Solving Inequality: -∞ < a < -3 ∨ -1 < a < ∞

[andreask]

Hi

I'm trying to solve this inequality

|1/(2+a)| < 1.

1/(2+a) < 1 ∨ 1/(2+a) > -1

1 < 2+a
a > -1

and

1 > -2-a
3 > -a
a > -3

I know that the boundaries are
-∞ < a < -3 ∨ -1 < a < ∞

What have I done wrong?

thanks in advance

 

[symbolipoint]

|1/(2+a)|-1<0, result of adding -1 to both sides.Examine two conditions separately. The case, 2+a>=0, and the case, 2+a<0. Follow each solution process separately.

I can't think in advance if they will be conjoint or disjoint, but you'll find out once both conditions are solved.

Be aware, the expression 2+a must not be allowed equal to zero, meaning a<>-2;
so -2 is a critical value.

 

[Mark44]

Hi

I'm trying to solve this inequality

|1/(2+a)| < 1.

1/(2+a) < 1 ∨ 1/(2+a) > -1

The connector above should be ∧ ("and") rather than "or".

If |x| < b, then an equivalent inequality is -b < x < b. This is what you should have done here.

1 < 2+a

When you multiply both sides by 2 + a, are you multiplying by a positive number or a negative number? It makes a difference as regards the inequality symbol.

a > -1

and

1 > -2-a
3 > -a
a > -3

I know that the boundaries are
-∞ < a < -3 ∨ -1 < a < ∞

What have I done wrong?

thanks in advance

 

[mathman]

The case with the negative sign should read 1/(2+a) < -1. Take absolute value on both sides is equivalent to changing sign on both sides, which requires reversing the inequality direction.

 

[statdad]

For an alternate approach
$$\begin{align*}<br /> -1 & < \frac 1 {2+a} < 1 \\<br /> -(2+a)^2 & < (2+a) < (2+a)^2 \\<br /> -a^2 - 4a - 4 & < 2+a < a^2 + 4a + 4 \\<br /> -a^2 - 5a - 6 & < 0 < a^2 + 3a + 2 \\<br /> -\left(a^2 + 5a + 6\right) < 0 < a^2 + 3a + 2 \\<br /> -(a+3)(a+2) < 0 < (a+1) (a+2)<br /> \end{align*}$$
Solve the right inequality, solve the left inequality. The solution is the intersection of the two parts.