Solving Inequalities by Grouping - Examples & Practice

Solving Inequalities by Grouping

There are two things we need to do to solve inequalities with variables on both sides: simplify and isolate. That is, reduce anything we can, and isolate the variable to one side. This is the exact approach taken when we solve equations with variables on both sides of the equals sign.

Let's look at an example:

$$\frac{x+3}{4} > 7x - 1$$

We need to isolate $x$ on one side of the inequality.

$$\begin{align} \frac{x+3}{4} &> 7x - 1\\ \\ x + 3 &> 4(7x - 1)\\ x + 3 &> 28x - 4\\ x &> 28x - 7\\ -27x &> - 7\\ x &< \frac{7}{27}\\ \end{align}$$

We flipped the inequality sign when we divided both sides by $-27$. Dividing or multiplying by a negative number requires this flip.

We can also write our answer in interval notation to show the range of values that $x$ can take.

Since we don't have $\leq$ or $\geq$ in our inequality, $x<\frac{7}{27},$ we will use regular parentheses instead of brackets.

$$\left(-\infty, \frac{7}{27}\right)$$

This tells us that $x$ can be any value less than $\frac{7}{27}$.

REMEMBER: It's a good idea to double check your work by plugging your solution back into the inequality.

(Videos) Solving Linear Inequalities with Variables on Both Sides

by PatrickJMT

This video by Patrick JMT goes through an example inequality with variables on both sides.

$\large{\text{Summary}}$

Patrick lists many of the same steps that you would use for solving "normal" equations:

1. Use the distributive property to deal with parentheses.

2. Group your variables.

3. Get your variable on its own on one side of the equation.

He starts with the following inequality: $$5(x+3) \le 2(x-5)$$

Let's distribute over the parentheses $$\begin{align}5(x+3) &\le 2(x-5)\\ 5\cdot x + 5\cdot 3 &\le 2\cdot x - 2\cdot 5\\ 5x+15 &\le 2x-10 \end{align}$$

Now we'll move all of the $x$ terms to the left side of the equation. $$\begin{align}5x+15 &\le 2x-10 \\ 5x {\color{#8925AE}{-2x}} + 15 &\le 2x{\color{#8925AE}{-2x}} -10\\ 3x+15 &\le -10 \end{align}$$

We solve the remaining inequality by isolating $x$. $$\begin{align}3x+15 &\le -10\\ 3x &\le -25\\ x &\le \frac{-25}{3} \end{align}$$

We divided both sides by a positive, so we didn't need to flip the sign.

Solving Inequalities: Grouping

When tackling linear inequalities, there are two simple steps to follow. First, simplify. Then, isolate the variable.

Let's look at a practice problem!

Solution

$$2x+2<3x-9$$ $$-x<-11$$ $$x>11$$