# Solving Inequality Equations

## 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...

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: