In the equation |6x+9| = |3x+25|, what is the sum of all possible valu

Hello...
I followed the Absolute value equation Strategies... got answer c - 14/9

Following is the explanation:

We need to test four cases overall: positive/positive, positive/negative, negative/positive, and negative/negative.
(1) The positive/positive case: (6x+9) = (3x+25)
(2) The positive/negative case: (6x+9) = -(3x+25)
(3) The negative/positive case: —(6x+9) = (3x+25)
(4) The negative/negative case: —(6x+9) = -(3x+25)

case (1) and case (4) yield the same equation. Likewise, case (2) and case (3) yield the same equation. Thus only need to consider two real cases: one in which neither expression changes sign, and another in which one expression changes sign.

CASE A: Same sign
6x+9 = 3x+25
6x-3x = 25-9
3x = 16
x=16/3

CASE B: Different Sign
-(6x+9) = (3x+25)
-6x-9 = 3x+25
-6x-3x=25+9
-9x=34
x=-34/9

Sum of all possible values of x = 16/3+(-34/9)
16/3-34/9 = 14/9 (Answer C)

Please, consider giving kudos if you find my answer helpful in any way

since it is MOD on both sides we can square both sides..
thereafter we get the two terms on ONE side and we will have a QUADRATIC equation..
In a Quadratic equation, the value of sum of roots /x is\(-\frac{(coeff of x}{coeff of x^2)}\)..
so lets concentrate ONLY on x^2 and x values..

\(|6x+9| = |3x+25|\)..
\((6x+9)^2 = (3x+25)^2\)..
\(36x^2 +108 x+ 9^2 = 9x^2+150x +25^2\)..
\((36-9)x^2 + (108-150)x +9^2+25^2\)..
sum of the roots =\(\frac{-b}{a} = -(\frac{-42}{27}) = \frac{14}{9}\)
C
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x<-25/3
- 6x - 9 = -3x - 25
=> 3x = 16
=> x = 16/3

-25/3 < x < -9/6
6x + 9 = -(3x+25)
=> 6x+9 = -3x - 25
=> 9x = -34
=> x = -34/9

x> -9/6
6x+9 = 3x+25
=> 3x = 16
=> x = 16/3

Sum of all possible values of a = 16/3 + (-34/9)
=48/9 - 34/9
=14/9

Answer C
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Hi

I added all 3 values and ended up getting 16/3 + 16/3 - 16/9
which resulted in to 80/9.

Probably it could be silly mistake from my end but can you please help me out here?

Anurag06
When x<-25/3, x = 16/3 is not possible.
So, you should add 2 values(16/3 and -34/9) only

for given eqn |6x+9| = |3x+25l
the possible values of x
6x+9 = 3x+25
x=16/3 and
6x+9=-3x-25
x=-34/9
its sum l
16/3-34/9 ; 14/9
OPTION C

Please correct me if I am wrong.
Though the answer is correct but the approach may be wrong as squaring produces more roots, which may not be same as of the original equation.
Here, it is mod, I think that's why it produced the same result.

|6x+9| = |3x+25|

\((6x+9)^2 = (3x+25)^2\)

\((6x+9)^2 - (3x+25)^2 = 0\)

(6x+9+3x+25) (6x+9 -3x-25) =0

(9x+ 34)(3x-16) = 0

x = \(\frac{-34}{9}\) or \(\frac{16}{3}\)

sum = \(\frac{16}{3}-(\frac{-34}{9})\) =\(\frac{14}{9}\)
IMO C

|6x+9| = |3x+25|
(6x+9)^2 –(3x+25)^2 = 0
Hint Difference of squares: a2 – b2 = (a + b)(a – b)
(6x+9+3x+25)( 6x+9-3x-25) = 0
Solve for x
(9x+34)(3x-16) = 0
X = -34/9 & X2 = 16/3
Sum of x is
(-34/9) + (16/3)
((3)(-34)+(9)(16)) / ((9)(3))
Sum = 14/9

Hope this helps!

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