If y = |6x - 18| + |-7x + 7| + |3x + 2|, for what x-value is the value

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If  y = |6x - 18| + |-7x + 7| + |3x + 2| , for what x-value is the value of y minimized?

A.  -\frac{2}{3} 
B. 1
C. 2
D. 3
E.  \frac{7}{2}

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lacktutor · Accepted Answer

|6x —18|+ |—7x+7|+ |3x+2| Is always greater than zero(inclusive)

—> the minimum value of y could be 0, if 6x—18=0, —7x+7=0, and 3x+2=0 at the same time. 
—> we have to check the values of y ( if x=3, x= 1, and x= —2/3) 
y(3)= 0+ 14+ 11= 25
y(1)= 12+ 0+ 5= 17
y(—2/3)= 22+ 35/3+ 0= 101/3

Minimum value of y should be 17, if x is equal to 1. 
The answer is B

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Archit3110 · Answer

value of y=|6x−18|+|−7x+7|+|3x+2| is minimised at x= 1 ; y = 17
IMO B

If y=|6x−18|+|−7x+7|+|3x+2|, for what x-value is the value of y minimized?

A. −23−23
B. 1
C. 2
D. 3
E. 7272

exc4libur · Answer

Replace each answer choice into y=|6x−18|+|−7x+7|+|3x+2| and the least value of y is when x=1.

Ans (B)

madgmat2019 · Answer

From options

A. -2/3.....33+
B. 1........17........ correct
C. 2........21
D. 3........25
E. 7/2......28+

OA:B

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CareerGeek · Answer

If y = |6x - 18| + |-7x + 7| + |3x + 2|, for what x-value is the value of y minimized?

We can substitute and check for which value of x from the option is the value of y least.

—> y = 6|x - 3| + 7|-x + 1| + |3x + 2|

A. -2/3 ; y = 6|-2/3 - 3| + 7|-(-2/3) + 1| + |3(-2/3) + 2| = 22 + 35/3 = 101/3

B. 1 ; y = 6|1 - 3| + 7|-1 + 1| + |3(1) + 2| = 12 + 0 + 5 = 17

C. 2 ; y = 6|2 - 3| + 7|-2 + 1| + |3(2) + 2| = 6 + 7 + 8 = 21

D. 3 ; y = 6|3 - 3| + 7|-3 + 1| + |3(3) + 2| = 0 + 14 + 11 = 25

E. 7/2 ; y = 6|7/2 - 3| + 7|-7/2 + 1| + |3(7/2) + 2| = 3 + 35/2 + 25/2 = 33

Least value of y is 17, when x = 1

IMO Option B

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eakabuah · Answer

y=|6x-18|+|7-7x|+|3x+2|
when x=-2/3
y=22+35/3+0=33+2/3
when x=1
y=12+5=17
when x=2
y=6+7+8=21
when x=3
y=14+11=25
Obviously, we can see that x=3.5 will yield a much larger value that x=3 since we can see a trend whereby when x increases, y also increases.
when x=3.5
y=3+17.5+12.5=31
The least value out of the options is when x=1

The answer is B.

freedom128 · Answer

Fastest and surest way to do this is to plug-in numbers (from choices) to the equation and see any inflection point among x values.

y = |6x - 18| + |-7x + 7| + |3x + 2|

A. x=-2/3 --> y= 22 + 2.3 + 0 = 24.3

B. x=1 --> y= 12 + 0 + 5 = 17

C. x=2 --> y= 6 + 7 + 8 = 21

D. x=3 --> y= 0 + 14 + 11

E. x=3.5 --> y= 3 + 17.5 + 12.5

For this question, we just need to calculate choices (A),(B),(C). We see clear inflection point at choice (B). . Calculation on choices (D) and (E) is shown for clarity purpose only.

Final answer is (B)

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unraveled · Answer

If y=|6x−18|+|−7x+7|+|3x+2|, for what x-value is the value of y minimized?

