Thread: 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?

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


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

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.

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?

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 $\displaystyle \displaystyle \begin{align*} x < -\sqrt{2} \end{align*}$.