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**sarah****Guest**

find the value(s) of m such that the equations mx+y=2, y-2x=4 have

1. No solution.

2. An infinite number of solutions.

3. One real solution.

I don't know how to solve these kinds of problems :s. please, help someone?

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

First, off the bat you should recognize mx+y=2 and y-2x=4 as equations of lines. Now, remember (or learn), that the "solution" to these equations is that line. That is, every (x, y) on that line will work out so that 0 = 0 when you plug it in. 1. When do two lines have no common solution? In other words, when do two lines share no points? 2. When do two lines share all points? And finally, 3. When do two lines share only one point?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**sarah****Guest**

I get it, I get it! :-D

lol, thanks a lot:)

**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

What are the answers sarah? I don't.

Aloha Nui means Goodbye.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

y-2x=4 ⇒y=2x+4 ⇒

mx+y=2 <=>mx+2x+4=2

(m+2)x+4=2

(m+2)x=-2

1. If m+2=0;m=2

0x=-2

No solution

2. If m+2≠0;m≠2

x=-2/(m+2)

So:

No solution-

m=2

One solution-

m≠2

Infinite many solutions-

m∈{}

sorry for the syntax, I don't know English well.

*Last edited by krassi_holmz (2005-12-04 03:46:05)*

IPBLE: Increasing Performance By Lowering Expectations.

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I wanted sarah to answer the question krassi

Aloha Nui means Goodbye.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Sorry Ricky. But Sarah can still answer the question and can verify it with mine.

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"1. If m+2=0;m=2

0x=-2

No solution"

Not true. Let m = 2, then you get the equations: 2x + y = 2, so y = -2x + 2, and y = 2x + 4. A common solution to these is the point (-1/2, 3).

"2. 2. If m+2≠0;m≠2

x=-2/(m+2)

So:

No solution-

m=2

One solution-

m≠2"

The logic here is right, if m = 2 had no solutions, than all lines without that slope intersect the line. Unforutunately, m = 2 has a solution.

"Infinite many solutions-

m∈{}"

Correct. The x-intercepts are off, and they only thing you can change is the slope.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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