A. -\(\frac{2}{3}\)
B. 1
C. 2
D. 3
E. \(\frac{7}{2}\)

Here x > 3 or x < 3 OR x > 1 or x < 1 OR x > -\(\frac{2}{3}\) or x < -\(\frac{2}{3}\)
Since y is sum of absolute values we have to find its least possible positive value which would be possible for only one value of x.

Checking for each value -
x = -\(\frac{2}{3}\), y = 24
x = 1, y = 17
x = 2, y = 21
x = 3, y = 25
x = \(\frac{7}{2}\), y = 33

Answer B.

Amlu95 · Answer

Greatest coefficient of X is 7. So how can we minimize that term. We'd easily land up with 1 which makes that 0.

metalhead2593 · Answer

y = |6x - 18| + |-7x + 7| + |3x + 2|
 = 6|x-3| + 7|1-x| + 3|x+2/3|
 = 6|x-3| + 7|x-1| + 3|x+2/3|

So we can feel that each time x move by z units, y will move by 6z + 7z + 3z and y is greater than equal than 0. So, 7|x-1| contributes the greatest value to y, thus making 7|x-1| = 0 is likely to cause y to reach minimum. Thus x = 1.

Kinshook · Answer

Asked: If  y = |6x - 18| + |-7x + 7| + |3x + 2| , for what x-value is the value of y minimized?

y = |6x - 18| + |-7x + 7| + |3x + 2| 
y = 6|x-3| + 7|x-1| + 3|x+2/3|

Let us check the value of y at x = 3 ; x = 1 & x=-2/3
At x = -2/3; y = 23 -16(-2/3) = 23 +32/3 ~ 34
At x = 1 ; y = 6(2) + 7(0) + 5 = 17
At x = 3 ; y = 7(2) + 11 = 25

y is least when x = 1

IMO B

catinabox · Answer

Is there any algebraic method to solve this question?

chacinluis · Answer

Going over several similar scenarios, by adding or subtracting more absolute value functions to y=....It seems that if we have an odd number of absolute value functions adding each other, the minimum happens at the middle value of the list of values that make each absolute value zero.

For example. y=|x-7|+|x-3|+|x+1|+|x+20|+|x+100|
we have an odd number of absolute value functions adding each other.
The list of values that make each zero is 7,3,-1,-20,-100
The middle value is -1
So the minimum of y occurs at -1

When y is an even sum of absolute value functions it seems to be that the minimum occurs everywhere between the two middle values (inclusive) of the list of values that make each absolute value zero.

For example. y=|x+2|+|x+1|+|x-20|+|x-30|
The list of values that make each absolute value function zero is -2, -1, 20, 30
So the minimum of y occurs everywhere in

AashishGautam · Answer

This can be solved in 30 seconds if the derivative of |x| is known.

Derivative of |x| = x/|x| ; x/|x| = 1 (if X>0) and x/|x| = -1 (if x<0)

We have three corner points of the function y, -2/3, 1,3
dy/dx = 6 (6x-18)/|6x-18| - 7 (-7x+7)/|-7x+7| + 3 (3x+2)/|3x+2|

x>3 6+7+3=16
1<x<3 -6+7+3=4
-2/3<x<1 -6-7+3 = -10
x<-2/3 -6+7-3 = -2

Hence value will occour in the range -2/3<x<1 ; checking x=1

Kimberly77 · Answer

I'm confused with as why are these values in Modulus only calculated with x>0 and not x<0 ? Could anyone clarify? Thanks

SergejK · Answer

This is not always true. If we would have had an expression like  6|x-54| that we would have added to get y, the rule wouldn't have worked.

SergejK · Answer

Look at what is asked. We want to have y min. Y must be positive, so we are checking when we can get there. So no need to test all the ranges. Just plug the answer choices to get to the correct answer.

akwenok · Answer

Bunuel , could you please provide the most time-efficient way to approach this one?

If y = |6x - 18| + |-7x + 7| + |3x + 2|, for what x-value is the value

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GMAT Problem Solving (PS) Questions

If y = |6x - 18| + |-7x + 7| + |3x + 2|, for what x-value is the value

Prep Toolkit

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