What values must x and y have to make the quadrilateral a parallelogram.

Question

Answer 1

The opposite angles of the parallelogram are equal.

Eqn. 1: 7y - 2 = 3x

Eqn. 2: 4x - 1 = 9y

Rearranging:

Eqn. 1: 3x - 7y = -2

Eqn. 2: 4x - 9y = -1

Multiplying Eqn. 1 by 4 and Eqn. 2 by -3 we get:

+12x - 28y = -8

-12x + 27y = 3

When added this becomes:

-y = -5

y = 5

Using Eqn. 1 we can get y:

3x - 7(5) = -2

3x = 33

x = 11

Answer 2

depending on how the problem is written,

if its angles are 9y, 3x, 7y-2 and 4x-1 degrees

then x=25, y=11

you could just graph the two lines

3x = 7y-2 and 9y =4x-1 and find where they intersect

use a graphing calculator

or

3x =7y-2

12x = 28y -8 **

4x =9y+1

12x = 27y + 3 ****

subtract equation ** from ****

0=y-11

y =11

3x=7(11)-2=77-2=75

x =9(11)/4 +1/4 =100/4= 25

since that's the only solution that comes out to integers

and not a messy complex irrational solution, it's probably what you want

although that means angles of 6, 174, 6 and 174 degrees

with the diagram very not to scale

but alternative interpretations, depending on how it's written, how well your eyesight is,

include

x= about -.57

y = about -.43

x =-13/(28-3sqr3)

y= -52/(84-9sqr3)

those above numbers are found by the method below, but by solving for y first

then for x

but below's calculations have an error somewhere

but they illustrate the general method. Somewhere there is an arithmetic mistake

opposite parallelogram angles are equal

bottom left angle = top right angle

9y^2+3x^2 = (7y+2)^2+(4x+1)^2 (it's hard to read, but this looks like the 2 angles)

top left = bottom right, so other 2 angles of the interior triangles are also equal

9y^2+(7y+2)^2 = 3x^2+ (4x+1)^2

combine like terms in 1st equation, y's on left, x's on right

9y^2 -(7y+2)^2 =-3x^2 + (4x+1)^2

add to 2nd equation

18y^2 = 2(4x+1)^2

divide by 2

9y^2= (4x+1)^2

take square roots

3y= 4x+1 or -4x-1

solve for x

x= 3y/4 - 1/4 or -1/4 -3y/4

9y^2+3x^2=(7y+2)^2+(4x+1)^2

9y^2 = (4x+1)^2

subtract to get

3x^2=(7y+2)^2

take the square root

xsqr3= 7y+2 or -7y-2

x= (7y+2)/sqr3= the 1st solution for x = 3y/4-1/4 or -1/4-3y/4

set the two equal

(7y+2)/sqr3= 3y/4 -1/4 or (7y+2)/sqr3 = -1/4-3y/4

multiply by sqr3

7y+2= 3ysqr3/4 -sqr3/4 or -sqr3/4 -3ysqr3/4

7y-3ysqr3/4= 2-sqr3/4 or 7y+3ysqr3/4 =2-sqr3/4

y(7-3sqr3/4)=2-sqr3/4 ory(7+3sqr3/4)= 2-sqr3/4

y=(2-sqr3/4)/(7-3sqr3/4) or (2-sqr3/4)/(7+3sqr3/4)

x=ysqr2/2 -1/4

x=(sqr2-sqr6/8)/(7-3sqr3/4)-1/4 or (sqr2-sqr6/8)/(7+3sqr3/4)

check the answers

y= about (2-2.45/8)/(7-3(1.732)/4) or (2-2.45/8)/(7+3(1.732/4)

= (2-.36)/(7-5.196/4) or (1.64)/(7+1.299)

= 1.64/(7-1.299) or 1.64/8.3

= 1.64/(7-1.3)=1.64/5.7 or

= .288 or ..198

x = about (1.414-2.45/8)/(7-1.3) -.25 or (1.414+2.45/8)/7+1.3)

= (1.414-.306)/5.7 or ( 1.72/8.3

= 1.108/5.7 or 1.72/8.3

=.189 or .207

9y^2 +3x^2 = about

9(.288)^2 + 3(.189)^2 or 9(.414)^2 +3(.207)^2

= .746+.107= .853 or 1.543+.129= 1.672

(7y+2)^2 +(4x+1)^2 = about

(7(.288)+2)^2 + (4(.189)+1)^2

= 4+3=7 which isn't close to .853

error(s) somewhere above

but if the problem really has (7x-2)^2 not (7x+2), then a very different answer

9y^2 + 3x^2 = (7x+2)^2 + (4x+1)^2

9y^2 -(7x-2)^2 = -3x^2 + (4x+1)^2

9y^2+(7x-2)^2 = 3x^2 +(4x+1)^2

add

18y^2 = 2(4x+1)^2

9y^2 = (4x+1)^2

3y = 4x+1

y = 4x/3 +1/3

3x^2 = (7y-2)^2

xsqr3 = 7y-2 = 7(4x/3 + 1/3)-2 = 28x/3 +7/3-2

xsqr3 -28x/3 = 7/3-6/3 = 1/3

x(3sqr3/3 -28/3) = 1/3

x(3sqr3 -28)/3 = 1/3

x = 1/(3sqr3 -28)=(3sqr3 +28)/(27-28^2)

y= 4/(3sqr3-28) +1/3 =

OR maybe the "2" exponent is really a degree symbol

then

9y+3x =7y+2 +4x+1

2y-x = 3

9y +7y+2 = 3x +4x+1

16y -7x = -1

16y -8x = 24

x =-25

y = x/2 +3/2 = -25/2+3/2

y = -22/2 = 11

opposite angle are then 99-75= 24 and 77-2-100-1=-26

which don't match and are impossible

or maybe the problem had angles 3x, 9y, 7y-2 and 4x-1 degrees

if so, then solution is:

y=11, x=25