UK Secondary (7-11)
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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=3x1 and $y=3x+6$y=3x+6 will never have a solution since they both have a gradient of $3$3.



Question 1

Consider the following linear equations:

$y=2x+2$y=2x+2 and $y=-2x+2$y=2x+2

  1. What are the gradient and $y$y-intercept of the line $y=2x+2$y=2x+2?

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

    $x$x-value of $x$x-intercept $\editable{}$
    $y$y-value of $y$y-intercept $\editable{}$
  3. Plot the lines of the 2 equations on the same graph.

    Loading Graph...

  4. State the values of $x$x and $y$y which satisfy both equations.

    $x$x = $\editable{}$

    $y$y = $\editable{}$


question 2

Consider the two equations $3x-y=5$3xy=5 and $2x+y-1=0$2x+y1=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


$3x-y$3xy $=$= $5$5
$3x-y-5$3xy5 $=$= $0$0
$y$y $=$= $3x-5$3x5

The gradient of $3x-y=5$3xy=5 is $3$3 and the $y$y-intercept is $-5$5

$2x+y-1$2x+y1 $=$= $0$0
$2x+y$2x+y $=$= $1$1
$y$y $=$= $-2x+1$2x+1

The gradient of $2x+y-1$2x+y1 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


I've graphed $3x-y=5$3xy=5 as green and $2x+y-1=0$2x+y1=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

Consider the following linear equations:

$y=5x-7$y=5x7 and $y=-x+5$y=x+5

  1. Plot the lines of the 2 equations on the same graph.

    Loading Graph...

  2. State the values of $x$x and $y$y which satisfy both equations.

    $x$x = $\editable{}$

    $y$y = $\editable{}$


Question 4

Consider the two equations $2x-6+y=0$2x6+y=0 and $15-2y=4x$152y=4x. Is there a solution that satisfies both?

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


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

$2x-6+y$2x6+y $=$= $0$0
$-6+y$6+y $=$= $-2x$2x
$y$y $=$= $-2x+6$2x+6
$15-2y$152y $=$= $4x$4x
$-2y$2y $=$= $4x-15$4x15
$y$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.



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