Question

1. in parallelogram defg, dh = x + 3, hf = 3y, gh = 2x - 5, and he = 5y + 2. find the values of x and y.

x = 36, y = 13

x = 39, y = 14

x = 14, y = 39

x = 13, y = 36

Explanation:

Step1: Use property of parallelogram diagonals

In a parallelogram, the diagonals bisect each other. So, $DH = HF$ and $GH=HE$.
We get the equations: $x + 3=3y$ and $2x−5 = 5y+2$.

Step2: Rearrange the first - equation

From $x + 3=3y$, we can express $x$ in terms of $y$ as $x=3y - 3$.

Step3: Substitute $x$ into the second - equation

Substitute $x = 3y-3$ into $2x−5 = 5y+2$.
$2(3y - 3)-5=5y + 2$.
Expand the left - hand side: $6y-6 - 5=5y + 2$.
$6y-11 = 5y + 2$.

Step4: Solve for $y$

Subtract $5y$ from both sides: $6y-5y-11=5y-5y + 2$.
$y-11 = 2$.
Add 11 to both sides: $y=2 + 11=13$.

Step5: Solve for $x$

Substitute $y = 13$ into $x=3y - 3$.
$x=3\times13-3=39 - 3=36$.

Answer:

$x = 36,y = 13$