1. ## Basic Inequality questions

Hi,

I want to check if my solution of the following question is correct :

1) $Solve : |x+1|=5$

A: $x \leq 4$

2) a) $|x-1| \leq 3$
b) $|x+1| \leq 3$

A: a) $x \leq 4 ,$
b) $x \leq 4$

3) $x < \frac{1}{x}$

I don't know how to solve the last one

2. Originally Posted by Musab
Hi,

I want to check if my solution of the following question is correct :

1) $Solve : |x+1|=5$

A: $x \leq 4$
Hi

$Solve : |x+1|=5$

Case 1 : $x \leq -1$
The equation becomes -(x+1)=5 or x=-6 and -6 < -1

Case 2 : $x \geq -1$
The equation becomes x+1=5 or x=4 and 4 > -1

2 solutions : -6 and 4

3. Originally Posted by Musab
Hi,

I want to check if my solution of the following question is correct :

1) $Solve : |x+1|=5$

A: $x \leq 4$

2) a) $|x-1| \leq 3$
b) $|x+1| \leq 3$

A: a) $x \leq 4 ,$
b) $x \leq 4$

3) $x < \frac{1}{x}$

I don't know how to solve the last one

1) $|x+1| = 5$

$x+1 = 5$ , $x + 1 = -5$

$x = 4$ , $x = -6$

2) $|x-1| \le 3$

$-3 \le x - 1 \le 3$

$-2 \le x \le 4$

3) $x < \frac{1}{x}$

$x - \frac{1}{x} < 0$

$\frac{x^2-1}{x} < 0$

critical values are $x = -1$ , $x = 0$ , and $x = 1$

for $x < -1$ ... original inequality is true

for $-1 < x < 0$ ... original inequality is false

for $0 < x < 1$ ... original inequality is true

for $x > 1$ ... original inequality is false

solution set is $x < -1$ or $0 < x < 1$

4. Originally Posted by skeeter

3) $x < \frac{1}{x}$

$x - \frac{1}{x} < 0$

$\frac{x^2-1}{x} < 0$

critical values are $x = -1$ , $x = 0$ , and $x = 1$

for $x < -1$ ... original inequality is true

for $-1 < x < 0$ ... original inequality is false

for $0 < x < 1$ ... original inequality is true

for $x > 1$ ... original inequality is false

solution set is $x < -1$ or $0 < x < 1$
Can you explain how you solved this question please.

First I have to find the critical points, and then check the values between these points against the original inequality ?

Is that right ?

5. Originally Posted by Musab
Can you explain how you solved this question please.

First I have to find the critical points, and then check the values between these points against the original inequality ?

Is that right ?
yes

6. Originally Posted by Musab
Can you explain how you solved this question please.

First I have to find the critical points, and then check the values between these points against the original inequality ?

Is that right ?
There is an easier way.

Assume x>0, then if we multiply both sides by x, the sign of the inequality is preserved.

$x^2<1$. Since we know that x is positive, the solution is $0

Now assume x is negative.... You try!