Solve for x
x=\frac{3y}{2}-z+\frac{15}{2}
Solve for y
y=\frac{2x}{3}+\frac{2z}{3}-5
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2x+2z=15+3y
Add 3y to both sides.
2x=15+3y-2z
Subtract 2z from both sides.
2x=3y-2z+15
The equation is in standard form.
\frac{2x}{2}=\frac{3y-2z+15}{2}
Divide both sides by 2.
x=\frac{3y-2z+15}{2}
Dividing by 2 undoes the multiplication by 2.
x=\frac{3y}{2}-z+\frac{15}{2}
Divide 15+3y-2z by 2.
-3y+2z=15-2x
Subtract 2x from both sides.
-3y=15-2x-2z
Subtract 2z from both sides.
-3y=15-2z-2x
The equation is in standard form.
\frac{-3y}{-3}=\frac{15-2z-2x}{-3}
Divide both sides by -3.
y=\frac{15-2z-2x}{-3}
Dividing by -3 undoes the multiplication by -3.
y=\frac{2x}{3}+\frac{2z}{3}-5
Divide 15-2x-2z 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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