New Zealand

Level 6 - NCEA Level 1

Lesson

So far we've had a look at what simultaneous equations are and at some of the ways to solve them. As with all algebraic expressions, simultaneous equations can also be expressed as graphs on a number plane. In coordinate geometry, we know that each graph represents ALL the possible solutions of a related equation. In other words, if a point is on a graph, it must solve its equation. In simultaneous equations, we mostly deal with linear equations, which can be represented as straight line graphs. Our aim is then to find a solution that solves BOTH equations, and graphically this means finding the point of intersection of the two straight lines.

Let's have a look at an example, where we want to find the solution to the simultaneous equations $y=5x$`y`=5`x` and $y=x+2$`y`=`x`+2. So then we would plot the two equations as graphs. Remember there are two ways to visualise linear equations as graphs: either through finding its intercepts or finding its gradient-intercept form. Here I have drawn the $y=5x$`y`=5`x` line as red and the $y=x+2$`y`=`x`+2 line as green:

We can then see that there is only one intersection point and it is $(0.5,2.5)$(0.5,2.5). Therefore the solution that solves the two equations must be when $x=0.5$`x`=0.5 and $y=2.5$`y`=2.5.

Do all pairs of simultaneous linear equations have a solution? Well let's think about this graphically: is it possible to graph two straight lines that never cross over? Of course, it happens when they're parallel! Let's remind ourselves that to find the gradient of a linear equation all we have to do is put it in the gradient intercept form $y=mx+b$`y`=`m``x`+`b` and $m$`m` will be our gradient. This means that for example, the simultaneous equations $y=3x-1$`y`=3`x`−1 and $y=3x+6$`y`=3`x`+6 will never have a solution since they both have a gradient of $3$3.

Consider the following linear equations:

$y=2x-4$`y`=2`x`−4 and $y=-2x-4$`y`=−2`x`−4

What are the gradient and $y$

`y`-intercept of the line $y=2x-4$`y`=2`x`−4?gradient $\editable{}$ $y$ `y`-value of $y$`y`-intercept$\editable{}$ What are the intercepts of the line $y=-2x-4$

`y`=−2`x`−4?$x$ `x`-value of $x$`x`-intercept$\editable{}$ $y$ `y`-value of $y$`y`-intercept$\editable{}$ Plot the lines of the 2 equations on the same graph.

Loading Graph...State the values of $x$

`x`and $y$`y`which satisfy both equations.$x$

`x`= $\editable{}$$y$

`y`= $\editable{}$

Consider the two equations $3x-y=5$3`x`−`y`=5 and $2x+y-1=0$2`x`+`y`−1=0

a) What are the gradients and $y$`y`-intercepts of the two equations?

Think: The gradient-intercept form looks like $y=mx+b$`y`=`m``x`+`b`, where $m$`m` is the gradient and $b$`b` the $y$`y`-intercept

Do:

$3x-y$3x−y |
$=$= | $5$5 |

$3x-y-5$3x−y−5 |
$=$= | $0$0 |

$y$y |
$=$= | $3x-5$3x−5 |

The gradient of $3x-y=5$3`x`−`y`=5 is $3$3 and the $y$`y`-intercept is $-5$−5.

$2x+y-1$2x+y−1 |
$=$= | $0$0 |

$2x+y$2x+y |
$=$= | $1$1 |

$y$y |
$=$= | $-2x+1$−2x+1 |

The gradient of $2x+y-1$2`x`+`y`−1 is $-2$−2 and the $y$`y`-intercept is $1$1.

b) Using the gradient-intercept form graph the two equations and find the solution that satisfies both

Think: Gradient means rise over run, and a solution that satisfies both equations will be the intersection of their graphs

Do:

I've graphed $3x-y=5$3`x`−`y`=5 as green and $2x+y-1=0$2`x`+`y`−1=0 as red. The intersection point is $(1.2,-1.4)$(1.2,−1.4). Therefore the solution to both equations is $x=1.2$`x`=1.2 and $y=-1.4$`y`=−1.4.

Consider the following linear equations:

$y=5x-7$`y`=5`x`−7 and $y=-x+5$`y`=−`x`+5

Plot the lines of the two equations on the same graph.

Loading Graph...State the values of $x$

`x`and $y$`y`which satisfy both equations.$x$

`x`= $\editable{}$$y$

`y`= $\editable{}$

Consider the two equations $2x-6+y=0$2`x`−6+`y`=0 and $15-2y=4x$15−2`y`=4`x`. Is there a solution that satisfies both?

Think: Parallel lines don't have a solution that satisfies both

Do:

Let's put both equations in $y=mx+b$`y`=`m``x`+`b` form to find their gradients:

$2x-6+y$2x−6+y |
$=$= | $0$0 |

$-6+y$−6+y |
$=$= | $-2x$−2x |

$y$y |
$=$= | $-2x+6$−2x+6 |

$15-2y$15−2y |
$=$= | $4x$4x |

$-2y$−2y |
$=$= | $4x-15$4x−15 |

$y$y |
$=$= | $-2x+\frac{15}{2}$−2x+152 |

The two equations have the same gradient of $-2$−2, so are parallel. Therefore there are no solutions that satisfy both equations.