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.

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

Relate graphs, tables, and equations to linear, quadratic, and simple exponential relationships found in number and spatial patterns

Apply algebraic procedures in solving problems

Investigate relationships between tables, equations and graphs