Answer :
To prove that a quadrilateral [tex]\(WXYZ\)[/tex] is a parallelogram, we need to show that either both pairs of opposite sides are equal or both pairs of opposite angles are equal, or one pair of opposite sides are both equal and parallel.
In this case, we are given the lengths of sides [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] with the equations:
[tex]\[ WC = 2x + 5 \][/tex]
[tex]\[ CY = 3x + 2 \][/tex]
For [tex]\(WXYZ\)[/tex] to be a parallelogram, opposite sides must be equal. Hence, we set the lengths [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] equal to each other:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]
To solve for [tex]\(x\)[/tex], we first isolate [tex]\(x\)[/tex] on one side of the equation. Start by subtracting [tex]\(2x\)[/tex] from both sides:
[tex]\[ 5 = x + 2 \][/tex]
Next, subtract 2 from both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ 3 = x \][/tex]
Therefore, to ensure that quadrilateral [tex]\(WXYZ\)[/tex] is a parallelogram, [tex]\(x\)[/tex] must be:
[tex]\[ x = 3 \][/tex]
In this case, we are given the lengths of sides [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] with the equations:
[tex]\[ WC = 2x + 5 \][/tex]
[tex]\[ CY = 3x + 2 \][/tex]
For [tex]\(WXYZ\)[/tex] to be a parallelogram, opposite sides must be equal. Hence, we set the lengths [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] equal to each other:
[tex]\[ 2x + 5 = 3x + 2 \][/tex]
To solve for [tex]\(x\)[/tex], we first isolate [tex]\(x\)[/tex] on one side of the equation. Start by subtracting [tex]\(2x\)[/tex] from both sides:
[tex]\[ 5 = x + 2 \][/tex]
Next, subtract 2 from both sides to solve for [tex]\(x\)[/tex]:
[tex]\[ 3 = x \][/tex]
Therefore, to ensure that quadrilateral [tex]\(WXYZ\)[/tex] is a parallelogram, [tex]\(x\)[/tex] must be:
[tex]\[ x = 3 \][/tex]
