High School

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Practice proving that a quadrilateral is a parallelogram.

In quadrilateral [tex]WXYZ[/tex], [tex]WC = 2x + 5[/tex] and [tex]CY = 3x + 2[/tex]. What must [tex]x[/tex] equal for quadrilateral [tex]WXYZ[/tex] to be a parallelogram?

[tex]x = \square[/tex]

Answer :

To determine what [tex]\( x \)[/tex] must be for quadrilateral [tex]\( WXYZ \)[/tex] to be a parallelogram, we need to use the property that opposite sides in a parallelogram are equal. The segments [tex]\( WC \)[/tex] and [tex]\( CY \)[/tex] represent parts of opposite sides, and thus, they must be equal in length.

We have the following given expressions for these segments:
- [tex]\( WC = 2x + 5 \)[/tex]
- [tex]\( CY = 3x + 2 \)[/tex]

Since [tex]\( WC \)[/tex] and [tex]\( CY \)[/tex] need to be equal, we set up the equation:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]

To solve for [tex]\( x \)[/tex], follow these steps:

1. Subtract [tex]\( 2x \)[/tex] from both sides to isolate terms involving [tex]\( x \)[/tex] on one side:
[tex]\[ 5 = x + 2 \][/tex]

2. Subtract 2 from both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ 5 - 2 = x \][/tex]
[tex]\[ x = 3 \][/tex]

Therefore, for [tex]\( WXYZ \)[/tex] to be a parallelogram, [tex]\( x \)[/tex] must equal 3.

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