# Some math assistance needed again

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## 24 Replies - 1492 Views - Last Post: 31 March 2010 - 09:56 AM

### #1 FrozenSnake

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# Some math assistance needed again

Posted 28 March 2010 - 12:53 PM

This is a new chapter for me and I have the key right here so finding the solution isnt hard (I just need to flip to the page with the key) but I wanna understand how I can find the solution myself.

This is the question
"Enter lines equations in the formula y = kx + m"

Check the attachments for pictures on my current task.

Anyone that can tell me how to find this out?
Simple calc wont do I need to understand each step you take to find the answer!

#### Attached image(s)

This post has been edited by FrozenSnake: 28 March 2010 - 12:57 PM

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## Replies To: Some math assistance needed again

• Saucy!

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## Re: Some math assistance needed again

Posted 28 March 2010 - 01:41 PM

Bearing in mind it's been 20 years or so since I've had to solve Cartesian plane questions, shouldn't you just need to pick two points on each line, calculate dy/dx to determine the slope and add the y-intercept? There's enough information in each pic to choose two points on each line.

### #3 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 01:44 PM

y = mx + b <- use these variables it's the standard

b = y intercept
m = slope

Use the equivalent equations to find 'm' and 'b'.

• Saucy!

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## Re: Some math assistance needed again

Posted 28 March 2010 - 01:47 PM

Looks like his book uses y = kx + m...must be a Swedish thing

EDIT: And actually, I'm not sure in the second one if 0,1 refers to the x-intercept of line d or the graph line at that point; my assumption is the latter.

### #5 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 01:49 PM

How dare they take our letters!

### #6 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 01:58 PM

The answers are suppose to be
a) y = 10x + 10
y = -5x - 5
c) y = 20x + 6
d) y = 60x - 2

If that is any help

• Saucy!

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:10 PM

I do believe I've done been bit by European decimal nomenclature! 0,1 == 0.1! Taking that into consideration, those are the answers I got.

### #8 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:22 PM

ok, but where do I start? Where each line meet the Y-axle and end where they meet?

### #9 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:27 PM

Oh you were still asking? I thought that was you informing us of your victory.

To find out slope:

slope = rise/run = (y2-y1)/(x2-x1)

So pick two points off of the graph

y intercept:

Plug in the point where x = 0, you'll get 'b'

In the picture on the left, the line with positive slope (up and to the right):

You can see the y-intercept right on the graph, it's 10.

Slope: two points (0,10) and (-1, 0)

= (10-0)/(0-(-1))
= 10

y = 10x + 10

### #10 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:31 PM

On C I took
Delta x = 0,2
Delta y = 4

4 / 0,2 = 20

I guess I did correct then?
But

2 / 0,1 = 20 too

This is very confusing, everyone seems to have their own way to do this.
But I think understand it a little now. Thanks

instead of making a new thread I keep this but with another problem
I have to find which on of these are parallell with eachother.

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

My calcs are

a )
-16 / 8 = -2
x = -2
-16 / 4 = -4
y = -4

b )
-8 / 8 = -1
x = -1
-8 / -2 = 4
y = 4

c )
-2 / 24 = -0,08333...
x = -0,08333...
-2 / -6 = 0,333...
y = 0,333...

d )
-18 / 6 = -3
x = -3
-18 / 3 = -6
y = -6

And the key say that 'a' and 'd' are parallell with eachother and 'b' and 'c' are parallell with eachother
So I guess I do something wrong.

This post has been edited by FrozenSnake: 28 March 2010 - 02:48 PM

### #11 macosxnerd101

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:43 PM

On C, this is what I did:
y-intercept: (0, 6)
Other Point (3, 10)

So m = (10-6)/(3-0) = 4/3

From there, I just use point slope formula: y - y1 = m(x-x1). So:
y - 10 = 4/3(x-3)

When you simplify, you get:
y - 10 = 4x/3 - 4
y = 4x/3 + 6

### #12 FrozenSnake

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:55 PM

macosxnerd101, on 28 March 2010 - 09:43 PM, said:

On C, this is what I did:
y-intercept: (0, 6)
Other Point (3, 10)

So m = (10-6)/(3-0) = 4/3

From there, I just use point slope formula: y - y1 = m(x-x1). So:
y - 10 = 4/3(x-3)

When you simplify, you get:
y - 10 = 4x/3 - 4
y = 4x/3 + 6

But the answer on C is suppose tp be 20x + 6 not '4x/3 + 6' maybe it's the samething?

### #13 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:57 PM

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.

### #14 macosxnerd101

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## Re: Some math assistance needed again

Posted 28 March 2010 - 02:58 PM

macosxnerd101, on 28 March 2010 - 05:43 PM, said:

On C, this is what I did:
y-intercept: (0, 6)
Other Point (2, 10)

So m = (10-6)/(2-0) = 4/3

From there, I just use point slope formula: y - y1 = m(x-x1). So:
y - 10 = 2(x-2)

When you simplify, you get:
y - 10 = 4x - 4
y = 4x + 6

Whoops- I counted over one too far on the x-axis for the point (3,10). I corrected my work, but I still didn't get the answer you are expecting. I'd like to see the work of whoever came up with that answer.

### #15 KYA

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## Re: Some math assistance needed again

Posted 28 March 2010 - 03:00 PM

macosxnerd101's points are off (x axis is in .1 increments)

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

Using (0,6) and (-.3, 0)

m = (6-0)/0-(-.3) = 6/.3 = 20

y - 6 = 20(x-0)
y = 20x + 6

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