Question

How do I solve the below square?

Answer 1

Help !!

Answer 2

What's the question?

Answer 3

Hmm, let me think for a moment. I'll let you know whether or not I can figure it out.

Answer 4

Thank you (:

Answer 5

So, I'm just going to get down what I know while I'm trying to solve, so anything that I type is me just trying to figure it out, and should not be taken as an answer to your question.

Answer 6

Thank you (:

Answer 7

Sorry, I can't figure it out. :(

Answer 8

sighh its fine

Answer 9

@mathmate

Answer 10

It seems to be a rectangle. If that's the case, we could equate the top and bottom sides first to get our y-value:
\[y + 3 = 2y - 4\]\[7 = y\]
Then, we could plug this y-value to the length and width equations to get our dimensions for its area:
\[2(7) + 4 = 18\]\[(7) + 3 = 10\]\[L \times W = 18 \times 10\]
The rectangular area is 18 units by 10 units.

Answer 11

It says square on the test..

Answer 12

You left out that important detail. (sigh)

Answer 13

The main question says how do I solve the square !! lmao

Answer 14

If it's a square like you said, then we equate two perpendicular sides:
\[y + 3 = 2y + 4\]\[-1 = y\]
Then we plug this y-value back into one of the three equations to get our dimensions:
\[(-1) + 3 = 2\]
\[L \times W = 2 \times 2 = 2^{2}\]
Our square is 2 units by 2 units.

Answer 15

Please specify geometrical shapes in the future. Otherwise, you may get reported.

Answer 16

THANK YOU THIS IS WHAT I NEEDED.

Answer 17

I did..

Answer 18

@badlandsarah
The figure is not a square, unless there are typos in your figure.
If it is a square, 2y+4=2y-4
after cancelling y, 4=-4 which is absurd.
I think your best bet is to talk to show that it is absurd as a square, and solve it as a rectangle. THEN report to your teacher about the question.

Answer 19

See, that was part of my problem. I couldn't figure it out as a square.