To solve for \(x\) and \(y\) in the parallelogram DEFG, where DH = \( x + 2 \), HF = \( 2y \), GH = \( 3x - 3 \), and HE = \( 5y + 1 \), we must use the properties of a parallelogram.
In a parallelogram, opposite sides are equal in length. Since DH and HE are diagonals splitting the parallelogram into two congruent triangles, the segments on each diagonal will be equal due to the properties of the diagonals bisecting each other. That means:
\[ DH = HF \]
\[ GH = HE \]
1. **Equate DH and HF:**
\[ x + 2 = 2y \]
2. **Equate GH and HE:**
\[ 3x - 3 = 5y + 1 \]
Now we can solve these two equations simultaneously:
From equation \(1\):
\[ x + 2 = 2y \]
So:
\[ x = 2y - 2 \tag{3} \]
Substitute equation (3) into equation (2):
\[ 3(2y - 2) - 3 = 5y + 1 \]
\[ 6y - 6 - 3 = 5y + 1 \]
\[ 6y - 9 = 5y + 1 \]
\[ 6y - 5y = 1 + 9 \]
\[ y = 10 \]
Now, substitute \( y = 10 \) back into equation (3):
\[ x = 2(10) - 2 \]
\[ x = 20 - 2 \]
\[ x = 18 \]
Thus, the values of \( x \) and \( y \) are:
\[ x = 18, y = 10 \]
So, the correct answer is:
\[ \boxed{x = 18, y = 10} \]
In parallelogram DEFG, DH = x + 2, HF = 2y, GH = 3x – 3, and HE = 5y + 1. Find the values of x and y.
x = 20, y = 11
x = 11, y = 20
x = 18, y = 10
x = 10, y = 18
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