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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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Skywalker18
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

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?
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Anurag06
Skywalker18
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

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

A. 4/9
B. 4/5
C. 14/9
D. 15/7
E. 17/3

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
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chetan2u
Bunuel
In the equation |6x+9| = |3x+25|, what is the sum of all possible values of x?

A. 4/9
B. 4/5
C. 14/9
D. 15/7
E. 17/3

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

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.
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|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
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|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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