Solve for x
x=\frac{3\left(y+4\right)}{2}
Solve for y
y=\frac{2\left(x-6\right)}{3}
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2x=3y+7+5
Add 5 to both sides.
2x=3y+12
Add 7 and 5 to get 12.
\frac{2x}{2}=\frac{3y+12}{2}
Divide both sides by 2.
x=\frac{3y+12}{2}
Dividing by 2 undoes the multiplication by 2.
x=\frac{3y}{2}+6
Divide 12+3y by 2.
3y+7=2x-5
Swap sides so that all variable terms are on the left hand side.
3y=2x-5-7
Subtract 7 from both sides.
3y=2x-12
Subtract 7 from -5 to get -12.
\frac{3y}{3}=\frac{2x-12}{3}
Divide both sides by 3.
y=\frac{2x-12}{3}
Dividing by 3 undoes the multiplication by 3.
y=\frac{2x}{3}-4
Divide -12+2x 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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