1. ## Help: Test tomorrow!

Determine the solution set and mention what theorem you used to solve the problem.

a) |2x - 7| > or = 3
answer: ( - inf., 2] U [5, inf.)
theorem: ?

b) |3 - 5x| < or = 12
theorem: ?

c) |6 - 4x| < or = 0
theorem: ?

d) (3x^2 + 4) / (5 - x ) < 0
theorem: ?

f) |3x - 2| = |5 + 2x|
theorem: ?

help thanks! i have a feeling most are wrong

2. Originally Posted by NeedHelp18
Determine the solution set and mention what theorem you used to solve the problem.

a) |2x - 7| > or = 3
answer: ( - inf., 2] U [5, inf.)
theorem: ?

b) |3 - 5x| < or = 12
theorem: ?

c) |6 - 4x| < or = 0
theorem: ?

d) (3x^2 + 4) / (5 - x ) < 0
theorem: ?

f) |3x - 2| = |5 + 2x|
theorem: ?

help thanks! i have a feeling most are wrong
a) Do you remember what the definition of the mod function (absolute value function) is?

$\displaystyle |x|=x$ if $\displaystyle x\geq 0$ and $\displaystyle |x|=-x$ if $\displaystyle x<0$.

So if $\displaystyle |2x-7|\geq 3$ then, by definition of a mod...

$\displaystyle 2x-7\geq 3$ if $\displaystyle 2x-7\geq 0$ and so $\displaystyle x\geq 5$ if $\displaystyle x\geq \frac{7}{2}$

AND

$\displaystyle -(2x-7)\geq 3$ if $\displaystyle 2x-7<0$, and so

$\displaystyle 2x-7 \leq -3$ if $\displaystyle 2x-7<0$, thus $\displaystyle x \leq 2$ if $\displaystyle x<\frac{7}{2}$.

So your answer is right, but you have to write the restrictions on x as well as the solution set. The theorem you would have used here is the definition of a mod.

b) and c) follow the exact same principle.

3. Originally Posted by NeedHelp18
Determine the solution set and mention what theorem you used to solve the problem.

a) |2x - 7| > or = 3
answer: ( - inf., 2] U [5, inf.)
theorem: ?

b) |3 - 5x| < or = 12
theorem: ?

c) |6 - 4x| < or = 0
theorem: ?

d) (3x^2 + 4) / (5 - x ) < 0
theorem: ?

f) |3x - 2| = |5 + 2x|
theorem: ?

help thanks! i have a feeling most are wrong
For d), you can solve an inequality exactly the same way as you would an equality, except that if you multiply or divide by a negative number, the inequality changes direction. Also you have to take into account the IMPLIED domains.

$\displaystyle \frac{3x^2 + 4}{5 - x}<0$.

First of all, note that the denominator can not be 0. Thus $\displaystyle x\neq 5$.

Now we solve exactly the same way as we would an equality...

$\displaystyle \frac{3x^2 + 4}{5 - x}<0$. Multiply both sides by the denominator.

$\displaystyle 3x^2+4 < 0$ Subtract 4 from both sides...

$\displaystyle 3x^2 < -4$ Divide both sides by 3...

$\displaystyle x^2 < -\frac{4}{3}$ Take the square root of both sides...

$\displaystyle |x| < \sqrt{-\frac{4}{3}}$.

But here we have the square root of a negative number, which gives a non-real answer.

So we say that there are not any real solutions for this inequality.

4. Originally Posted by NeedHelp18
Determine the solution set and mention what theorem you used to solve the problem.

a) |2x - 7| > or = 3
answer: ( - inf., 2] U [5, inf.)
theorem: ?

b) |3 - 5x| < or = 12
theorem: ?

c) |6 - 4x| < or = 0
theorem: ?

d) (3x^2 + 4) / (5 - x ) < 0
theorem: ?

f) |3x - 2| = |5 + 2x|
theorem: ?

help thanks! i have a feeling most are wrong
For f) we have to remember that there are 2 situations for each mod, so there can be up to 4 solutions to this equality.

Case 1: $\displaystyle 3x - 2 = 5 + 2x$ if $\displaystyle 3x-2 \geq 0$ and $\displaystyle 5+2x \geq 0$.

This gives $\displaystyle x = 7$ if $\displaystyle x \geq \frac{2}{3}$ and $\displaystyle x \geq -\frac{5}{2}$. Putting the inequalities together we get $\displaystyle x = 7$ if $\displaystyle x \geq \frac{2}{3}$.

Case 2: $\displaystyle 3x-2 = -(5+2x)$ if $\displaystyle 3x-2 \geq 0$ and $\displaystyle 5+2x < 0$.

Just by looking at the inequalities, we see that x has to be both $\displaystyle \geq \frac{2}{3}$ and $\displaystyle \leq -\frac{5}[2}$. This can not happen, so we can not find a solution for this case.

Case 3: $\displaystyle -(3x-2) = 5 + 2x$ if $\displaystyle 3x - 2 < 0$ and $\displaystyle 5 + 2x \geq 0$. This gives...

$\displaystyle x=-\frac{3}{5}$ if $\displaystyle -\frac{5}{2}\leq x<\frac{2}{3}$.

Case 4: $\displaystyle -(3x-2) = -(5+2x)$ if $\displaystyle 3x-2 < 0$ and $\displaystyle 5+2x < 0$.

This gives $\displaystyle x=7$ if $\displaystyle x<\frac{2}{3}$ and $\displaystyle x<-\frac{5}{2}$. For both these inequalities to occur, $\displaystyle x<-\frac{5}{2}$.

So putting all the cases together we get...

$\displaystyle x=7$ if $\displaystyle x<-\frac{5}{2}$ or $\displaystyle x \geq \frac{2}{3}$ and $\displaystyle x=-\frac{3}{5}$ if $\displaystyle -\frac{5}{2} \leq x<\frac{2}{3}$.

Once again the theorem used is the definition of a mod.