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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]\[ x = 5 \][/tex]

Sagot :

GinnyAnswer
To determine the value of [tex]\( x \)[/tex] that makes quadrilateral [tex]\( WXYZ \)[/tex] a parallelogram, given the equations [tex]\( WC = 2x + 5 \)[/tex] and [tex]\( CY = 3x + 2 \)[/tex], we need to understand one fundamental property of parallelograms: opposite sides are equal.

In our problem, the segments [tex]\( WC \)[/tex] and [tex]\( CY \)[/tex] are opposite sides of the parallelogram. Therefore, we set their lengths equal to each other to form an equation and find [tex]\( x \)[/tex].

1. Set the expressions for the sides equal to each other:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]

2. Next, we need to isolate [tex]\( x \)[/tex]. First, subtract [tex]\( 2x \)[/tex] from both sides of the equation to eliminate [tex]\( x \)[/tex] from the left side:
[tex]\[ 5 = x + 2 \][/tex]

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

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