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Answer :
To determine what [tex]\( x \)[/tex] must equal for quadrilateral [tex]\( WKYZ \)[/tex] to be a parallelogram, we use the property that in a parallelogram, the opposite sides are equal in length.
Given:
- [tex]\( WC = 2x + 5 \)[/tex]
- [tex]\( CY = 3x + 2 \)[/tex]
For [tex]\( WKYZ \)[/tex] to be a parallelogram, we need the opposite sides to be equal, so we set the lengths equal to each other:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]
Now, let's solve for [tex]\( x \)[/tex].
1. Move the terms involving [tex]\( x \)[/tex] to one side of the equation. This can be done by subtracting [tex]\( 2x \)[/tex] from both sides:
[tex]\[ 5 = 3x - 2x + 2 \][/tex]
Simplifying the right side gives:
[tex]\[ 5 = x + 2 \][/tex]
2. Next, isolate [tex]\( x \)[/tex] by subtracting 2 from both sides of the equation:
[tex]\[ 5 - 2 = x \][/tex]
[tex]\[ 3 = x \][/tex]
Therefore, [tex]\( x \)[/tex] must equal 3 for quadrilateral [tex]\( WKYZ \)[/tex] to be a parallelogram.
Given:
- [tex]\( WC = 2x + 5 \)[/tex]
- [tex]\( CY = 3x + 2 \)[/tex]
For [tex]\( WKYZ \)[/tex] to be a parallelogram, we need the opposite sides to be equal, so we set the lengths equal to each other:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]
Now, let's solve for [tex]\( x \)[/tex].
1. Move the terms involving [tex]\( x \)[/tex] to one side of the equation. This can be done by subtracting [tex]\( 2x \)[/tex] from both sides:
[tex]\[ 5 = 3x - 2x + 2 \][/tex]
Simplifying the right side gives:
[tex]\[ 5 = x + 2 \][/tex]
2. Next, isolate [tex]\( x \)[/tex] by subtracting 2 from both sides of the equation:
[tex]\[ 5 - 2 = x \][/tex]
[tex]\[ 3 = x \][/tex]
Therefore, [tex]\( x \)[/tex] must equal 3 for quadrilateral [tex]\( WKYZ \)[/tex] to be a parallelogram.
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