03Hsimineq

toc = Simultaneous Inequations =

When graphing an inequation we are finding a region. Any point in the required region makes the inequation true.

When we solve simultaneous equations, we find the coordinates of the point where the two graphs intersect. At that point, both equations are true.

When we ** solve simultaneous inequations **, we find the region where any point in the region makes __both__ inequations true.

We do this by graphing the two inequations on the one set of axes.
 * we shade out the area not required by each of the inequations
 * the region that is left __unshaded__ is the required region.
 * the unshaded region is the intersection of the solutions of the individual inequations.

To graph the inequation:
 * draw the boundary line
 * use a broken line for
 * use a solid line for __<__ or __>__
 * shade out the area __not__ required

** Example 1 **
Solve the following simultaneous equations by drawing their graphs. math \\ . \qquad \centerdot \quad 3x + 2y > 6 \\ \\ . \qquad \centerdot \quad y \leqslant 2x - 1 math

__**Solution:**__ Graph first inequation: ** 3x + 2y > 6 **
 * ** x-intercept: ** let y = 0
 * gives x = 2 so ** (2, 0) **
 * ** y-intercept: ** let x = 0
 * gives y = 3 so ** (0, 3) **
 * "** > **" so broken line
 * ** (0, 0) ** gives **False** so shade __includes__ (0, 0)

Graph second inequation: ** y __<__ 2x – 1 **
 * ** c = –1 ** so ** (–1, 0) **
 * ** m = 2 ** (rise = 2, run = 1)
 * so ** (1, 1) **
 * "** __<__ **" so solid line
 * ** (0, 0) ** gives **False**, so shade __includes__ (0, 0)


 * Required region is the area left unshaded. **

__**Checking solution:**__

If we select __any__ point in the unshaded region, it should make both inequations true. For example, selecting ** (4, 1) **
 * First inequation: ** 3x + 2y > 6 **
 * 12 + 2 > 6 **(true)**
 * Second inequation: ** y __<__ 2x – 1 **
 * 1 __<__ 8 – 1 **(true)**

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