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(A) 4 < x < 8 x = 7 => |7 - 5| + |7 - 7| < 4 => |2| + 0 < 4 => TRUE => Possible
(C) 3 < x < 5 x = 4 => |4 - 5| + |4 - 7| < 4 => |-1| + |-3| = 4 NOT LESS THAN 4 => FALSE
(D) -8 < x < -4 x = -5 => |-5 - 5| + |-5 - 7| < 4 => |-10| + |-12| < 4 => 10 + 12 NOT LESS THAN 4 => FALSE
Method 2: Algebra
|x - 5| + |x - 7| < 4 Let's find the points where the values inside the Absolute Value will change signs => x-5 = 0 => x = 5 => x-7 = 0 => x = 7
So, we can divide the number line into three parts
x < 5, 5 <= x < 7 , x > 7
Attachment:
5 to 7.JPG [ 19.92 KiB | Viewed 1727 times ]
Case 1: x < 5
|x - 5| = -(x-5) (as x-5 will be negative) |x - 7| = -(x-7) (as x-7 will be negative) => -(x - 5) + (-(x - 7)) < 4 => -x + 5 -x +7 < 4 => -2x + 12 < 4 => -2x < 4-12 => -2x < -8 => x > \(\frac{-8}{-2}\) => x > 4 But the condition was x < 5 So, intersection is 4 < x < 5
Case 2: 5 <= x < 7
|x - 5| = x-5 (as x-5 will be non-negative) |x - 7| = -(x-7) (as x-7 will be negative) => x - 5 + (-(x - 7)) < 4 => x - 5 -x +7 < 4 => 2 < 4 Which is always true => 5 <= x < 7
Case 3: x >= 7
|x - 5| = x-5 (as x-5 will be non-negative) |x - 7| = x-7 (as x-7 will be non-negative) => x - 5 + x - 7 < 4 => -x + 5 + x -7 < 4 => 2x - 12 < 4 => 2x < 4+12 => 2x < 16 => x < \(\frac{16}{2}\) => x < 8 But the condition was x >= 7 So, intersection is 7 <= x < 8
So, solution is 4 < x < 5, 5 <= x < 7, 7 <= x < 8 Which is nothing but 4 < x < 8
Method 3: Graphical Method
| x - 5 | is the distance between x and 5 | x - 7 | is the distance between x and 7
Case 1: If we take a value of x between 5 and 7
Attachment:
between 5 and 7.JPG [ 15.76 KiB | Viewed 1719 times ]
Then, the distance between x and 5 + Distance between x and 7 = Distance between 5 and 7 = 2 < 4 => All values between 5 and 7 inclusive are possible
Case 2: If we take a value of x < 5
Attachment:
x lt 5.JPG [ 17.1 KiB | Viewed 1732 times ]
Then Distance between x and 7 = Distance between x and 5 + Distance between 5 and 7 => Distance between x and 7 = Distance between x and 5 + 2 So, for Distance between x and 5 + Distance between x and 7 < 4 => Distance between x and 5 + Distance between x and 5 + 2 < 4 => 2 * Distance between x and 5 < 4-2 => Distance between x and 5 < 1 So, If x is at a distance of less than 1 from 5 on the left hand side then it is possible => 4 < x <= 5
Case 3: If we take a value of x > 7
Attachment:
x gt 7.JPG [ 19.27 KiB | Viewed 1723 times ]
Then Distance between x and 5 = Distance between x and 7 + Distance between 5 and 7 => Distance between x and 5 = Distance between x and 7 + 2 So, for Distance between x and 5 + Distance between x and 7 < 4 => Distance between x and 7 + 2 + Distance between x and 7 < 4 => 2 * Distance between x and 7 < 4-2 => Distance between x and 7 < 1 So, If x is at a distance of less than 1 from 7 on the right hand side then it is possible => 7 =< x < 8
So, solution is 4 < x < 5, 5 <= x < 7, 7 <= x < 8 Which is nothing but 4 < x < 8
So, Answer will be A Hope it helps!
Watch the following video to learn How to Solve Absolute Value Problems
Re: If |x - 5| + |x - 7| < 4, what is the range of values of x ? (A) 4 <
[#permalink]
15 Nov 2021, 06:25
1
Carcass wrote:
If |x - 5| + |x - 7| < 4, what is the range of values of x ?
(A) 4 < x < 8 (B) -4 < x < 8 (C) 3 < x < 5 (D) -8 < x < -4 (E) -4 < x < 4
For these kinds of questions, I typically find that the easiest/fastest approach is to simply test numbers.
For example, is x = 0 a solution to |x - 5| + |x - 7| < 4? Plug x = 0 into the inequality to get: |0 - 5| + |0 - 7| < 4 Simplify: 5 + 7 < 4. Doesn't work! So x = 0 is NOT a solution, which means we can eliminate B and E, since they say that x = 0 IS a solution.
Now let's test x = 4 Plug x = 4 into the inequality to get: |4 - 5| + |4 - 7| < 4 Simplify: 1 + 3 < 4. Doesn't work! Since x = 4 is NOT a solution, we can eliminate C, since it says x = 4 IS a solution.
We're down to A and D. We'll test x = 6 to get: |6 - 5| + |6 - 7| < 4 Simplify: 1 + 1 < 4. WORKS! Since x = 6 is a solution, we can eliminate D, since it says x = 6 is NOT a solution.
15 Nov 2021, 03:13 
88% (01:31) correct
