is this right?

kaebun

Junior Member
Joined
Sep 11, 2005
Messages
135
solve the equation graphically by converting it to an equivalent equation with 0 on the right hand side an then finding the x-intercepts

|2x-5|=4-|x-3|
i thin the first step is to make it 2 equations -2x+5=4+x+3 and
2x-5x=4-x-3
 
kaebun said:
solve the equation graphically by converting it to an equivalent equation with 0 on the right hand side an then finding the x-intercepts

|2x-5|=4-|x-3|
i thin the first step is to make it 2 equations -2x+5=4+x+3 and
2x-5x=4-x-3

Hi, kaebun. I think I can help, but first I have to know if those parenthesis around the 2x-5 and x-3 or absolute value signs?
 
they are absolute value signs bu um i just relized i don't have to do that problem :oops: sorry
 
but could you help me with this one

Squar root of(x+6)=6-2(square root of(5-x))
 
Square both sides, and simplify. Isolate the remaining radical. Square both sides again. Can you proceed from that point?

Eliz.
 
Yep, stapel beat me to it, but that's how it's done . . .

Just for laughs and giggles, I tried to tackle

|2x-5|=4-|x-3|

and couldn't. I passed it off to Mrs. Fresh who is a math genius. She didn't do it algebraically, she just started plugging numbers in until she came up with the answer. She asked me to ask you how you can graph an equation with only one variable. Is there any more to the problem?

Also, I'm interested in how to go about solving the first posted problem algebraically. Is there a way, or is it strictly plug and solve? Thanks.
 
Linty Fresh said:
I'm interested in how to go about solving the first posted problem algebraically.

. . . . .|2x - 5| = 4 - |x - 3|

If 2x - 5 < 0 (that is, if x < 2.5), then |2x - 5| = -(2x - 5) = 5 - 2x. Otherwise, |2x - 5| = 2x - 5.

If x - 3 < 0 (that is, if x < 3), then |x - 3| = -(x - 3) = 3 - x. Otherwise, |x - 3| = x - 3.

These inequalities divide the number line into three intervals: x < 2.5, 2.5 < x < 3, and x > 3. Solve the equation on each interval. For instance:

. . . . .x < 2.5:

. . . . .2x - 5 < 0, so |2x - 5| = 5 - 2x
. . . . .x - 3 < 0, so |x - 3| = 3 - x

. . . . .|2x - 5| = 4 - |x - 3|
. . . . .5 - 2x = 4 - (3 - x)
. . . . .5 - 2x = 4 - 3 + x
. . . . .5 - 2x = 1 + x
. . . . .4 = 3x
. . . . .4/3 = x

. . . . .Since 4/3 < 2.5, then this solution is valid.

Now you do the other two intervals.

Eliz.
 
thanx

how bout this one y1=4x+5
y2=x^3+2x^2-x-3
y3+-x^3-2x^2+5x+

A) write an equation that can be solved to find the points on intersection for y1 and y2
 
Please post new questions as new threads, not as replies to old threads, where they tend to be overlooked.

When you post this new question as its own thread, please include all the steps you have tried so far. Thank you.

Eliz.
 
Top