Solve for x
x=-\frac{2y}{3}+z
Solve for y
y=\frac{3\left(z-x\right)}{2}
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3x=3z-2y
Subtract 2y from both sides.
\frac{3x}{3}=\frac{3z-2y}{3}
Divide both sides by 3.
x=\frac{3z-2y}{3}
Dividing by 3 undoes the multiplication by 3.
x=-\frac{2y}{3}+z
Divide 3z-2y by 3.
2y=3z-3x
Subtract 3x from both sides.
\frac{2y}{2}=\frac{3z-3x}{2}
Divide both sides by 2.
y=\frac{3z-3x}{2}
Dividing by 2 undoes the multiplication by 2.
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
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Limits
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