Parallelograms

LN1

LN1

L - N - 1
So I had this Geometry homework, most of it done. However, I am still without the answer to this last two parallelograms.
7S8Q5Hc.png

(Vague drawing, sorry for terrible paint skills. If any clarity is needed on what goes where, I can tell so.)

Can anybody help get the x and y of each parallelogram(the x and y of the two parallelograms are not necessarily the same.)?

Thanks in advance.
 
For the bottom one:

2X + Y = 36
Y = 36 - 2X

4X - Y = 24
4X - (36 - 2X) = 24
4X - 36 + 2X = 24

6X = 60
X = 10

Then, Y = 16

I've got my own math to study, so I don't have time to help with the other one.
 
4x-y = 24
2x + y = 36

Isolate the variables.

For the second one

2x + y = 36
y = -2x + 36

So plug in the variable and solve for x.

4x-y = 24
4x -(-2x + 36) =24
4x + 2x - 36 = 24
6x - 36 = 24
6x = 60
x = 10

And then plug in x to solve y.

2x + y = 36
2(10) + y = 36
20 + y = 36
y = 16

Then double check to make sure.

2x + y = 36
2(10) + (16) = 36
20 + 16 =36
36 = 36

4x-y = 24
4(10) - (16) = 24
40 - 16 = 24
24 = 24

If you have any questions, you can always ask.

Next time, use the line tool to make your own parallelograms. It really helps.


Since the second one is already answered, I'll attempt the first one.

Here, I laid out some additional numbers for you. If you're wondering how I got those numbers, you can ask. But, otherwise, try solving it yourself.

fXp5hB0.png


But it could be nice if you provided clarity for where the (x-y)° actually goes.
 
(x-y)° does go in the 85° area of the triangle formed by a 45° angle, a 85° angle, and a 50° angle. Or if you want some technical terms, let's call the parallelogram PQRS( the problem doesn't actually give it a name, it just supplies the parallelogram), where P is the top left point, Q is the top right point, R is the bottom right point, and S is the bottom left point. (x-y)° is the measurement of the angle PQR. I will say I have already solved the other parallelogram myself, so thanks to both you and M4E.
 
Oooh, that makes it a bit easier.

Since (x-y) deals with the entire angle, you can do this:

(x + y) + (x-y) = 180
2x = 180
x = 90

Solving for y, I'm thinking what I can do.
 
But didn't you just tell me that (x - y) is the entire angle? And since (x + y) is another entire angle, this means that they're supplementary angles. And supplementary angles add to 180 degrees.
 
I guess I misworded it, so sorry. (x-y) is a part of the angle, and (x+y) is part of another. They both are part of the same triangle that has the exterior angle 130. Either way, I had solved that one already, but I have another one that I think I did something wrong because I somehow got 15y+3=15y+3(ergo, y=All Real Numbers, which I believe is not a valid answer for the measurement of the angles of a parallelogram).
UcECY4y.png


(The problem is technically "What must the value of x and y be for the quadrilateral to be a parallelogram?")
 
LuigiNummerEins said:
I guess I misworded it, so sorry. (x-y) is a part of the angle, and (x+y) is part of another. They both are part of the same triangle that has the exterior angle 130. Either way, I had solved that one already, but I have another one that I think I did something wrong because I somehow got 15y+3=15y+3(ergo, y=All Real Numbers, which I believe is not a valid answer for the measurement of the angles of a parallelogram).
UcECY4y.png


(The problem is technically "What must the value of x and y be for the quadrilateral to be a parallelogram?")
Click to expand...

If it's a parallelogram then:

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

So
3x-7y = -2 (L1)
-4x+9y = 1 (L2)

4(L1)+3(L2): -y = -5 so y = 5.
Then 3x =33 so x = 11.
 
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