# Some math assistance needed again

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## 24 Replies - 1520 Views - Last Post: 31 March 2010 - 09:56 AM

### #16 macosxnerd101

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:03 PM

Yah, it looked like a comma to me. That's why my answer was off.

Quote

Damn you European comma where you need a period conventions! *shakes fist*

```for(int i = 0; i < Double.POSITIVE_INFINITY; i++)
System.out.pritnln("This");

```

This post has been edited by macosxnerd101: 28 March 2010 - 03:04 PM

### #17 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:07 PM

KYA, on 28 March 2010 - 09:57 PM, said:

For the parallel question, all you need to do is find the slope and compare them.

Quote

Parallel lines and their slopes are easy. Since slope is a measure of the angle of a line from the horizontal, and since parallel lines must have the same angle, then parallel lines have the same slope — and lines with the same slope are parallel.

'a' and 'd' are parallell because each have twice the size?
a = -2 and -4
d = -3 and -6

And 'b' and 'c' I must I have done something wrong?
b's x = -1 doesnt look close to c's x = -0,08333...

So how can I "see" this?

### #18 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:10 PM

No, the slope has to be the same. Graph y = x and y = 2x for a visual, they are not parallel.

### #19 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:13 PM

KYA, on 28 March 2010 - 10:10 PM, said:

No, the slope has to be the same. Graph y = x and y = 2x for a visual, they are not parallel.

So
a ) x = -2, y = -4
d ) x = -3, y = -6
aren't parallell?

The key say they are but without numbers

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:16 PM

No, no. The formula is:

y = slope * x + y-intercept

So you need to take these:

Quote

a ) 8x + 4y + 16 = 0
b ) 8x - 2y + 8 = 0
c ) 24x - 6y + 2 = 0
d ) 6x + 3y + 18 = 0

and find what the slope is for each. If the slopes are the same, then the lines are parallel.

For example a, solving for y:
8x + 4y + 16 = 0
4y = -8x - 16 (subtract 8x and 16 from both sides of the equation)
y = -2x - 4 (divide each side by 4 to solve for y)

### #21 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:19 PM

a ) 8x + 4y + 16 = 0

8x + 16 = -4y
y = -2x - 4

b ) 8x - 2y + 8 = 0

2y = 8x + 8
y = 4x + 4

c ) 24x - 6y + 2 = 0

6y = 24x + 2
y = 4x + (1/3)

d ) 6x + 3y + 18 = 0

-3y = 6x + 18
y = -2x - 6

Check the slopes

y = mx + b, m is the slope

### #22 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:24 PM

JackOfAllTrades, on 28 March 2010 - 10:16 PM, said:

No, no. The formula is:

y = slope * x + y-intercept

So you need to take these:

Quote

a ) 8x + 4y + 16 = 0
b ) 8x - 2y + 8 = 0
c ) 24x - 6y + 2 = 0
d ) 6x + 3y + 18 = 0

and find what the slope is for each. If the slopes are the same, then the lines are parallel.

For example a, solving for y:
8x + 4y + 16 = 0
4y = -8x - 16 (subtract 8x and 16 from both sides of the equation)
y = -2x - 4 (divide each side by 4 to solve for y)

Oh, there was a example that was like this in the book:
Descide where the linje 4x - 2y + 20 = 0 cuts the coordinations axles.
1) The line cuts the y axle when x = 0
4 * 0 - 2y + 20 = 0
-2y + 20 = 0
20 = 2y
y = 10

The intersection point is (0, 10)
Why do they move -2y?

2) The line cuts the x axle when y = 0
4x - 2 * 0 + 20 = 0
4x + 20 = 0
4x = -20
x = -5

This was what I did, maybe I implement it in the wrong place?
This is very very confusing, but I must go to bed now have school in 8h and need to wake up in 6.
I check this thread again after school

This post has been edited by FrozenSnake: 28 March 2010 - 03:26 PM

### #23 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:25 PM

When you're looking for a specific variable you have to isolate it. The issue with those questions is that they aren't in the right form to check out the slope. Plugging in values to get points is too much work. Just rearrange the variables/values.

### #24 NeoTifa

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## Re: Some math assistance needed again

Posted 29 March 2010 - 07:44 AM

There are a few ways you can solve these.

y - y1 = m(x - x1)
Basically, for this you would find a point, and plug it in, solve for m, then plug it in to the y = mx + b equation, then solve for b.

m = (y2-y1)/(x2-x1) or m = rise/run
Find slope, plug it in, solve for b.

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This post has been edited by NeoTifa: 29 March 2010 - 08:00 AM

### #25 Meister_Chicho

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## Re: Some math assistance needed again

Posted 31 March 2010 - 09:56 AM

O_O O_O O_O O_O O_O

Are you comfortable with trigonometry? It can be solved with that too:

Find the tangents (which is "m" -slope- KYA is talking about) and compare. Equal tangents -> Parallel

Are you comfortable with algebra?

Treat the equations as if they were vectors. Create the basis-matrix and look how many rows are a combination of the others.

Anyway, you can solve that problem in many ways, it depends on how you treat the variables, but the best you can do if you don't know mathematics is to do what Kya and Neotifa told you: y-y1 = m(x - x1)