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The figure is parallelogram D E F G with diagonals that intersect at point H
In parallelogram DEFG, DH = x + 1, HF = 3y, GH = 3x − 4,
and HE = 5y + 1. Find the values of x and y. The diagram is not drawn to scale.
(2 points)
Since DHEG is a parallelogram with diagonals that intersect at point H, we can use the properties of parallelograms to find the values of x and y.
In a parallelogram, opposite sides are equal. Therefore, we have the following equations:
DH = GH
x + 1 = 3x - 4
Simplifying the equation, we get:
2x = 5
x = 5/2
Now, let's look at the equation for the other pair of opposite sides:
HE = GF
5y + 1 = 3x + 4
Substitute the value of x we found:
5y + 1 = 3(5/2) + 4
5y + 1 = 15/2 + 8/2
5y + 1 = 23/2
Subtract 1 from both sides:
5y = 23/2 - 2/2
5y = 21/2
Divide both sides by 5:
y = (21/2) / 5
y = 21/10
Therefore, x = 5/2 and y = 21/10.
To find the values of x and y, we can use the fact that the diagonals of a parallelogram bisect each other.
Step 1: Set up two equations using the given information.
Since the diagonals bisect each other, we have:
DH = HG
HF = HE
Substituting the given values, we get:
x + 1 = 3x - 4 (Equation 1)
3y = 5y + 1 (Equation 2)
Step 2: Solve Equation 1 for x.
Subtract x from both sides:
1 = 2x - 4
Add 4 to both sides:
5 = 2x
Divide both sides by 2:
x = 5/2
Step 3: Solve Equation 2 for y.
Subtract 5y from both sides:
-2y = 1
Divide both sides by -2:
y = -1/2
Step 4: Check the solution.
Substitute the values of x and y back into the equations to check if they satisfy the original conditions.
For Equation 1:
x + 1 = 3x - 4
5/2 + 1 = 3(5/2) - 4
7/2 = 15/2 - 4
7/2 = 7/2
For Equation 2:
3y = 5y + 1
3(-1/2) = 5(-1/2) + 1
-3/2 = -5/2 + 1
-3/2 = -3/2
Since both equations are satisfied, the values of x and y are:
x = 5/2
y = -1/2
To find the values of x and y in parallelogram DEFG, we can use the properties of diagonals in a parallelogram.
Given that DH = x + 1, HF = 3y, GH = 3x - 4, and HE = 5y + 1, we can set up the following equations:
1) DH + HG = DG (property of diagonals in a parallelogram)
(x + 1) + (3x - 4) = DG
2) HF + FE = HE (property of diagonals in a parallelogram)
3y + (5y + 1) = HE
By solving these equations simultaneously, we can find the values of x and y.
1) DH + HG = DG
(x + 1) + (3x - 4) = DG
4x - 3 = DG (Simplify)
2) HF + FE = HE
3y + (5y + 1) = HE
8y + 1 = HE (Simplify)
Now, since DEFG is a parallelogram, its opposite sides are congruent. Therefore, DG = HE.
So, we can set up the equation:
4x - 3 = 8y + 1
Simplifying this equation, we get:
4x - 8y = 4
Now, let's substitute the value of DG (which is equal to HE) from equation 2 into equation 1:
4x - 3 = 8y + 1
4x = 8y + 4
x = 2y + 1
Now we have a system of equations:
4x - 8y = 4
x = 2y + 1
To solve this system, we can substitute the value of x from the second equation into the first equation:
4(2y + 1) - 8y = 4
8y + 4 - 8y = 4
4 = 4
Since the equation is true, this means that the system is dependent and has infinitely many solutions. This means that there are infinitely many possible values for x and y that satisfy the given conditions.