Solve for x
x=\frac{3y+5z}{2}
Solve for y
y=\frac{2x-5z}{3}
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3x+3y-5x=-5z
Subtract 5x from both sides.
-2x+3y=-5z
Combine 3x and -5x to get -2x.
-2x=-5z-3y
Subtract 3y from both sides.
-2x=-3y-5z
The equation is in standard form.
\frac{-2x}{-2}=\frac{-3y-5z}{-2}
Divide both sides by -2.
x=\frac{-3y-5z}{-2}
Dividing by -2 undoes the multiplication by -2.
x=\frac{3y+5z}{2}
Divide -5z-3y by -2.
3y=5x-5z-3x
Subtract 3x from both sides.
3y=2x-5z
Combine 5x and -3x to get 2x.
\frac{3y}{3}=\frac{2x-5z}{3}
Divide both sides by 3.
y=\frac{2x-5z}{3}
Dividing by 3 undoes the multiplication by 3.
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
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Limits
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