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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

3. Tests for Parallelograms

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Exercise 21 Page 501

Recall the theorem referenced in the book that states if a quadrilateral is a parallelogram, then its diagonals bisect each other.

x=11
y=7

Practice makes perfect

We want to find the values of x and y for which quadrilateral is a parallelogram, using the given algebraic expressions for the segment lengths.

Recall the Parallelogram Diagonal Theorem states that if a quadrilateral is a parallelogram, then its diagonals bisect each other. Therefore, the following segments are congruent. 3y+5=2x+4 and 2y+3=4y-11 Let's create a system of equations. 3y+5=2x+4 & (I) 2y+3=4y-11 & (II) We will solve the system by using the Substitution Method. Let's start by solving Equation (II).

3y+5=2x+4 & (I) 2y+3=4y-11 & (II)

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Solve by elimination

(II): LHS-2y=RHS-2y

3y+5=2x+4 3=2y-11

(II): LHS+11=RHS+11

3y+5=2x+4 14=2y

(II): .LHS /2.=.RHS /2.

3y+5=2x+4 7=y

(II): Rearrange equation

3y+5=2x+4 y=7

We found that y=7. Now we will find the value of x by substituting this value in Equation (I).

3y+5=2x+4 y=7

(I): y= 7

3( 7)+5=2x+4 y=7

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Solve for x

(I): Multiply

21+5=2x+4 y=7

(I): Add terms

26=2x+4 y=7

(I): LHS-4=RHS-4

22=2x y=7

(I): .LHS /2.=.RHS /2.

11=x y=7

(I): Rearrange equation

x=11 y=7

Subchapter links

Exercises p.499-503