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ineedyouubiebs
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Attempt to solve for x! (4,y) and (7,-6); slope:-4
\[\frac{ -6-y }{ 7-4 }=-4\] \[\frac{ -6-y }{ 3 }=-4 \] I just set it up like the normal slope equation, (y2-y1)/(x2-x1). Now just see if you can solve for y.
Olay, let me try it out for myself to see if I understand it
Wait, I get -6 / 3 how did you get -4?
You said the slope = -4, right? Well, the result of (y2-y1)/(x2-x1) is slope. But we already have the slope, so I just set the equation equal to -4.
Ahhhhhh okay I see how you set it up!
And then you subtract the bottom, correct?
Yeah, which left me with \[\frac{ -6-y }{ 3 }=-4 \]Would you know how to use that to solve for y?
My guess is that I multiply -6 and 3
Just multiply both sides by 3 to start.
That will cancel the 3 out of the denominator of the fraction.
So would it be 18+3y=-4
Nah. Okay, so we multiply both sides by 3 and it looks liek this: \[\frac{ 3(-6-y) }{ 3 }=-4(3)\]From here, the 3's on the left cancel out and -4 and 3 multiplies to get: \[-6-y = -12 \] Kinda see why?
Yea aha I sorta do, it's because I got -18-3y=-12 when I multiplied it by 3
Yeah, but you dont multiply the -6 and the -y by 3. Because you have a 3 on top and bottom, they cancel out and become 1. That is the whole purpose of multiply by 3 is because itll get rid of the denominator. It's just like if we had: \[\frac{ x }{ 3 }=3 \]Because x is divided by 3, to get it by itself we do the opposite operation and multiply by 3. |dw:1378605836732:dw| Same thing with our problem: |dw:1378605874087:dw|
Ahhhhhh okay I see now! And then after I get -6-y=-12 I add 6 to both sides
When I divide y to both sides, would I be dividing -y or just y?
Well if you have -y = -6, it does no good to divide by y or -y. If you divided by -y you get: \[1=\frac{ 6 }{ y } \] You just want to divide by -1 is all to get y by itself.
And the final answer will be y=6
Bingo. Can even test it :3 \[\frac{ -6-6 }{ 7-4 }=\frac{ -12 }{ 3 }=-4 \] So yep, works :3
Thank you much, you explained very well unlike most people here :D do you have time to help me with another 3 problems I had trouble with?