The Graphical Method

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=5x 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=5x 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.

 

Parallel lines

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=mx+b and $m$m will be our gradient. This means that for example, the simultaneous equations $y=3x-1$y=3x−1 and $y=3x+6$y=3x+6 will never have a solution since they both have a gradient of $3$3.

 

Examples

Question 1

 

question 2

Consider the two equations $3x-y=5$3x−y=5 and $2x+y-1=0$2x+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=mx+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$3x−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$2x+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$3x−y=5 as green and $2x+y-1=0$2x+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. 

 

question 3

 

Question 4

Consider the two equations $2x-6+y=0$2x−6+y=0 and $15-2y=4x$15−2y=4x. 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=mx+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.