Ix+14I+3>17 pre calc?

Question

need help please. i tried to figure this out and i have no idea. the example in the book didn't help either. the answer is x<-28, x>0

Answer 1

okay to solve this equation first you have to isolate the variable:

|x + 14| + 3 &gt; 17

|x + 14| &gt; 14

now, remember that the absolute value graph includes both positive and negative values for x; so, you have to solve for values of x that are both greater than 14 and less than -14.

x + 14 &gt; 14, x + 14 &lt; -14

x &gt; 0, x &lt; -28

Answer 2

|x + 14| + 3 &gt; 17

To solve this absolute value inequality, isolate the absolute value.

|x + 14| &gt; 14

Split into two inequalities.  For absolute inequalities in the form

|z| &gt; c, then

z &gt; c OR z &lt; -c

And for the form

|z| &lt; c, then

-c &lt; z &lt; c

We have the first form, obviously, so,

|x + 14| &gt; 14

becomes

x + 14 &gt; 14

OR

x + 14 &lt; -14

Solve each inequality by isolating x.

x &gt; 0

OR

x &lt; -28

Answer 3

There&#039;s more then one way to do this problem. I&#039;ll do a couple and you can choose which one you like:

——————————————————————————————————————

(Option 1)

Absolute value does one of two things:

(a) If the contents are positive or zero, it does nothing.

(b) If the contents are negative, it changes the sign (multiplies by negative 1).

Possibility (a), it does nothing:

( x + 14 ) + 3 &gt; 17

x + 17 &gt; 17

x &gt; 0

Possibility (b), it changes the sign:

-( x + 14 ) + 3 &gt; 17

-x - 14 + 3 &gt; 17

-x - 11 &gt; 17

-x &gt; 28

Remember to change the direction of the equality when you multiply or divide by a negative number:

-x &gt; 28

x &lt; -28

If you look at these on a number line, it becomes clear that they are on opposite sides of two points:

&lt;===)----(===&gt;

.... -28 .. 0

As an interval, this would be:

( -∞, -28 ) U ( 0, ∞ )

——————————————————————————————————————

(Option 2)

Isolate the absolute value:

| x + 14 | + 3 &gt; 17

| x + 14 | &gt; 14

The contents of the absolute value must have a distance from zero greater than 14.

That means that the contents must be &gt;14 or &lt;-14

Express this on one line:

-14 &gt; x + 14 &gt; 14

Solve for x:

-14 -14 &gt; x + 14 -14 &gt; 14 - 14

-28 &gt; x &gt; 0

It&#039;s not &quot;squeezed&quot; between two values (less than a bigger one and greater than the smaller one), which makes this a somewhat &quot;odd&quot; inequality, but it&#039;s a bit easier to do than the other perhaps.

Thus, you have the separate solutions:

x &lt; -28 or x &gt; 0

---

If, through either of these methods, you came up with a solutions such as:

x &gt; -2 and x &lt; 1

-2 &lt; x &lt; 1

1 &gt; x &gt; -2

As these would graph on a number line like:

&lt;---(====)---&gt;

.... -2 ..... 1

You have the values confined &quot;between&quot; two values.

Thus your answer would be:

-2 &lt; x &lt; 1

or in interval notation:

(-2, 1)

Answer 4

|x + 14| + 3 &gt; 17

Subtract 3 from both sides

|x + 14| &gt; 14

Write as 2 inequalities

x + 14 &gt; 14

x + 14 &lt; -14

Subtract 14 from both sides and the answer is:

x &gt; 0

x &lt; -28