### Linear formulas and equations: Linear equations and inequalities

### Linear inequalities

Aside from knowing when two linear formulas are equal to each other, it can also be interesting to know when one linear formula is greater than the other. Therefore, we will take a look at how to solve linear inequalities.

We are interested in the case when the formula # \blue{y=3 \cdot x+1}# (solid) is less than the formula # \green{y=x-3}# (dashed). We write this as follows:

\[\blue{3 \cdot x+1} < \green{x-3} \]

This means we are interested in when the blue graph is below the green graph.

We see from the graph that that is the solution to the inequality: \[\orange{x\lt -2}\]

Just as with linear equations, when working with linear inequalities we have a couple of steps we can perform, while keeping the inequality equivalent.

With the help of some rules, we can reduce inequalities to the form #x\lt a#; #a\leq a#; #x \gt a# or #x \geq a#. These rules are very similar to the reduction rules with equations. We will take a look at them one by one.

Two inequalities are equivalent if you multiply both sides by the same **negative** number and switch signs, or if you divide both sides by the same **negative** number and switch the sign.

By switching signs we mean that the sign becomes the opposite sign. Hence:

**Example**

\[\begin{array}{rcl}-3x &\lt& 6 \\ \dfrac{-3x}{-3}&\gt &\dfrac{6}{-3} \\ x&\gt& -2 \end{array}\]

With these rules we can reduce linear inequalities. In the examples below we will show how that is done.

#\begin{array}{rcl} 3x+3 &\lt & -8x+4\\&&\phantom{xxx}\blue{\text{given}} \\

11x &\lt& 1 \\&&\phantom{xxx}\blue{\text{subtracted }3-8x \text{ on both sides}}\\

x &\lt& {{1}\over{11}}\\&&\phantom{xxx}\blue{\text{divided by } 11 \text{ on both sides, because }11 \text{ is positive, the inequality sign does not change}}\\

\end{array}#

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