract $6$ from each side of the equation to isolate constants on the right side of the equation: $3x = 11 + 4 - 6$ Simplify the right side of the equation by adding or subtracting from left to right: $3x = 9$ $x = 3$