oh, okay! that makes a little bit more sense, so i gave it another try and heres what i got this time
y= 1 1/3 x -2
x= 1/4 y + 1/2
and this time, im pretty sure of myself....though i probably did mess up somewhere a little....
oh, okay! that makes a little bit more sense, so i gave it another try and heres what i got this time
y= 1 1/3 x -2
x= 1/4 y + 1/2
and this time, im pretty sure of myself....though i probably did mess up somewhere a little....
I don't see what problem you are solving - where did you get those.Originally Posted by c_ortiz92
If you are solving your original problem - then those are well "totally misguided".
“... mathematics is only the art of saying the same thing in different words” - B. Russell
Why are you solving for x and y? All you need is to get your original equation in standard form and plug x = 0 and y = 0 to find your y and x intercepts. Subhotosh Khan showed you how to put it in standard form for you. Now all you need to do is plug in x = 0 and y = 0.
really? well can you thoroughly explain the original problem to me, which is
8x-3y=6-4x
supposed to find x and y intercept
Again, I'll go back to my example (which was not your problem).Originally Posted by c_ortiz92
13y - 9x = -3y - x - 12
4y - 2x + 3 = 0 <--- standard form
x-intercept(y=0)
0 - 2x + 3 = 0
x = 3/2
So x-intercept is at (1.5,0)
y-intercept (x =0)
4y - 0 + 3 = 0
y = -3/4
So y-intercept is at (0,-0.75)
“... mathematics is only the art of saying the same thing in different words” - B. Russell
Do you understand what I mean when I say "set x = 0" to find the y-intercept? Let's try an example question and see if you can apply it to yours.
7x = 8y - 32
For finding the y-intercept, we let x = 0:
7(0) = 8y - 32
0 = 8y - 32
32 = 8y
y = 4
So for the equation 7x = 8y - 32, it crosses the y-intercept at y = 4. We do the same thing for the x-intercept except we set y = 0 instead.
For putting it into standard form (Ax + Bx + C = 0), we move everything to the other side:
7x - 8y + 32 = 8y - 32 - 8y + 32
7x - 8y + 32 = 0
And that's it.
okay, so first of all, i think im getting this roght this time,....how many times have i said that? anyways.
i have figured that the standard is
4x-3y-6=0
so
x= 3/4 y + 6/4
Can you show us your steps? Something's wrong when you're moving your variables around. Also, why do you solve for x?
shouldnt i be solving for x if im
1)writing the equation in standadrd form
2)determining the x intercept
3)determining the y intercept
?
well heres what ive done so far
the original problem is
8x-3y=6-4x
next i get
4x-3y-6=0
i got this because if you have a negative 4x and you add it to a positive 8x, you are gonna get 4x. and then, because you are moving the 6 to the other side of the equal sign, you are going to change it to a negative...
and i keep this as the standard form....
4x-3y-6=0
as for determining the x intercept
you have the standard form, which is
4x-3y-6=0
so you are going to get rid of the 4 that is attatched to the x, which means that you are going to do this:
4x=3y+6
4........4
and so as the answer, you get
x=3/4 y + 6/4
thats what i did,
8x - 3y = 6 - 4x
To get rid of the -4x on the right side, you ADD 4x to both sides. Not subtract.
8x + 4x - 3y = 6 - 4x + 4x
12x - 3y = 6
12x - 3y - 6 = 0
Ok. When finding the x-intercept, you don't solve for x. You take your equation (in this case 12x - 3y - 6 = 0) and set y = 0:
12x - 3(0) - 6 = 0
12x - 6 = 0
And you solve for x.
Then, for finding the y-intercept, you don't solve for y. You take your equation again and set x = 0 and THEN solve for y. Don't waste your time doing unnecessary algebra.
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