absolute value

I getting a different answer...don't know why

|3-4x|≥2

what i did was=====> i made it in to 2 parts ===>3 - 4x ≥ 2 and 3 - 4x ≥-2

my final answer that i got is
1/4≥x (first part) and (second part) 5/4 ≥ x

but the teacher got
1/4≥x and 5/4 ≤ x
why did the sign change for the last one? i dont get it

2. Originally Posted by Candy101
absolute value

I getting a different answer...don't know why

|3-4x|≥2

what i did was=====> i made it in to 2 parts ===>3 - 4x ≥ 2 and 3 - 4x ≥-2

my final answer that i got is
1/4≥x (first part) and (second part) 5/4 ≥ x

but the teacher got
1/4≥x and 5/4 ≤ x
why did the sign change for the last one? i dont get it
Redo the second problem: 3-4x≥-2, first subtract 3: 3-3-4x≥-5 = -4x≥-5
then, divide by NEGATIVE 4 = 5/4≤ x (or you could say x≥5/4). The reason the sign changed is because you divided by a NEGATIVE number. Any time you divide by a negative number, the direction of the sign changes.

3. assuming you get that:

$2 \leq |3x - 4|$
$\rightarrow 2 \leq (3x - 4 )$ and $2 \leq -(3x - 4)$

Using the second inequality and multiplying both sides by -1

$2 \leq -(3x -4) \rightarrow -2 \geq (3x-4)$

Note: When you multiply both sides of an inequality by -1, you have to flip the direction of inequality

For example: $1 < 2 \rightarrow -1 > -2$

So $-2 \geq (3x -4) \rightarrow \frac{5}{4} \leq x$

4. Originally Posted by Candy101
absolute value

I getting a different answer...don't know why

|3-4x|≥2

what i did was=====> i made it in to 2 parts ===>3 - 4x ≥ 2 and 3 - 4x ≥-2

my final answer that i got is
1/4≥x (first part) and (second part) 5/4 ≥ x

but the teacher got
1/4≥x and 5/4 ≤ x
why did the sign change for the last one? i dont get it
Hi Candy101,

$|3-4x| \le 2$

Understand this basic principal:

If |a| > b, then a > b or a < -b

Therefore,

$3-4x \ge 2 \ \ or \ \ 3-4x \le -2$

$-4x \ge -1 \ \ or \ \ -4x \le -5$

$x \le \frac{1}{4} \ \ or \ \ x \ge \frac{5}{4}$

5. how to graph absolute value

how to graph this

y=3|x+2|-9

it says graph: vertex and intercepts

by the way how do u find the vertex of any problem?

thanks

6. Originally Posted by Candy101
how to graph this

y=3|x+2|-9

it says graph: vertex and intercepts

by the way how do u find the vertex of any problem?

thanks
HI

you can start by sketching y=3|x+2| first . Find its intercepts

when x=0 , y=6

when y=0 , x=-2

so there will be a line passing through 6 to -2 , then the line is reflected since there is a modulus , it cant take negative values of y .

After this , shift the graph down for 9 units .

HI

you can start by sketching y=3|x+2| first . Find its intercepts

when x=0 , y=6

when y=0 , x=-2

so there will be a line passing through 6 to -2 , then the line is reflected since there is a modulus , it cant take negative values of y .

After this , shift the graph down for 9 units .
how did you get 6 and -2?
can u please show or explain ur steps?
thanks,

8. Originally Posted by Candy101
how did you get 6 and -2?
can u please show or explain ur steps?
thanks,
Put x and y =0 respectively to get the intercepts (points where the line cuts the x and y axis )