6 + 5 | 2 r - 3 | ? 4: can somebody help me solve this plz?

Re: 6 + 5 | 2 r - 3 | ? 4

Can you solve: 6 + 5(2r - 3) = 4 ?
 


Like Denis, I also wonder where you're stuck and how much you already know. Posting an exercise with no work shown, and no questions asked about specific steps, makes it hard to know where to start helping.

To remove the absolute value symbols from inequalities, we use the following rules.

If the inequality symbol is "greater than", then we're working with something that looks like

|expression| > constant

Because of the absolute value symbols, this statement actually means two different inequalities:

expression > constant

expression < -constant

If the given inequality symbol is "less than", then we've got

|expression| < constant

This statement also means two inequalities.

expression < constant

expression > -constant

We can write these last two as a compound inequality:

-constant < expression < constant

Here's two examples to show these rules using values.

|7x - 3| > 40

Remove the absolute value symbols as explained above, and write two new inequalities.

7x - 3 > 40

7x - 3 < -40


|22t + 5| < 9

Remove the absolute value symbols as explained above for |expression|<constant.

-9 < 22t + 5 < 9

The given inequality in your exercise is not ready to apply the rule because |2r-3| is not by itself on one side of the inequality symbol.

Carry out the two operations needed to get |2r - 3| by itself on the left-hand side, and then apply the rule to get rid of the absolute value symbols.

Solve the two resulting inequalities, and go from there.

Please show whatever work you can accomplish, if you would like more help with this exercise. Try to say something about WHY you're stuck, so that I might determine where to continue helping.

By the way, if you understand the perception of absolute value as a distance from zero on the Real number line, and you realize that there are two possible directions to move away from zero on the Real number line, then these rules are easy to remember.

|x| > 10

This statement tells us that valid numbers for x are all located more than 10 units away from zero on the Real number line. But we can move away from zero toward positive 10 or we can move away from zero toward -10. In order for a value to be more than 10 units away from zero in either direction, it must be more than 10 or less than -10.

x > 10

x < -10

Likewise, |x| < 10 tells us that valid numbers for x are within 10 units distance from zero. They cannot be 10 or more units to the right of zero, and they cannot be 10 or more units to the left of zero. They are inbetween.

-10 < x < 10

This is the same as writing the union of the following two sets.

x < 10

x > -10
 
Re: 6 + 5 | 2 r - 3 | ? 4

If the problem is \(\displaystyle 6 + 5\left| {2r - 3} \right| \geqslant 4\), then there is nothing to solve.
It is true for all r.
Note
\(\displaystyle \begin{gathered} \left| {2r - 3} \right| \geqslant 0 \hfill \\ 5\left| {2r - 3} \right| \geqslant 0 \hfill \\ 6 + 5\left| {2r - 3} \right| \geqslant 6 > 4 \hfill \\ \end{gathered}\)
 
As PKA noted, the solution to this inequality is all Real numbers.

-oo < r < oo

(Maybe, I should have actually worked this exercise before posting a general approach.)

6 is obviously greater than 4. It's clear that increasing 6 (by adding 5 times some positive amount) will make it even larger than 4.

If this analysis is not clear, then start with the general approach that I posted: isolate the absolute value expression.

After subtracting 6 from both sides and dividing both sides by 5, the absolute value expression is isolated.

|2r - 3| >= -2/5

No further calculations are required, at this point; simple analysis of the signs on each side is enough. Since the absolute value of any expression is greater than a negative quantity, the left-hand side will always be greater than negative 2/5ths, no matter what value r takes on.



As an example of the general approach, here's a different exercise.

6 - 5|2r - 3| >= 4

Isolate the absolute value expression by subtracting 6 and dividing by -5.

-5|2r - 3| >= -2

|2r - 3| <= 2/5

Remove the absolute value symbols to get two new inequalities.

2r - 3 <= 2/5

2r - 3 >= -2/5

Solving each of these gives

r <= 17/10 and r >= 13/10, respectively.

The union of these two sets is the solution.

13/10 <= r <= 17/10

 
You must log in or register to reply here.
Top