Posted by Maria on Thursday, September 14, 2006 at 7:24pm.
I'm not quite sure about these problems. I know how to do them, but it seems that I get the wrong answer. Maybe I'm supposed to get 'no solution' but I'm not sure. Could someone please help me with them?
565x=15x35
23x7=10x8
1/44x+7=8x+16
Ok for
565x=15x35
first simplify it by dividing both sides by 5
Then check the intervals where
65x>0 then where 65x<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
65x=3x7 when you divide the 5's out. Then we look where
65x=0 which is 6/5
When x>6/5 the LHS(left hand side) is +. Setting
65x=3x7 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 65x=3x7
(65x)=3x7 so
6+5x=3x7 , 2x=1 and x=1/2
Again, 3x7>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 3:02am
5x6+3x7

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