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**lisathedork****Member**- Registered: 2013-10-24
- Posts: 36

How do I graph these?

1. y = (2/3)x - 1

y = -x + 4

2. x + y = 0

3x + y = -4

3. 4x + 3y = -15

y = x + 2

4. x + 2y = -4

4y = 3x + 12

5. y = 2x

x + y = 3

6. x = 3 - 3y

x + 3y = -6

7. y = -2x + 1

y = x - 5

8. y = (1/2)x - 3

y = (3/2)x - 1

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi lisa;

Welcome to the forum.

You make a table of x, y values and plot the points on graph paper. Start with the first one:

1. y = (2/3)x - 1

Now just plot point (0,-1) and (1,-1/3) in blue.

y = -x + 4

Click on the drawing below. Now you try the second one.

Now just plot point (0,4) and (1,3) in red.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**lisathedork****Member**- Registered: 2013-10-24
- Posts: 36

Shouldn't it be like that?

2x - 3y = -2

2(0) - 3y = -2

-3y = -2

y = 2/3

(0, 2/3)

2x - 3(0) = -2

2x = -2

x = -1

(-1, 0)

4x + y = 8

4(0) + y = 8

y = 8

(0, 8)

4x + y = 8

4x + 0 = 8

4x = 8

x = 2

(2, 0)

these are just examples.

Cause I know I have to found the y = mx + b

*Last edited by lisathedork (2013-10-24 08:00:25)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,979

hi lisathedork

Welcome to the forum.

You certainly can put x=0 and get y; and then y=0 and get x.

Any pair of points will be enough to draw the line.

Q5 y = 2x.

Here using zero will only get one point (0,0) so you'll have to choose another value as well.

eg x = 5 .... y = 2 times 5 = 10 .......(5,10)

You don't need to get them in the format y = mx + b for this

eg

Q5 again

x + y = 3 any two numbers that add to 3 will do so (0,3) (1,2) (2,1) (3,0) (4,-1) 5,-2) .............

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**lisathedork****Member**- Registered: 2013-10-24
- Posts: 36

I totally understand now. Thanx.

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

bobbym wrote:

Hi lisa;

Welcome to the forum.

...

First time bobbym calls someone by a shortened name!

Welcome Lisa.

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