## Proportions!

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Tue May 29, 2012 7:36 pm
thestorygirl says...

I realize this is kind of stupid question to ask since I didn't pay attention in Algebra, but I have absolutely no idea how to solve proportions. My math teacher won't answer me and no website will give me a straight answer.

(BTW, if you don't know what I mean by proportion I mean something like this.
1/2 = x/6)

I know how to find the answer in the problem above, but I can't do it when the variable is pn both sides of the equal sign. Help please!
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Tue May 29, 2012 8:26 pm
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Kale says...

I can't do it when the variable is pn both sides of the equal sign.

An example would be helpful.
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Thu May 31, 2012 2:41 pm
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barefootrunner says...

I think I may be able to help. I don't know what level of maths you have but I'll try to explain. If you mean like this:

2x + 3 = 5x - 1

What you want to do is to get all the xes on the same side. To do that, you have to remember: you can do anything to manipulate the proportion, but what you do on the one side of the =, you have to do on the other side.

Solve for x:

2x + 3 = 5x - 1

To remove the xes on the one side, neutralise the amount of xes on one side, on both sides:

2x + 3 - 2x = 5x - 1 - 2x (I chose to remove the 2x by subtracting it on both sides)

Then you get:

3 = 3x - 4

The stray xes are gone!

Now get the constants (numbers) on the same side:

3 + 4 = 3x - 4 + 4

Then you get:

7 = 3x

And then you solve it finally by dividing on both sides with the coefficient of the xes:

7 ÷ 3 = 3x ÷ 3

And you get:

7/3 = x

And that is the answer!

In your example, though, you had fractions. Let's give it a go.

Solve for x:

x/5 + 3x/2 = 3/4 - x/5

First thing is, remove the fractions. Remember, you can do whatever you like as long as you do it on both sides of the proportion.

To remove the fractions, find the smallest number that all the denominators can divide into, that is to say, the smallest common denominator.

In this case, you have the denominators 5, 2 and 4. Find the smallest number that is a multiple of all three of these. The answer is 20.
(Why 20:
1—not a multiple of any;
2—only of 2;
3—none;
4—only of 2 and 4;
5—only of 5;
6—only of 2;
7—none;
8—only of 2 and 4;
9—none;
10—only 2 and 5;
11—none;
12—2 and 4;
13—none;
14—only 2;
15—only 5;
16—only 2 and 4;
17—none;
18—only 2;
19—none;
20—2, 4 and 5!)

Now that you have found it, you can attempt mission 1: terminate the fractions!

x/5 + 3x/2 = 3/4 - x/5

Multiply straight through with that common denominator to remove the fractions:

(x/5 + 3x/2) x 20 = (3/4 - x/5) x 20

Let's look at each individual fraction:

x/5 x 20/1 = 20x/5 (now simplify) = 4x/1 = 4x

Fraction one terminated! Next:

3x/2 x 20/1 = 60x/2 (simplify) = 30x/1 = 30x

Fraction two obliterated! Keep going with the other two and you will get:

15 and 4x. Now put it all together in the proportion:

4x + 30x = 15 - 4x

And now it is just a regular proportion! Solve it by doing the same thing on both sides. (But you can simplify it first. Do you see how?):

4x + 30x = 15 - 4x (add the xes together)
34x = 15 - 4x (now remove the - 4x)
34x + 4x = 15 - 4x + 4x (do the math...)
38x = 15 (divide on both sides by the coefficient of the x)
x = 15/38

Mission accomplished! Basically, remove fractions, separate variables and constants, and solve for x.

I hope that was what you asked, and that it helped. If not, give an example like Kyllorac suggested and we might be able to help.
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Sat Jan 25, 2014 6:03 am
Ventomology says...

I know that it's been a while since this was posted, but I figured that maybe someone else would find it helpful.
You probably have something like this: 1/x=x/36, right?
In this case, you are looking for a number to satisfy both equations while keeping them equal, and I'm going to say that while the above equation could be solved through logical guessing (it equals 6), you're probably looking for a system for solving it.

Truth be told, no one actually covers this in Algebra 1! It's a Geometry subject used for finding the lengths of the altitudes of right triangles, among other things regarding hypotenuses and legs. (Dumb, I know, but school is wacky like that.)

So I solve proportions through cross multiplication. I'll demonstrate with one x first.
2/3=x/9 Then I multiply by 9 on both sides.
18/3=x Then we multiply by three on both sides.
18=3x Then divide
6=x

Proportions with the variable on both sides work exactly the same way, but with evil things like square roots.
1/x=x/36 First, cross-multiply.
36/x=x
36=x(x) Basically thirty-six equals x squared. Then, to get rid of x squared, find the square root of both sides.
x=6 Easy peasy! So what happens when we get icky numbers that aren't perfect squares? Here's an example:
2/x=x/9 First we cross-multiply
18/x=x
18=x squared Then, just like earlier, find the square root of both sides. 18 isn't a perfect square, but nine is, so you have to use the handy-dandy radical sign. When simplifying imperfect squares, we take divide the number by the square that fits in it, so 9x2=18 And then find the root of the square that we took out and put a radical over what's left over.
So in the problem we just did, x=3radical2
Yippee!
I hope you found this helpful.
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Thu Feb 06, 2014 2:54 pm
lostthought says...

I would've times the known number by (amount) and then times where the x is at by that number as well. Made no sense? Well, example_

2/5-x/15

/5 * 3 = /15

2/ * 3 = /6

Still makes no sense? Well, I'm at a loss at that
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Thu May 22, 2014 12:38 pm
1Weirdsituation says...

1/2 = x/6
All you do is add and multiply. 1 x 6 divided by 2. x= 3
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Thu May 22, 2014 12:39 pm
1Weirdsituation says...

maybe there's more to it but I learned that in the beginning of the year.
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