solving eqautions using graphs

Showing 1 to 9 of 9

any1 kno how to solve eqautions via graphs pls,

cos i sure dont. please help. :)

Posted: 21-04-11 00:26 by symone

Do you mean when you get an equation (like y=x+1) and have to draw the line on?

When we've done it in school we've made tables where, using the example up there, you substitute in a number for x and see what y would be with each.

so if x=2, and y being equal to x+1, y would be 3. Is this making sense?

I don't know if that's exactly what you mean, if this isn't right you can let me know and I'll try and explain the other equation/graph stuff :)

Posted: 21-04-11 15:25 by Em Summerfield

yeah, thats what i ment,but sometimes the line on the graph is already drawn, then they ask you find for example"y=5x-2" on the grap.. do u know anything about that?

thanks alot thoe :)

Posted: 21-04-11 16:01 by symone

It's no problem :) Is that to do with finding out the gradient and y-intercept? If it is...

The gradient tells you the steepness of the line, and which direction it goes. If the gradient is a positive number, the line goes up from left to right- if it's negative it goes down from left to right.

The gradient is the coefficient of x. So like in your example, the gradient is 5x. That mean for every one across it goes UP 5 squares. So that's going to be a steep upward line.

The y intercept is the other number, so that's -2. So the line will be steep, diagonally up and will go through -2 on the y axis.

Is that what you mean? It would be easier to explain with a diagram or something... :P

Hope that helps? :)

Posted: 22-04-11 11:32 by Em Summerfield

Ok :P right, I've put that in a word document with a diagram so you can see if I'm explaining what you mean or something completely different :P It's loaded on my getrevising profile thingy, so take a look :) If I've got it wrong, let me know :)

Posted: 22-04-11 11:49 by Em Summerfield

aww thanks for your help, im going to look at it now and get back to you x

Posted: 25-04-11 08:42 by symone

alright then :) I hope it's ok!

Posted: 27-04-11 18:19 by Em Summerfield


Posted: 27-05-11 14:16 by bek

Draw the graphs, either straight lines or curves.  Where they cross is the solution or solutions

Posted: 30-05-11 17:57 by Arron Beckett