thnx for your help
2006-08-19 12:15:46 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
y^2 + 3yz - 8z -4x = 0
First get the 2 z's on the same side. We accomplish this by subtracting y^2 and adding 4x to both sides. They cancel out on the left. Now we have:
3yz -8z = -y^2 + 4x
Next we factor out a z so we have just one z in the equation. We see both 3yz and 8z has a z so we are able to factor one out.
z(3y - 8) = (-y^2 + 4x)
Now we divide by both sides by 3y - 8 so that the 3y-8 on the left side will cancel out leaving the z by itself. We do this and we get:
z = (-y^2+4x)/(3y-8)
Hope this helped!
2006-08-19 12:23:15 · answer #1 · answered by Elim 5 · 0⤊ 0⤋
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RE:
Solve for Z: (y^2) + 3yz-8z-4x = 0?
thnx for your help
2015-08-24 22:45:40 · answer #2 · answered by Waneta 1 · 0⤊ 0⤋
y^2 + 3yz - 8z - 4x = 0
y^2 - 4x = -3yz + 8z
y^2 - 4x = z(-3y + 8)
z = (y^2 - 4x) / (-3y + 8)
2006-08-19 12:24:34 · answer #3 · answered by Benny 2 · 1⤊ 0⤋
y^2 + 3yz - 8z - 4x = 0
3yz - 8z = -y^2 + 4x
z(3y - 8) = -y^2 + 4x
z = (-y^2 + 4x)/(3y - 8)
ANS : (-y^2 + 4x)/(3y - 8) or -(y^2 - 4x)/(3y - 8)
2006-08-19 15:28:02 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
3yz - 8z = 4x - y^2
z(3y -8) = 4x - y^2
z = (4x - y^2) / (3y -8)
2006-08-19 13:01:04 · answer #5 · answered by sonicwingmode 2 · 0⤊ 0⤋
3yz - 8z = 4x - y^2
z(3y -8) = 4x - y^2
z = (4x - y^2) / (3y -8)
2006-08-19 12:23:09 · answer #6 · answered by ___ 4 · 1⤊ 0⤋
Solve For Z
2016-11-08 06:36:43 · answer #7 · answered by Anonymous · 0⤊ 0⤋
whatever you do to one side of the equal sign, you do to the other side.
You are trying to get z all by itself on one side of the equal sign.
Show your work.
Try that.
2006-08-19 12:20:45 · answer #8 · answered by Anonymous · 0⤊ 1⤋
I think one would need more data to actually solve it...
2006-08-19 12:24:30 · answer #9 · answered by andreicnx 3 · 0⤊ 1⤋
pink
2006-08-19 12:54:20 · answer #10 · answered by Tommy 4 · 0⤊ 1⤋