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Answer :
To prove that quadrilateral [tex]\(WXYZ\)[/tex] is a parallelogram, we need to ensure that its opposite sides are equal. In this case, we are looking at sides [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex].
We are given:
- [tex]\(WC = 2x + 5\)[/tex]
- [tex]\(CY = 3x + 2\)[/tex]
Since these two sides must be equal for [tex]\(WXYZ\)[/tex] to be a parallelogram, we set the expressions for [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] equal to each other:
[tex]\[
2x + 5 = 3x + 2
\][/tex]
Now, let's solve for [tex]\(x\)[/tex]:
1. Subtract [tex]\(2x\)[/tex] from both sides of the equation to get the terms with [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]\[
3 = x
\][/tex]
So, for quadrilateral [tex]\(WXYZ\)[/tex] to be a parallelogram, [tex]\(x\)[/tex] must equal 3. This ensures that the opposing sides are equal, satisfying the conditions for a parallelogram.
We are given:
- [tex]\(WC = 2x + 5\)[/tex]
- [tex]\(CY = 3x + 2\)[/tex]
Since these two sides must be equal for [tex]\(WXYZ\)[/tex] to be a parallelogram, we set the expressions for [tex]\(WC\)[/tex] and [tex]\(CY\)[/tex] equal to each other:
[tex]\[
2x + 5 = 3x + 2
\][/tex]
Now, let's solve for [tex]\(x\)[/tex]:
1. Subtract [tex]\(2x\)[/tex] from both sides of the equation to get the terms with [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]\[
3 = x
\][/tex]
So, for quadrilateral [tex]\(WXYZ\)[/tex] to be a parallelogram, [tex]\(x\)[/tex] must equal 3. This ensures that the opposing sides are equal, satisfying the conditions for a parallelogram.
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