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
Consider the following linear equations:
$y=2x-4$y=2x−4 and $y=-2x-4$y=−2x−4
What are the gradient and $y$y-intercept of the line $y=2x-4$y=2x−4?
gradient $\editable{}$ $y$y-value of $y$y-intercept $\editable{}$ What are the intercepts of the line $y=-2x-4$y=−2x−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 = $\editable{}$
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
Consider the following linear equations:
$y=5x-7$y=5x−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 = $\editable{}$
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.