Simultaneous equations part four

Here is a graph with two straight lines plotted on it, $y = 2x + 1$ and $y = 3x - 1$

We have looked at the properties of $y = 2x + 1$ . The stepsize (or gradient) is 2 and the line crosses the $y$ axis at $y = 1$ . We can tell this just by looking at the equation. We also worked out that the line crosses the x-axis (when $y = 0$ ) at $x = -\frac{1}{2}$ .

For $y = 3x - 1$ , the gradient is 3 and the line crosses the $y$ axis at $y = - 1$ . We can work out where the line crosses the x-axis by rearranging the equation for x:

$y = 3x - 1$

$3x - 1 = y$

$3x =y + 1$

$x =\frac{y + 1}{3}$

Then substitute $y =0$ in the equation to get

$x =\frac{1}{3}$

Looking at the graph this looks right.

We can also read the answer to the simultaneous equations $y = 2x + 1$ and $y = 3x - 1$ from the graph.

$y = 2x + 1$ and, simultaneously, $y = 3x - 1$ at the point at which the two straight lines cross. We can see from the graph that this is when $x = 2$ and $y = 5$ .

We could also shown this by substitution or elimination as we saw earlier in the course.

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