#2**+1 **

The table of x- and y-values definitely represent a linear equation because the increase from one point to another is constant. This is a direct indicator of a linear function.

If you cannot figure it by trial and error, then you can do this:

\(y=mx+b\)

We know that all linear functions are in this form. Let's plug in a point into this equation. I'll use the first one in the table, (2,8)

\(y=mx+b\) | Plug in 2 for x and 8 for y. |

\(8=2m+b\) | |

Let's do this again for a separate point. In this case, I'll use (3,11). It does not matter which point you choose. The only criteria is that it must be different from the one you previously chose.

\(y=mx+b\) | Plug in the coordinate (3,11) into the equation. |

\(11=3m+b\) | |

Look at this! We have a system of equations that we can solve for both m and b. I'll use substitution for this demonstration because it is easier to showcase on an online format. I'll solve for b in the first equation, 8=2m+b

\(8=2m+b\) | Subtracting 2m from both sides is the easiest method to isolate one variable. |

\(8-2m=b\) | |

Now, plug 8-2m in for b in the second equation.

\(11=3m+b\) | Plug in the aforementioned value for b. |

\(11=3m+8-2m\) | Simplify the right hand side of the equation. |

\(11=m+8\) | Subtract 8 on both sides. |

\(m=3\) | |

Now, substitute m=3 into either original equation. I'll do 8=2m+b

\(8=2m+b\) | Plug in 3 for m. |

\(8=6+b\) | Subtract 6 from both sides. |

\(b=2\) | |

Therefore, the equation that satisfies that is \(y=3x+2\). You can confirm this by testing out each coordinate. Of course, it works.

TheXSquaredFactor Oct 10, 2017

#2**+1 **

Best Answer

The table of x- and y-values definitely represent a linear equation because the increase from one point to another is constant. This is a direct indicator of a linear function.

If you cannot figure it by trial and error, then you can do this:

\(y=mx+b\)

We know that all linear functions are in this form. Let's plug in a point into this equation. I'll use the first one in the table, (2,8)

\(y=mx+b\) | Plug in 2 for x and 8 for y. |

\(8=2m+b\) | |

Let's do this again for a separate point. In this case, I'll use (3,11). It does not matter which point you choose. The only criteria is that it must be different from the one you previously chose.

\(y=mx+b\) | Plug in the coordinate (3,11) into the equation. |

\(11=3m+b\) | |

Look at this! We have a system of equations that we can solve for both m and b. I'll use substitution for this demonstration because it is easier to showcase on an online format. I'll solve for b in the first equation, 8=2m+b

\(8=2m+b\) | Subtracting 2m from both sides is the easiest method to isolate one variable. |

\(8-2m=b\) | |

Now, plug 8-2m in for b in the second equation.

\(11=3m+b\) | Plug in the aforementioned value for b. |

\(11=3m+8-2m\) | Simplify the right hand side of the equation. |

\(11=m+8\) | Subtract 8 on both sides. |

\(m=3\) | |

Now, substitute m=3 into either original equation. I'll do 8=2m+b

\(8=2m+b\) | Plug in 3 for m. |

\(8=6+b\) | Subtract 6 from both sides. |

\(b=2\) | |

Therefore, the equation that satisfies that is \(y=3x+2\). You can confirm this by testing out each coordinate. Of course, it works.

TheXSquaredFactor Oct 10, 2017