fiftrombone

07-16-2006, 05:16 PM

if you have like l ax+b l = l cx+d l do you jsut solve it like ax+b = cx+d then solve also for ax+b = -(cx+d)?

View Full Version : absolute value on both sides of the equal sign

fiftrombone

07-16-2006, 05:16 PM

if you have like l ax+b l = l cx+d l do you jsut solve it like ax+b = cx+d then solve also for ax+b = -(cx+d)?

daon

07-16-2006, 05:20 PM

if you have like l ax+b l = l cx+d l do you jsut solve it like ax+b = cx+d then solve also for ax+b = -(cx+d)?

You got it!

You got it!

fiftrombone

07-16-2006, 05:34 PM

thanks a lot.

fiftrombone

07-16-2006, 05:38 PM

wait. that didnt work when i tried to solve my problem. it is the same variable on both sides of the equation. and it is an absolute value on both sides. how do i solve for x?

jonboy

07-16-2006, 05:56 PM

So this is exactly what we are trying to solve for \L x:

\L \left| {ax + b} \right| = \left| {cx + d} \right|

\L \left| {ax + b} \right| = \left| {cx + d} \right|

skeeter

07-16-2006, 05:58 PM

|ax + b| = |cx + d|

case 1 ...

ax + b = cx + d

ax - cx = d - b

x(a - c) = d - b

x = (d - b)/(a - c)

case 2 ...

ax + b = -(cx + d)

ax + b = -cx - d

ax + cx = -b - d

x(a + c) = -(b + d)

x = -(b + d)/(a + c)

case 1 ...

ax + b = cx + d

ax - cx = d - b

x(a - c) = d - b

x = (d - b)/(a - c)

case 2 ...

ax + b = -(cx + d)

ax + b = -cx - d

ax + cx = -b - d

x(a + c) = -(b + d)

x = -(b + d)/(a + c)

fiftrombone

07-16-2006, 06:08 PM

ok thank you. i think when i worked my problem through the first time i just dod some math wrong becasue it does work. thank all so much for your help.

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