Linear Equations I

Hong Kong

Stage 1 - Stage 3

Lesson

Recall that a table of values is what we call a table that shows the values of two quantities (usually represented by $x$`x` and $y$`y`) that are related in some way. As an example, a table of values might look like:

$x$x |
$3$3 | $6$6 | $9$9 | $12$12 |
---|---|---|---|---|

$y$y |
$10$10 | $19$19 | $28$28 | $37$37 |

Let's construct our own table of values using the following equation:

$y=3x-5$`y`=3`x`−5

The table of values for this equation connects the $y$`y`-value that result from substituting in a variety of $x$`x`-values. Let's complete the table of values below:

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |

To substitute $x=1$`x`=1 into the equation $y=3x-5$`y`=3`x`−5, we want replace all accounts of $x$`x` with $1$1.

So for $x=1$`x`=1, we have that:

$y$y |
$=$= | $3\left(1\right)-5$3(1)−5 |

$=$= | $3-5$3−5 | |

$=$= | $-2$−2 |

So we know that $-2$−2 must go in the first entry in the row of $y$`y`-values.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |
$-2$−2 |

Next let's substitute $x=2$`x`=2 into the equation $y=3x-5$`y`=3`x`−5.

For $x=2$`x`=2, we have that:

$y$y |
$=$= | $3\left(2\right)-5$3(2)−5 |

$=$= | $6-5$6−5 | |

$=$= | $1$1 |

So we know that $1$1 must go in the second entry in the row of $y$`y`-values.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |
$-2$−2 | $1$1 |

If we substitute the remaining values of $x$`x`, we find that the completed table of values is:

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |
$-2$−2 | $1$1 | $4$4 | $7$7 |

Each column in a table of values may be grouped together in the form $\left(x,y\right)$(`x`,`y`). We call this pairing an ordered pair. Let's return to our table of values:

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |
$-2$−2 | $1$1 | $4$4 | $7$7 |

The table of values has the following ordered pairs:

$\left(1,-2\right),\left(2,1\right),\left(3,4\right),\left(4,7\right)$(1,−2),(2,1),(3,4),(4,7)

We can plot each ordered pair as a point on the $xy$`x``y`-plane.

We can plot the ordered pair $\left(a,b\right)$(`a`,`b`) by first identifying where $x=a$`x`=`a` along the $x$`x`-axis and $y=b$`y`=`b` along the $y$`y`-axis.

Take $\left(3,4\right)$(3,4) as an example. We first identify $x=3$`x`=3 along the $x$`x`-axis and draw a vertical line through this point. Then we identify $y=4$`y`=4 along the $y$`y`-axis and draw a horizontal line through that point. Finally we plot a point where two lines meet, and this represents the ordered pair $\left(3,4\right)$(3,4).

Now that we have drawn the ordered pairs from the table of values, we can draw the graph that passes through these points.

In the example above, the line that passes through these points is given by:

This straight line is the graph of $y=3x-5$`y`=3`x`−5 which we used to complete the table of values.

To draw a line from a table of values, it is useful to plot the significant points and draw the line that passes through them.

For example, consider the equation:

$y=3x-6$`y`=3`x`−6

And the table of values:

$x$x |
$0$0 | $1$1 | $2$2 | $3$3 |
---|---|---|---|---|

$y$y |
$-6$−6 | $-3$−3 | $0$0 | $3$3 |

There are two significant ordered pairs, namely the $x$`x`-intercept and the $y$`y`-intercept.

- The $x$
`x`-intercept has the form $\left(a,0\right)$(`a`,0) which is a point that lies on the $x$`x`-axis. - The $y$
`y`-intercept has the form $\left(0,b\right)$(0,`b`) which is a point that lies on the $y$`y`-axis.

The $x$`x`-intercept in our example is $\left(2,0\right)$(2,0) and the $y$`y`-intercept is $\left(0,-6\right)$(0,−6).

We can draw the line of $y=3x-6$`y`=3`x`−6 which passes through these two points.

Consider the equation $y=4x$`y`=4`x`. A table of values is given below.

$x$x |
$-2$−2 | $-1$−1 | $0$0 | $1$1 |
---|---|---|---|---|

$y$y |
$-8$−8 | $-4$−4 | $0$0 | $4$4 |

Plot the points in the table of values.

Loading Graph...Is the graph of $y=4x$

`y`=4`x`linear?Yes

ANo

B

Consider the equation $y=3x+1$`y`=3`x`+1.

Complete the table of values below:

$x$ `x`$-1$−1 $0$0 $1$1 $2$2 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the points in the table of values.

Loading Graph...Draw the graph of $y=3x+1$

`y`=3`x`+1.Loading Graph...

Consider the equation $y=-\frac{x}{7}$`y`=−`x`7.

Complete the table of values below:

$x$ `x`$-7$−7 $-4$−4 $-3$−3 $0$0 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Draw the graph of $y=-\frac{x}{7}$

`y`=−`x`7.Loading Graph...