Thursday

November 27, 2014

November 27, 2014

Posted by **Maria** on Thursday, September 14, 2006 at 7:24pm.

5|6-5x|=15x-35

2|3x-7|=10x-8

1/4|4x+7|=8x+16

Ok for

5|6-5x|=15x-35

first simplify it by dividing both sides by 5

Then check the intervals where

6-5x>0 then where 6-5x<0.

In other words, find where the expression in the absolute values is 0 and divide the interval into two parts there. Since the other 2 problems are similar I'll walk through the first one.

You should get

|6-5x|=3x-7 when you divide the 5's out. Then we look where

6-5x=0 which is 6/5

When x>6/5 the LHS(left hand side) is +. Setting

6-5x=3x-7 means 8x=13 or x=13/8

However, the RHS is positive when

X>7/3 so there is no solution when x>6/5

If x<6/5 then |6-5x|=3x-7

-(6-5x)=3x-7 so

-6+5x=3x-7 , 2x=-1 and x=-1/2

Again, 3x-7>0 means x>7/3 so no solution here either.

You could also verify this by graphing too.

The LHS and RHS can sometimes be zero at the same time, so check the valuse that make them both 0 too. For this problem the LHS is 0 at 6/5 but the RHS isn't. Conclude no solution for this one.

- Algebra -
**shelly**, Saturday, May 30, 2009 at 3:02am5x-6+-3x-7

- Algebra -
**tom**, Tuesday, July 16, 2013 at 6:59pm(-16)^-2/3

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