Problem Solving

Which sets includes ALL of the solutions of x that will satisfy the eqn: |x-2|-|x-3|=|x-5|

A. (-6, -5, 0 1 7 8 )
B. (-4 -2 0 10/3 4 5)
C. (-4 0 1 4 5 6)
D. (-1 10/3 3 5 6 8)
E. (-2 -1 1 3 4 5)

oA: C

ashishsj wrote:Which sets includes ALL of the solutions of x that will satisfy the eqn: |x-2|-|x-3|=|x-5|

A. (-6, -5, 0 1 7 8 )
B. (-4 -2 0 10/3 4 5)
C. (-4 0 1 4 5 6)
D. (-1 10/3 3 5 6 8)
E. (-2 -1 1 3 4 5)

oA: C

Rewrite equation as |x-5| + |x-3|-|x-2| = 0

for x > 5

x - 5 + (x -3) - (x - 2) = 0 --> x = 6 --> only (c) or (d) feasible sets

for 3 < x < 5

-(x - 5) + (x -3) - (x - 2) = 0 --> x = 4 --> eliminate (d) --> (c) is the answer

Thanks xyz21.

Your method seems to be an easy approach.

Here is what i did i squared both the sides os the eqn and solved for X

same i got X =6,4

agree with ans.C

To all:

for range: 2<x<3 I am getting x = 10/3

Please respond

Understood this partially. Little more help on this,please.

xyz21 wrote:

ashishsj wrote:Which sets includes ALL of the solutions of x that will satisfy the eqn: |x-2|-|x-3|=|x-5|

A. (-6, -5, 0 1 7 8 )
B. (-4 -2 0 10/3 4 5)
C. (-4 0 1 4 5 6)
D. (-1 10/3 3 5 6 8)
E. (-2 -1 1 3 4 5)

oA: C

Rewrite equation as |x-5| + |x-3|-|x-2| = 0

for x > 5

x - 5 + (x -3) - (x - 2) = 0 --> x = 6 --> only (c) or (d) feasible sets

for 3 < x < 5

-(x - 5) + (x -3) - (x - 2) = 0 --> x = 4 --> eliminate (d) --> (c) is the answer

maihuna wrote:To all:

for range: 2<x<3 I am getting x = 10/3

Please respond

i agree!?

why is 10/3 incorrect? whats the smartest way to solve this?

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

A. (-6, -5, 0 1 7 8 )
B. (-4 -2 0 10/3 4 5)
C. (-4 0 1 4 5 6)
D. (-1 10/3 3 5 6 8)
E. (-2 -1 1 3 4 5)

If we try 0 for x , then | -2| - |-3| = |-5| => 2 -3 = -1 != 5 so zero is not the solution and can rule out A,B,C.

D,E ;
try x as 4 : 2 - 1 = 1 == 1 so E is the answer.

But again i am able to find one element from the sets of each option that the equation fails to hold good.

try 3 , 1 - 0 = 2 (false) so 3 is not the soltuion so neither D or E can be the answer.

Can some one explain this Q as to what does it mean by solution of X?