#1**+1 **

Let's think about this problem. There are 2 solutions to \(y!=1\). It is 0 and 1. Therefore, we need to set \(3x-2\) equal to both 0 and 1:

\(3x-2=0\) | Add 2 to both sides. |

\(3x=2\) | Divide by 3 on both sides. |

\(x=\frac{2}{3}\) | |

Let's do the other one now!

\(3x-2=1\) | Add 2 to both sides. |

\(3x=3\) | Divide by 3 on both sides. |

\(x=1\) | |

Therefore, x=1 and 2/3

TheXSquaredFactor Aug 15, 2017