Solve for x
x=\frac{7y}{6}-\frac{4}{3}
Solve for y
y=\frac{6x+8}{7}
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6x+8=7y
Add 7y to both sides. Anything plus zero gives itself.
6x=7y-8
Subtract 8 from both sides.
\frac{6x}{6}=\frac{7y-8}{6}
Divide both sides by 6.
x=\frac{7y-8}{6}
Dividing by 6 undoes the multiplication by 6.
x=\frac{7y}{6}-\frac{4}{3}
Divide 7y-8 by 6.
-7y+8=-6x
Subtract 6x from both sides. Anything subtracted from zero gives its negation.
-7y=-6x-8
Subtract 8 from both sides.
\frac{-7y}{-7}=\frac{-6x-8}{-7}
Divide both sides by -7.
y=\frac{-6x-8}{-7}
Dividing by -7 undoes the multiplication by -7.
y=\frac{6x+8}{7}
Divide -6x-8 by -7.
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\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
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Limits
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