Linear Equations I

Hong Kong

Stage 1 - Stage 3

Lesson

We've been learning about linear relationships

We can say that an equation is linear if it fits in the form $y=mx+b$`y`=`m``x`+`b` or if it can be drawn graphically as a straight line.

It's pretty straight forward to tell if an equation is linear from a graph. All you need to determine is whether or not the graph is a straight line. If it is a straight line, it's a linear equation.

We can say that an equation is linear if it fits in any of the linear equation forms we have learnt about. They are:

- Gradient-intercept form: $y=mx+b$
`y`=`m``x`+`b` - General form: $ax+by+c=0$
`a``x`+`b``y`+`c`=0

It must fit into these forms, even if you have to rearrange the equation.

Careful!

The $x$`x` value in a linear relationship will never have a power.

We can't just assume the points in a table of values are linear (make a straight line). So when given a table of values, we must first confirm that the values in the table form a linear relationship.

You can do this a couple of ways, depending on what sort of information is provided to you in the table.

What is it that makes a straight line? It is the fact that for each 1 unit change in $x$`x`, the $y$`y` value changes by the same amount each time (i.e. the gradient is always the same).

If the table shows you values of $y$`y`, for consecutive values of $x$`x` (like this one) then it is quite easy to both check for a linear relationship and find the gradient of that relationship.

Some things to note about this table: the $x$`x` values go up by 1 each time and it doesn't matter what number the table starts at.

Notice in the above table, that for each 1 unit increase in $x$`x`, $y$`y` increases by $3$3 each time. This is a linear relationship, and what's more, this constant change in $y$`y` is the * gradient* of the line.

So we can check for a linear relationship by looking for a common difference between the $y$`y` values. If each successive $y$`y` value has the same difference then it is linear.

From here we not only have the gradient, we have 4 possible points to choose from and we can use the point gradient formula to find the equation of the line.

$y-y_1=m\left(x-x_1\right)$`y`−`y`1=`m`(`x`−`x`1)

Using gradient = $3$3 and the point $\left(3,12\right)$(3,12):

$y-12$y−12 |
$=$= | $3\left(x-3\right)$3(x−3) |

$y-12$y−12 |
$=$= | $3x-9$3x−9 |

$y$y |
$=$= | $3x-9+12$3x−9+12 |

$y$y |
$=$= | $3x+3$3x+3 |

If the table shows you values of $y$`y`, for non-consecutive values of $x$`x` (like this one) then we need another way to check for a linear relationship.

Some things to note about this table: the $x$`x` values do not go up by 1 each time.

We can check for a linear relationship by either plotting accurately, and looking to see if it is a straight line, or by checking the gradient between all the points. We will need to do this 3 times in this table.

$m=\frac{y_2-y_1}{x_2-x_1}$`m`=`y`2−`y`1`x`2−`x`1

Gradient between $\left(-3,25\right)$(−3,25) and $\left(2,0\right)$(2,0):

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{0-25}{2-\left(-3\right)}$0−252−(−3) |

$m$m |
$=$= | $\frac{-25}{5}$−255 |

$m$m |
$=$= | $-5$−5 |

Gradient between $\left(2,0\right)$(2,0) and $\left(10,-40\right)$(10,−40):

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{-40-0}{10-2}$−40−010−2 |

$m$m |
$=$= | $\frac{-40}{8}$−408 |

$m$m |
$=$= | $-5$−5 |

Looking good so far. We now have shown that the $3$3 points $\left(2,0\right)$(2,0), $\left(10,-40\right)$(10,−40) and $\left(-3,25\right)$(−3,25) are collinear which means they are all on the one line. We have to confirm that the last point is also on this line.

Gradient between $\left(2,0\right)$(2,0) and $\left(12,-50\right)$(12,−50):

(see how I used the $\left(2,0\right)$(2,0) point again, this is because 0 values make evaluating the gradient easier)

$m$m |
$=$= | $\frac{y_2-y_1}{x_2-x_1}$y2−y1x2−x1 |

$m$m |
$=$= | $\frac{-50-0}{12-2}$−50−012−2 |

$m$m |
$=$= | $\frac{-50}{10}$−5010 |

$m$m |
$=$= | $-5$−5 |

Got it! A linear relationship.

Now let's check out some more examples and see if you can spot the linear relationship.

$y$`y` is equal to $7$7 less than $2$2 groups of $x$`x`.

a) Write the statement above as a mathematical equation.

b) Is the equation linear?

Would the following equations be classified as linear?

a) $y=x^{10}+5x+26$`y`=`x`10+5`x`+26

b) $y=-9x+\frac{11}{2}$`y`=−9`x`+112

c) $y+26x-\frac{11}{2}=10$`y`+26`x`−112=10

Would the following table of values be represent a linear graph?

a)

x | 3 | 6 | 9 | 12 | 15 |
---|---|---|---|---|---|

y | -7 | -14 | -21 | -28 | -35 |

b)

x | 7 | 21 | 35 | 49 | 63 |
---|---|---|---|---|---|

y | 15/2 | 15 | 45/2 | 45 | 135/2 |

Consider the graph plotted below. Is the equation of this line linear?

Changing the subject of a formula is an important skill to learn. It can come in very handy when you know the value of one algebraic symbol but not another. Basically, the subject of an equation is the pronumeral that is by itself on one side on the equals sign and it usually is at the start of the formula.

For example, in the formula $A=pb+y$`A`=`p``b`+`y`, $A$`A` is the subject because it is by itself on the left hand side of the equals sign.

We sort of started changing the subject of equations when we learnt to solve equations because we took steps to get the pronumeral by itself, for example we made $x$`x` the subject of equations. We can make any term in an equation the subject, even if it starts off as the denominator of a fraction. When we're changing the subject of a formula, we often have more than one pronumeral but we still use a similar process.

- Group any like terms
- Simplify using the inverse of addition or subtraction.
- Simplify further by using the inverse of multiplication or division.

Make $x$`x` the subject of the following equation:

$y=\frac{x}{4}$`y`=`x`4

Make $m$`m` the subject of the following equation:

$\frac{m}{y}=gh$`m``y`=`g``h`

Make $m$`m` the subject of the following equation:

$y=6mx-9$`y`=6`m``x`−9