Two Questions :)

[Courtneydb]

First off, I'm not sure which section of the forum these questions should be in, or even if they should be in the same section, but I took my best guess, sorry if I'm wrong.

1. Solve for x and y. Both variables must satisfy the equation.

Part One: 3^{(3(x+2y)-2(x-2y)-y)} = 81
I simplified it like this:
3^{(3x+6y-2x+4y-y)} = 81
3^(x+9y) = 81

Part Two: 9^{((2x+y)-(x+2y))} = 243
I simplified it like this:
9^(2x+y-x-2y) = 243
9^(x-y) = 243

I'm fairly sure that for this, I'm supposed to use elimination OR substitution, but if anyone can just explain to me a way to understand this, I'd really appreciate it. I don't need the answer, it's in the back of the textbook, but I really want to understand how to do this. I'm pretty lost at this point, I'm not sure where to go from here. Thanks
(By the way, if you want it, the answer is x=53/20 and y=3/20)


2. This question isn't from my textbook, so I don't have the answer, but I'd really appreciate help with it Here is the picture of the shape. I had to copy it down off the blackboard, so it IS NOT to scale. Side FG is equal to side GA. Side AB is 10m. Side FE is 20m. Line CD is the diagonal line in the middle.

The total area is 280m2
the area of shape AGFED = the area of shape ABCD (so, 140m2)

What is the length of CD?

So far, I'm fairly lost with this question. I've tried a bunch of different ways, and I've also spoken to a lot of my classmates. Nobody has figured out what this is.

Thank you

 

[Courtneydb]

First off, these are good problems for this forum. However, please post only 1 problem per thread.

Your first problem is solved most easily by recognizing that 81 = 3⁴. But a^b = a^c means b = c unless a = 0 or a = 1.
So 3^x+9y = 3⁴ means x + 9y = 4, or x = 4 - 9y.

So, 9^x-y = 9^4-9y-y = 9^4-10y = 243 = 3 * 81 = 3 * 9². Divide both sides by 9².

So 9^2-10y = 3 = 9^1/2.

10y = 2 - (1/2) = 3/2.
y = 3/20.
x = 4 - (9 * 3)/20 = (80 - 27)/20 = 53/20.

The whole point of the problem is to turn a problem involving exponents into a linear problem by finding a common base (IF THERE IS ONE).

I suspect you can solve your other problem if you extend the line GA to the line CE at point H. If that hint is not enough, please come back and say what you have done and where you are stuck.

Thank you for helping me with the first question I think I understand it now.

I tried to extend that line, turning that space into a rectangle and a right angle triangle. The formula I came up with was 10x+(xy/2)=140 x being the base (Line GA to point H) and Y being the height of the triangle (point H to point D). From there, I'm not sure where to go.

 

[Deleted member 4993]

First off, I'm not sure which section of the forum these questions should be in, or even if they should be in the same section, but I took my best guess, sorry if I'm wrong.

1. Solve for x and y. Both variables must satisfy the equation.

Part One: 3^{(3(x+2y)-2(x-2y)-y)} = 81
I simplified it like this:
3^{(3x+6y-2x+4y-y)} = 81
3^(x+9y) = 81

Part Two: 9^{((2x+y)-(x+2y))} = 243
I simplified it like this:
9^(2x+y-x-2y) = 243
9^(x-y) = 243

I'm fairly sure that for this, I'm supposed to use elimination OR substitution, but if anyone can just explain to me a way to understand this, I'd really appreciate it. I don't need the answer, it's in the back of the textbook, but I really want to understand how to do this. I'm pretty lost at this point, I'm not sure where to go from here. Thanks
(By the way, if you want it, the answer is x=53/20 and y=3/20)


2. This question isn't from my textbook, so I don't have the answer, but I'd really appreciate help with it Here is the picture of the shape. I had to copy it down off the blackboard, so it IS NOT to scale. Side FG is equal to side GA. Side AB is 10m. Side FE is 20m. Line CD is the diagonal line in the middle.
View attachment 1383
The total area is 280m2
the area of shape AGFED = the area of shape ABCD (so, 140m2)

What is the length of CD?

So far, I'm fairly lost with this question. I've tried a bunch of different ways, and I've also spoken to a lot of my classmates. Nobody has figured out what this is.

Thank you

First calculate the length of FG (= x)

20*(10+x) - 10*x = 280 → x = 8

Now - I would extend BA to FE to meet at H and assign ED = y.

Continue further....

 

[Courtneydb]

First calculate the length of FG (= x)

20*(10+x) - 10*x = 280 → x = 8

Now - I would extend BA to FE to meet at H and assign ED = y.

Continue further....

Alright, thank you. So that makes FG and GA 8m each, which makes that little square part to the left 64m². Total area is 280, so 280-64=216
Now, there's just a rectangle here. It's 12*18, which is 216 so that works. There's the smaller rectangle on the bottom, which is 10*12, so that's 120. 216-120=96. The top rectangle is 96m2. Here's where I get lost again, I'm not sure where to go from here. Am I on the right track?
CE is 18m, so CD is shorter than 18...

 

[Deleted member 4993]

Area AGFEDA = Square AGFH + Trapezoid AHED = 140