# Thread: Solving compound inequalities problem

1. ## Solving compound inequalities problem

I have read through the solving inequalities pdf and am still struggling with this problem.

$\displaystyle -7<\frac{1}{x}\leq1$

I've tried taking the $\displaystyle -7<\frac{1}{x}$ and the $\displaystyle \frac{1}{x}\leq1$ on their own and combining the solutions but my answers are obviously wrong. Could someone show me how to solve these kinds of inequalities?

2. ## Re: Solving compound inequalities problem

Hello, BobRoss!

This one is tricky and dangerous.
There are several approaches.
I'll show you mine . . .

$\displaystyle \text{Solve: }\:\text{-}7\:<\:\frac{1}{x}\:\le\:1$

Like you, I split it into two inequalities: .$\displaystyle -7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1$
Note the "and"; we want both to be true.

We'd like to multiply through by , but we must be very careful!
What we get depends on whether $\displaystyle x$ is positive or negative.

Suppose $\displaystyle x$ is positive.
Multiply both inequalities by $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}$
. . . and: .$\displaystyle 1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1$

To make both statements true, we use: .$\displaystyle \boxed{x\:\ge\:1}$

Suppose $\displaystyle x$ is negative.
Multiply both inequalities by negative $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}$
. . . and: .$\displaystyle 1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1$

To make both statements true, we use: .$\displaystyle \boxed{x \:<\:\text{-}\tfrac{1}{7}}$

Solution: .$\displaystyle \left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)$

n . . . . . . . $\displaystyle \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$

. . . . . $\displaystyle \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$

3. ## Re: Solving compound inequalities problem

Originally Posted by Soroban
Hello, BobRoss!

This one is tricky and dangerous.
There are several approaches.
I'll show you mine . . .

Like you, I split it into two inequalities: .$\displaystyle -7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1$
Note the "and"; we want both to be true.

We'd like to multiply through by , but we must be very careful!
What we get depends on whether $\displaystyle x$ is positive or negative.

Suppose $\displaystyle x$ is positive.
Multiply both inequalities by $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}$
. . . and: .$\displaystyle 1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1$

To make both statements true, we use: .$\displaystyle \boxed{x\:\ge\:1}$

Suppose $\displaystyle x$ is negative.
Multiply both inequalities by negative $\displaystyle x.$

We have: .$\displaystyle \text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}$
. . . and: .$\displaystyle 1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1$

To make both statements true, we use: .$\displaystyle \boxed{x \:<\:\text{-}\tfrac{1}{7}}$

Solution: .$\displaystyle \left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)$

n . . . . . . . $\displaystyle \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)$

. . . . . $\displaystyle \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}$
Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.

4. ## Re: Solving compound inequalities problem

Thanks so much, that helps a lot! Although I don't quite understand this part:

Originally Posted by Prove It
Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.
What is the new restriction and could you possibly give me an example of when I might see that?

5. ## Re: Solving compound inequalities problem

Originally Posted by BobRoss
Thanks so much, that helps a lot! Although I don't quite understand this part:

What is the new restriction and could you possibly give me an example of when I might see that?
The new restriction comes from saying "if we think about positive values for x", we are immediately making the restriction x > 0, and if we say "if we think about negative values for x", we are immediately making the restriction x < 0.

Take for example, the inequality \displaystyle \displaystyle \begin{align*} \frac{x^2 + 3x -2}{x} < 3 \end{align*}.

In order to solve this, we need to think of two cases, the first where x < 0, and the second where x > 0.

In case 1, with x < 0, we have

\displaystyle \displaystyle \begin{align*} \frac{x^2 + 3x - 2}{x} &< 3 \\ x^2 + 3x - 2 &> 3x \\ x^2 - 2 &> 0 \\ x^2 &> 2 \\ |x| &> \sqrt{2} \\ x < -\sqrt{2} \textrm{ or } x &> \sqrt{2} \end{align*}

So by solving the inequality we have \displaystyle \displaystyle \begin{align*} x < -\sqrt{2} \end{align*} or \displaystyle \displaystyle \begin{align*} x > \sqrt{2} \end{align*}, BUT to get this, we originally had to restrict \displaystyle \displaystyle \begin{align*} x < 0 \end{align*}. So that means that the solution for this case is just \displaystyle \displaystyle \begin{align*} x < -\sqrt{2} \end{align*}.