## Inequality Equations; what are they?

It's an equation written with a "less than" (\(<, \leq\)) or "greater than" (\(>, \geq \)) symbol where you might normally see an equals (\(=\)) sign. Here we're going to look at how to solve these for a few different cases. You will need to be comfortable with:

## Solving a Linear Inequality

Let's take the equation \(2x - 6 = 4\). You might be able to do this in your head, but let's write out the steps: \[\begin{align} 2x-6&=4 \\2x&=10 \\x&=5 \end{align}\] Now, if we were to be asked to solve \(2x - 6 > 4\), you might notice some similarities! \[\begin{align} 2x-6&>4 \\2x&>10 \\x&>5 \end{align}\]

## Rules and Warnings

Easy right?! Well let's run through some things to be careful of:
**1) **\(x=5\) and \(5=x\) are the same. \(x>5\) and \(5>x\) are not!
**2) **\(x=5\) and \(-x=-5\) are the same. We have to flip the symbol with inequalities when multiplying by \(-1\).
Let's carefully rearrange an example to see why:
\[\begin{align}
x&>5
\\x - 5 &> 5-5
\\x - 5 - x &> 0 - x
\\-5 &> -x
\\-x &< -5
\end{align}\]
This becomes especially important later on with quadratics...

## Quadratic Inequality Worked Solution

Leading on from the above example, we're going to solve a harmless looking quadratic
\[x^2 + 3x + 2 \leq 0\]
If we were to solve this as if it was an equality, we could by factorising:
\[\begin{align}
x^2 + 3x + 2 &= 0
\\(x+2)(x+1) &= 0
\\x = -2 \text{ or } x &= -1
\end{align}\]
So where does the \(\leq\) symbol fit in? Do we just stick it in to both solutions maybe so
\(x \leq -2\) and \(x \leq -1\)? **Nope! We need to do a sketch**.

The inequality in this question means "Less than or equal to". So the whole equation means
"The parts of the function \(y = x^2 + 3x + 2\) where \(y\) is less than or equal to 0". From our sketch, we can
see that it goes below \(y = 0\) (the \(x\) axis) *between* our two root values! These are highlighted red.
So the actual answer is:
\[-2\leq x \leq -1\]

## Quadratic Inequality Worked Solution 2

Question: Find values for \(x\) which satisfy \[x^2 + x + 2 \geq 4\] Answer: First we solve this as if there was an equals not an inequality. \[\begin{align} x^2 + x + 2 &= 4 \\x^2 + x - 2 &= 0 \\(x-1)(x+2) &= 0 \\x = 1 \text{ or } x &= -2 \end{align}\] Now we sketch \(x^2 + x - 2 = 0\):

We know that \(x^2 + x + 2 \geq 4\) so \(x^2 + x - 2 \geq 0\) On our sketch, we see that the equation has values greater than 0 outside of our roots, on the "arms" of the curve as highlighted in red. Our solution: \[x \leq -2 \text{ or } x \geq 1\] Note: We can't combine these into one set like we did in the previous example!

### Sketching Inequality Region

You may be asked specifically to do a sketch, and then to shade in the region which satisfies the equation. This is just a small extension to what we've already done, it's just a case of marking the solution on top of our sketch, like so: