Aly's ~ Help Me With Math Thread ; ;

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Where's my literature thread? Harumph.

10+7= a number that has yet to exist. I'm out.

*** limits

That's basically trial and error. If y = 46 you'd spend several minutes trying to figure out what to sub in for y instead of just rearranging the equation and solving for it. It's fairly obvious (to those of us comfortable with math), but to someone learning algebra it's neither obvious nor reliable.

y/5 + 3/4 = y/2 + 3/4

3/4 is equal to 3/4, so y/5 must equal y/2. (if a + b = c + b, then a = c)

so we get y/5 = y/2, which only makes sense if y = 0. (EDIT: otherwise, it doesn't make sense that a fifth of one number is the same as a half of that number.)

just adding words to the ideas that were already expressed by others

Still not the best way of explaining it, I think.

y/5 + 3/4 = y/2 + 3/4. Move all instances of Y to the left side, everything else goes to the right.

y/5 - y/2 = 3/4 - 3/4. When you move things to the other side of the equation, their sign changes.

y/5 - y/2 = 0. Now we need a common denominator. The least common denominator here is 10 (figure out the smallest number that all denominators will divide into evenly). If you're comfortable working with fractions, postpone this step as long as possible, so you have fewer terms to manipulate. You can do this at the start, as Kalilla did, but then you're working with larger numbers.

2y/10 - 5y/10 = 0. Now do the subtraction to get down to one y term.

-3y/10 = 0. Now we need to clear the denominator by multiplying both sides of the equation by 10.

10 x -3y/10 = -3y, 0 x 10 = 0. Therefore, we now have -3y = 0. Now we need to clear that coefficient. We need to divide both sides of the equation by -3.

-3y/-3 = y, 0/-3 = 0. Therefore, we're left with y = 0.

This is my reasoning and fastest in solving linear algebra equations.

y/5 + 3/4 = y/2 + 3/4

Shift like terms to one side and the rest the other side:

y/5 - y/2 = 0

LCD or Lowest denominator is 10 so multiply accordingly:

(2/10)*y - (5/10)*y = 0
-3/10*y = 0

Multiply both sides by -10/3

y = 0

I think this is the easiest way to approach it. When you get to use to it you can skip steps and solve in your head.

The other approach is trial and error by inserting numbers but this can be tedious unless you are given answer choices already.

It use to annoy people in my GRE class for the math section because the instructor would ask the class what the answer and say the answer out load real quick and people would ask how do you get it so fast. I stopped doing that but people have their pro's and cons. My verbal sucks so it evens out with the math.

Math is not fun! Took me three tries to pass 1st quarter calc! Had to grovel to my dean to bypass the calc requirement

Now Organic Chemistry! That's is FUN!

just because you don't like one way of doing the problem doesn't mean one way is better than the other. i just explained, in a few words, one method.

Lorzy it may seem your logic looks easy to you but explaining that to a student would get them confused.
y/5 = y/2 (can not just say it make sense). I can plug in any number and say y/5 =/= y/2

You really should simplify it reduced terms like:
-(3/10)*y = 0
Multiply both sides by -10/3 to cancel terms:

One should obtain:
y = 0*(-10/3) = 0

Whether or not I like your approach is irrelevant. That approach boils down to "well obviously y is 0 because y/5 has to = y/2." That's not helpful to someone trying to learn algebra. That's a method of trial and error.

You can't say "it only makes sense that y = 0" to someone trying to learn algebra, because it probably doesn't make sense to them. How did you get to your solution?

Had the answer not been posted with the question (which is how I advise the OP to post next time, to avoid the "cheating" we see here), it's likely we'd see far fewer posts reverse-engineering the question to arrive at the answer.

3y - 46 = y/7 + 18. Can you do this in two steps?

Ok that equation was already done :/

But here is a new one I'm confused on...

Find the perimeter and area for each figure (use pi = 3.14)



I have the answers for both but I only need to understand how to get the perimeter. I have the details for the area answer, so dont need that.

Answers:
A = 48cm^2

P= 28cm

When I did the perimeter...I got P=4s...= 4(6) = 24cm but it doesnt seem right if I just add 4 from the middle of the triangle.

So how do you get 28cm?

I'm so confused ; ;

the question was about the specific problem. i explained one way to solve that specific problem. yes, you can do it in other ways, which people explained, and i never said that this method was better than any others. i did add the edit in after you posted, because i think the idea does make sense, but maybe not immediately obvious.

i think it's good that people posted standard procedures for single variable equations, but textbook-type problems may also be geared towards noticing similar trends.

edit: i disagree with the "trial and error" part. you may not like it, and it's hopefully clearer with my edit, but it just doesn't make any sense for a fifth of any nonzero number to be the same as a half of said number.

the perimeter is only the outside. you add 6 for 3 sides of the square and 5 for the two sides of the triangle.

edit: too lazy to upload an mspaint so
5/\ 5
6|_| 6
6
edit: i reread your post, and saw your P=4s. that's the formula for the perimeter of a square. just think of the perimeter as the distance around the outside. it's easier to think that way imo than to use a bunch of fancy formulas.

