In a parallelogram, consecutive angles are supplementary. Thus,

$\displaystyle \measuredangle B + \measuredangle C = 180^{\circ}$

$\displaystyle \left ( 2x + 15\right ) + x = 180$

$\displaystyle 3x + 15 = 180$

$\displaystyle 3x = 165$

$\displaystyle x = 55$

In a parallelogram, consecutive angles are supplementary (i.e. add to $\displaystyle 180^{\circ}$) and opposite angles are congruent (i.e. equal). Using these properties, we can write a system of equations.

1. $\displaystyle \measuredangle B = \measuredangle D$

2. $\displaystyle \measuredangle C + \measuredangle D = 180^{\circ}$

Starting with equation 1.,

$\displaystyle 2x + 3 = y + x$

$\displaystyle x + 3 = y$

$\displaystyle x = y - 3$

Now substituting into equation 2.,

$\displaystyle y + \left ( y + x\right ) = 180$

$\displaystyle 2y + x = 2y + (y - 3) = 180$

$\displaystyle 3y - 3 = 180$

$\displaystyle 3y = 183$

$\displaystyle y = 61$

Finally, because opposite angles are congruent, we know that $\displaystyle \measuredangle A = \measuredangle C$.

$\displaystyle z = y$

$\displaystyle z = 61$

In a parallelogram, consecutive angles are supplementary and opposite angles are congruent. Using these properties, we can write a system of equations.

1. $\displaystyle \measuredangle B =\measuredangle D$

2. $\displaystyle \measuredangle B + \measuredangle C = 180^{\circ}$

3. $\displaystyle \measuredangle A = \measuredangle C$

Starting with equation 1.,

$\displaystyle x = 4y - x$

$\displaystyle 2x = 4y$

$\displaystyle x = 2y$

Substituting into equation 2.,

$\displaystyle x + y = 180$

$\displaystyle 2y + y = 180$

$\displaystyle 3y = 180$

$\displaystyle y = 60$

Using equation 3.,

$\displaystyle z = y$

$\displaystyle z = 60$

In a parellelogram, consecutive angles are supplementary.

$\displaystyle \measuredangle C + \measuredangle D = 180^{\circ}$

$\displaystyle 75^{\circ} + \measuredangle D = 180^{\circ}$

$\displaystyle \measuredangle D = 105^{\circ}$

In a parallelogram, opposite angles are congruent.

$\displaystyle \measuredangle B = \measuredangle D$

$\displaystyle \measuredangle B = 125^{\circ}$

In a parallelogram, consecutive angles are supplementary and opposite angles are congruent.

$\displaystyle \measuredangle A + \measuredangle D = 180^{\circ}$

$\displaystyle z + \left ( 2z + 9\right ) = 180$

$\displaystyle 3z + 9 = 180$

$\displaystyle 3z = 171$

$\displaystyle z = 57$

$\displaystyle \measuredangle A = \measuredangle C$

$\displaystyle z = y$

$\displaystyle 57 = y$

In the above parallelogram, $\displaystyle \measuredangle A$ and $\displaystyle \measuredangle B$ are consecutive angles (i.e. next to each other). In a parallelogram, consecutive angles are supplementary, meaning they add to $\displaystyle 180^{\circ}$.

$\displaystyle \measuredangle A + \measuredangle B = 180^{\circ}$

$\displaystyle 45^{\circ} + \measuredangle B = 180^{\circ}$

$\displaystyle \measuredangle B = 135^{\circ}$

In parallelogram $\displaystyle ABCD$, $\displaystyle \measuredangle B$ and $\displaystyle \measuredangle D$ are opposite angles. In a parallelogram, opposite angles are congruent. This means these two angles are equal.

$\displaystyle \measuredangle D = \measuredangle B$

$\displaystyle \measuredangle D = 117^{\circ}$

In a parallelogram, consecutive angles are supplementary.  $\displaystyle \measuredangle A$ and $\displaystyle \measuredangle D$ are consecutive, so their sum is $\displaystyle 180^{\circ}$.

In a parallelogram, consecutive angles are supplementary and opposite angles are congruent. Using these properties, we can write a system of equations. Because we have three variables, we will need three equations.

1. $\displaystyle \measuredangle A + \measuredangle B = 180^{\circ}$

2. $\displaystyle \measuredangle B = \measuredangle D$

3. $\displaystyle \measuredangle C + \measuredangle D = 180^{\circ}$

Start with equation 1.

$\displaystyle z + \left (x + z \right ) = 180$

$\displaystyle x + 2z = 180$

Now simplify equation 2.

$\displaystyle x + z = y + 2x$

$\displaystyle z = y + x$

Finally, simplify equation 3.

$\displaystyle \left ( y + 20\right ) + \left ( y+ 2x\right ) = 180$

$\displaystyle 2y + 2x + 20 = 180$

$\displaystyle 2y + 2x = 160$

$\displaystyle y + x = 80$

Note that we can plug this simplified equation 3 directly into the simplified equation 2 to solve for $\displaystyle z$.

$\displaystyle z = y + x = 80$

Now that we have $\displaystyle z$, we can solve for $\displaystyle x$ using equation 1.

$\displaystyle x + 2z = 180$

$\displaystyle x + 2\left ( 80\right ) = 180$

$\displaystyle x + 160 = 180$

$\displaystyle x = 20$

With $\displaystyle x$, we can solve for $\displaystyle y$ using equation 3.

$\displaystyle y + x = 80$

$\displaystyle y + 20 = 80$

$\displaystyle y = 60$

Now that we have $\displaystyle y$ and $\displaystyle z$, we can solve for $\displaystyle y + z$.

$\displaystyle y + z = 60 + 80 = 140$