I am trying to work out a solution to a question and am having a difficulty that is probably so easy to most people out there...
I think that it is a linear equation thing, and i have read up on them on a very useful site given to me by one of the members! (THANKS!)...
It could be my working out, but i keep coming up with decimals in my T Chart...
i'm going to try and explain it, so here goes... (please don't feel rude for telling me i'm an idiot, and i am doing the completley wrong thing)...
I have the two equations...
3x + 4y = 78
2x + 6y = 72
I have to put these onto a graph, and find the value of x & y by the point of intersection.
I rearranged the two equations to find the value of x and got...
x = -(6/2)(y)+36
x = -(4/3)(y)+26 ... (Feel free to tell me if i am wrong allready)
When putting these into a T Chart, and enterring a value for y, i keep coming up with decimals...
eg... y = 2
(-4 devided by 3) (2) + 26 = 23.33 recurring.
... the first equation i have comes up perfect everytime... please tell me where i have gone wrong, so i can put this information into a graph!! (Crying)
March 2nd 2010, 04:36 PM
The 0 intercepts?
If you are trying to find where they intercept the axis you are on the right track... No need to graph though.
These two equations represent two different lines. They will intersect at different points.
Let's use what you have written already and find the x intercept. What is the value of y at the x axis?
So, set y to 0 in your equation and you will have the x-intercept. Now do the same but setting x to 0 so you will find the y-intercept.
Is this what you are after?
March 2nd 2010, 06:41 PM
Well, i definatley need to graph, because that is the next question...
it says... Draw the graphs of the two equations on the same set of x and y axes and read off the co-ordinates to the point of intersections to find the values of x & y...
So i just took that information and made as much sense out of it as i could...
I did set the y value to 0 at one stage, which was great because i didn't get a decimal! =p... but obviously i didn't get the same value for both equations, which meant i couldn't put them onto a graph...
I hope i don't sound like a total moron, lol. I didn't do math B at school and now it is coming back to bite me in the but. =p
I'm not after answers, so please don't feel that i am trying to get someone else to do my work, i just really can't understand how to do it.
March 2nd 2010, 06:49 PM
here's what i would do:
(1) 3x + 4y = 78
(2) 2x + 6y = 72
setting y=0 gives your x-intercept, and x=0 gives your y-intercept
So for (1) x=26 and y=39/2
take the equations and isolate for y:
They are now in y=mx+b form where m is the slope and b is the y intercept.
March 2nd 2010, 08:09 PM
Hello ! You have two lines modelled by the equations :
Now that's pretty difficult to represent such lines, so we are going to put these equations in intercept form (this is where you went wrong).
The first one : , so , and therefore , this being equivalent to .
The second one : , so , and therefore , this being equivalent to .
So your two lines put in intercept form have the equations :
This can be graphed, and the coordinates can be read easily (they are integers :)). Do a big graph though, you're going to need space to see the intersection.
March 3rd 2010, 04:34 AM
Why do you believe that "coming up with decimals" means you are doing something wrong?
If all you want to do is graph them:
For 3x+ 4y= 78, if x=0, then 4y= 78, y= 78/4= 39/2= 19.5. If y= 0 then 3x= 78, x= 78/3= 26. The x and y intercepts are (26, 0), and (19.5, 0). Mark those points on the axes and draw the line between them.
For 2x+ 6y= 72, if x= 0, then 6y= 72, y= 72/6= 12. If y= 0 then 2x= 72, x= 72/2= 36. The x and y intercepts are (36, 0) and (0, 12). Mark those points on the axes and draw the line between them.
If you really need the equations in "point, slope" form, 4y= 78- 3x so . 2x+ 6y= 72 so 6y= 72- 2x, .
The equations are neater in fraction form. It is not really necessary to put the numbers in "decimal" form.