This looks just like that "ICanLearn" computer math class *** my college uses..

Math makes this pussy moist tween my legs!

But anyway, a good accountant will tell you 2+2=4.
A better accountant will ask 2 what plus 2 what...
But the best accountant will ask you what number you were looking for on your business ledger!

Hilarious comment*!

A solution for one specific problem is next to worthless. Unless that specific question shows up on their test, the OP is hosed. If you try to solve the equation I gave you using your method it won't work. What's far more useful is teaching the person how to step through the problem and solve it regardless of the numbers.

if a + c = b + c, then a must = b is true, but if the question is 5 + 4 = 1 + 4, what does the OP do? It's far more effective to teach them how to step through it themselves in a structured manner, and once they're comfortable with what they're doing, they can start skipping steps and using pattern recognition to expedite the process. Until then, your solution is about as much help to them as the answer in the back of the textbook is.

As for OP's new question...

The perimeter of any polygon is equal to the sum of its outer sides. The base of the triangle is not on the outer edge of the polygon, so it's excluded from the calculation. What you end up with is:

P = 6 + 6 + 6 + 5 + 5 = 28.

You omit the 4 in the middle of the triangle because that's the height of the triangle, not an exterior side.

To calculate the area of a polygon, break it down into shapes you know how to calculate the area of. In this case, you can see the polygon can be broken down into a square and a triangle. So, to calculate the area of the polygon, we need the area of the square and the area of the triangle.

A(square) = s^2. The area of a square is equal to the side of the square, squared.

A = 6^2 = 36 cm squared.

The area of a triangle is equal to the product of the base of the triangle multiplied by the height of the triangle, divided by two.

A(triangle) = (b x h)/2. The base of the triangle is 6 cm, the height of the triangle is 4 cm.

A = (6 x 4)/2
A = 24/2
A = 12

So, the area of the polygon is equal to the sum of the area of the square plus the area of the triangle.

A(polygon) = A(square) + A(triangle)
A = 36 + 12
A = 48 cm squared.

i'm not quite sure why you've been attacking my posts; i did note the worth of your post, and did try to make my own post clearer. there are different ways to do a problem, and while valuable to have a standard way to approach problems, i disagree that it's worthless to have an unorthodox way.

the a + c = b + c thing is a standard rule of mathematics, so i'm not sure why that part is an issue. the second part of my original post, i admitted might be hard, but i did my best to explain it.

i fully realize the value in knowing how to do the standard procedure, but that was already explained many times since the beginning of the thread. i don't see any problem with offering an additional alternative solution.

i still don't understand why offering an alternative is a problem.

itt: my way or the highway

When calculating the area, it usually helps to separate the object into multiple smaller, simpler objects. for instance, with your picture, separate it into the the triangle and the square. Simply put, the equation for the area of that shape is (area of the triangle) + (area of the square) = (area of the whole).

I wont give you the equations for area :3

Perimeter is much simpler. the fact that Pi = 3.14 is irrelevant here, since we're not dealing with a circle. Perimeter is the outside rim of the object, so only the walls (if you will) that are facing out are counted. how many surfaces are there? when you figure that out and their lengths, the perimeter is simply the sum.

edit: beaten by a lot QQ

If I appear hostile, I apologize. That's not how I want to come across. It just seems to me that you aren't understanding why I said it wasn't the best way of explaining the solution to the OP. Your post, to me, was the equivalent of a baseball coach on the first day of practice picking up a bat and crushing a home run and then telling his players to do that. That's great, but first you need to teach the kids how to hit the ball. Crushing the home run comes later. To put it another way, you gave them a torch, but you didn't show them how to light a fire.

It's not worthless to have multiple ways to solve a problem (different techniques are better for different problems), but I do think it's worthless to show people those alternate methods when they rely on an understanding of the method those alternate approaches are built upon (the "standard" approach), which the OP clearly lacks. For instance, you need to understand addition before you can do multiplication.

Start with your triangles.

You'll need the Pythagorean Theorem

A²+B²=C² For Left and Right Triangle

A²+4²cm=5²cm

A²+16cm=25cm

A²=25cm - 16cm

A²= The Square Root of 9

A= 3

Now Add All the Sides of Each Triangle and the square to get the area.

5 + 4 + 3= 12
5 + 4 + 3= 12

6 + 6 + 6 + 6 or 6 x 4 = 24.

To get the entire objects perimeter.

5 + 5 + 6 + 6 + 6= 28

that is not how to find the area.

She said she was looking for the perimeter...

Not even close, and I'm not even sure what you were trying to do here, because all relevant sides of both polygons are already given to you.

you don't need to do any calculations to find what the lengths of the sides are. they're written on the figure.

HAHAHA I didn't mean area.... lol. I had A in my head.

And yes they were all given but I guess I went farther than I needed too. But those are the perimeters.