# Problem with equation systems in math

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## 7 Replies - 899 Views - Last Post: 29 April 2010 - 04:48 PM

### #1 FrozenSnake

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# Problem with equation systems in math

Posted 29 April 2010 - 05:34 AM

I have a test tomorrow and most of it works great but some stuff I do not understand this is one of them. I will ask another question as well instead of posting a new thread on the problem. More problems might come up if it does I will post it as a "reply" and not in this topic-head.

First problem (attachment 1, if you cannot view the picture):

What am I doing wrong?

Second problem (solution in attachment 2, if you cannot view the picture):

Draw in the same coordination system the graphs for:
a ) y = 2|3x + 2 <-- clarification: the | should be _ with 2 above it and 3 under it
b ) y = -1|2x
c ) y = 2 <-- this one I understand

I have a problem with understanding how I know how to draw the line (right, left, steep, etc). Anyone that can try and clarify this?
My first on a) was like this (attachment 3)

#### Attached image(s)

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## Replies To: Problem with equation systems in math

### #2 LaFayette

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 07:21 AM

In your first problem you should subtract one of the equations from the other and not add them. Otherwise you won't zero out y.

In your other problem, when finding the graph of a first order polynomial just find two points and then draw a straight line going through both of them. A natural thing would be to find the first point by figuring out what y is when x is zero, and then the other by finding x when y is zero.

This post has been edited by LaFayette: 29 April 2010 - 07:24 AM

### #3 FrozenSnake

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 07:59 AM

Thats why I multiply them so I can remove both. If I subtract 2 i get -0.5y If I remove 1,5 i get 0.5y. so Y is still there.
If I multiply them both get 30 (in this case) and I can remove them. just like on the pic. Oh! I think I understand how you ment now
I should get a -30 instead of a +30 in one of them?

well I know that m = 2 but I don't know how to know how to draw k = 2|3x.

This post has been edited by FrozenSnake: 29 April 2010 - 08:04 AM

### #4 LaFayette

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 09:08 AM

Quote

Oh! I think I understand how you ment now
I should get a -30 instead of a +30 in one of them?

Yes, that's the idea behind what I meant. I seems to me like you are copying a solution from a previous problem without really understanding the idea. My suggestion is the really convince yourself why this work.

(Another thing, for future endevours, since x is alone on both sides here it's probably easier to do substitution and solve 2*y+5 = 1.5*y + 7.5 for y. Don't mind this if you find it confusing.)

Quote

well I know that m = 2 but I don't know how to know how to draw k = 2|3x.

Yes, m=2 since when x=0 y becomes 2. This tells you that the line will go through (0,2).
Secondly figure out what x is when y = 0 :
0 = (2/3)*x + 2
(2/3)*x = -2
x = (-2*3)/2 = -3

So (-3,0) is another point. Just line this two points up with your ruler and that's how you draw your k.

This post has been edited by LaFayette: 29 April 2010 - 09:10 AM

### #5 NeoTifa

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 09:33 AM

Whoa! Comma's and decimal points?! Wtf Sweden?

Okay as for drawing graphs of lines, the coefficient of the x is the slope, right? Slope is just rise over run. So for your 2/3 slope, you'd rise up 2 units, then run in the positive direction 3 units. The plus 2 is just where the line intersects the y-axis, which is positive 2. So 3 guaranteed coordinates would be (0, 2), (3, 4), and (1, 2/3)

For you -1/2 slope, the rise is -1, so you'd rise in a negative direction (or fall) 1 unit. Then you run 2 spaces in the positive direction. Since there is no addition at the end (or y-intercept), that is basically the same thing as (-1/2)x + 0, so the line would cross through the origin. So 3 guaranteed coordinates would be (0, 0), (2, -1), and (1, -1/2).

Hope that helped a little. ^__^

This post has been edited by NeoTifa: 29 April 2010 - 09:36 AM

### #6 FrozenSnake

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 11:24 AM

Yah, I think I understand thanks both of you
In Sweden we use 4,5 instead of 4.5 I don't know why I used both, we mainly uses ','.

### #7 FrozenSnake

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 11:35 AM

"Determine the intersections between the line and the coordinate axes"
A) x - 2y - 10 = 0
x = 10
y = -5
So far I have come but I did this (10, -5) and checked the key. And the answer is this way
(10, 0) and (0, -5) in separate () WHY?!

### #8 NeoTifa

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## Re: Problem with equation systems in math

Posted 29 April 2010 - 04:48 PM

Okay, for the equation I simplified it down to y = (1/2)x - 5. Therefore, you were correct in having those two coordinates. They're separate because they're two different intersections. When x = 0, y = -5, and when y = 0, x = 10.