Created 6 years ago2013-10-12 14:41:57 UTC by ninja defuse

Posted 6 years ago2013-10-12 14:41:57 UTC
Post #316202

Posted 6 years ago2013-10-12 14:53:29 UTC
Post #316203

:@

Posted 6 years ago2013-10-12 15:31:53 UTC
Post #316204

Both expressions are equal to Y, so set them equal to each other and solve for x.

Set both equations equal to each other:

2x + 2 = -(1/3)x + 2(2/3)

Collect the x's on one side, and the constants on the other. To keep the equation equal, if you subtract or add something from/to one side of the equation you have to do it to the other:

2x + (1/3)x = 2(2/3) - 2

Multiply out the 2(2/3):

2x + (1/3)x = (4/3) - 2

Re-express everything in terms of thirds so we can add/subtract them easily.

(6/3)x + (1/3)x = (4/3) - (6/3)

Add/subtract:

(7/3)x = -(2/3)

Divide both sides of the equation by 7/3 to get x:

x = -(2/7)

Set both equations equal to each other:

2x + 2 = -(1/3)x + 2(2/3)

Collect the x's on one side, and the constants on the other. To keep the equation equal, if you subtract or add something from/to one side of the equation you have to do it to the other:

2x + (1/3)x = 2(2/3) - 2

Multiply out the 2(2/3):

2x + (1/3)x = (4/3) - 2

Re-express everything in terms of thirds so we can add/subtract them easily.

(6/3)x + (1/3)x = (4/3) - (6/3)

Add/subtract:

(7/3)x = -(2/3)

Divide both sides of the equation by 7/3 to get x:

x = -(2/7)

Posted 6 years ago2013-10-12 16:01:24 UTC
Post #316205

can u help me doing this in graphic mode?

Posted 6 years ago2013-10-12 16:10:45 UTC
Post #316206

What do you mean graphic mode?

If you mean to do this problem graphically, just plot both lines on a grid and then mark the x value that the two lines meet up at.

Edit: Also, careful! I don't think your work makes sense! Plug your results back into the original equations and see if they equal each other.

If you mean to do this problem graphically, just plot both lines on a grid and then mark the x value that the two lines meet up at.

Edit: Also, careful! I don't think your work makes sense! Plug your results back into the original equations and see if they equal each other.

Posted 6 years ago2013-10-12 16:16:54 UTC
Post #316207

i think its OK i hope...

Posted 6 years ago2013-10-13 02:21:17 UTC
Post #316211

lol MSPaint.

Posted 6 years ago2013-10-13 07:47:06 UTC
Post #316212

@up

Gimp

Gimp

Posted 6 years ago2013-10-13 08:27:35 UTC
Post #316213

You must be in your early high-school years ninja?

I know everything seems hard at first. But trust me, those are the easy stuff :). Try KhanAcademy, it's a good source of math tutorials. Although they usually tackle the easy stuff when it comes to uni math.

I know everything seems hard at first. But trust me, those are the easy stuff :). Try KhanAcademy, it's a good source of math tutorials. Although they usually tackle the easy stuff when it comes to uni math.

Posted 6 years ago2013-10-13 14:33:10 UTC
Post #316216

well i have to refresh my knownledge from past...

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