Help: Test tomorrow!

**NeedHelp18** · September 11th 2008, 05:11 PM

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
answer: [-9/5, 9]
theorem: ?

c) |6 - 4x| < or = 0
answer: (3,9)
theorem: ?

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

f) |3x - 2| = |5 + 2x|
answer: {-3/5, 7, -3}
theorem: ?

help thanks! i have a feeling most are wrong

**Prove It** · September 11th 2008, 05:24 PM

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
answer: [-9/5, 9]
theorem: ?

c) |6 - 4x| < or = 0
answer: (3,9)
theorem: ?

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

f) |3x - 2| = |5 + 2x|
answer: {-3/5, 7, -3}
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?

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

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

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

AND

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

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

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.

**Prove It** · September 11th 2008, 05:31 PM

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
answer: [-9/5, 9]
theorem: ?

c) |6 - 4x| < or = 0
answer: (3,9)
theorem: ?

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

f) |3x - 2| = |5 + 2x|
answer: {-3/5, 7, -3}
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.

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

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

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

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

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

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

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

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

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.

**Prove It** · September 11th 2008, 05:50 PM

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
answer: [-9/5, 9]
theorem: ?

c) |6 - 4x| < or = 0
answer: (3,9)
theorem: ?

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

f) |3x - 2| = |5 + 2x|
answer: {-3/5, 7, -3}
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: $3x - 2 = 5 + 2x$ if $3x-2 \geq 0$ and $5+2x \geq 0$ .

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

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

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

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

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

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

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

So putting all the cases together we get...

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

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

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